The Mittag-Le ffler function · 2020. 11. 13. · The Mittag-Le ffler function P. Van Mieghem ∗...

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The Mittag-Le er function P. Van Mieghem in honor of Gösta Mittag-Leer for Huijuan v1. 23 May 2020 v2. 22 August 2020 Abstract We review the function theoretical properties of the Mittag-Leer function H d>e (}) in a self- contained manner, but also add new results; more than half is new! 1 Introduction We investigate the Mittag-Leer function, H d>e (})= X n=0 } n Γ (e + dn) (1) introduced by Gösta Mittag-Leer [27, 26] in 1903 with e =1, which he denoted as H d (})= P n=0 } n Γ(1+dn) = H d>1 (}). We consider the broader denition H d>e (}) and not H d (}), because the functional relations for H d>e (}) are closed and expressed in terms of H d>e (}), whereas connement to H d (}) only, deprives the analysis from a complete and more elegant picture. There exist generalizations 1 of the Mittag-Leer function H d>e (}), which are beyond the present scope, but discussed by Haubold et al. [22] and also covered in the recent book by Goreno et al. [16] on Mittag-Leer functions and their applications. The Mittag-Leer function H d (}) is treated by Erdelyi et al. [10, Sec. 18.1 on p. 206-211] and by Sansone and Gerretsen [35, Sec. 6.13 on p. 345-348]. Our aim 2 here is to deduce the most relevant functional properties of the Mittag-Leer function H d>e (}) dened in (1). Since about half of the results have been established before, the manuscript in the form of articles (art.) as in our book [40], is more a review, without detailed historical citations as in [16], but enriched with new results: art. 5, 8 10, 12, 14, 15, 16, 17, 18, 19, 23, 24, 26, 27, 28, 36, 37, 39, 56, 57 and part of art. 20, 31, 34, 38, Delft University of Technology, Faculty of Electrical Engineering, Mathematics and Computer Science, P.O Box 5031, 2600 GA Delft, The Netherlands; email : [email protected] 1 The generalized Mittag-Leer function is dened as Hd>e>f (})= S n=0 Γ(f+n) Γ(f) } n Γ(e+dn) , where Hd>e>0(})= Hd>e (}). 2 During my sabattical at Stanford in 2015, I encountered the rather exotic Mittag-Leer distribution, which I intended to include in my Performance Analysis book [41] as another example of a power law-like distribution. However, the functional properties of the Mittag-Leer function H d>e (}) require attention rst, before one can turn to probability theory. 1

Transcript of The Mittag-Le ffler function · 2020. 11. 13. · The Mittag-Le ffler function P. Van Mieghem ∗...

  • The Mittag-Leffler function

    P. Van Mieghem∗

    in honor of Gösta Mittag-Lefflerfor Huijuan

    v1. 23 May 2020v2. 22 August 2020

    Abstract

    We review the function theoretical properties of the Mittag-Leffler function ( ) in a self-contained manner, but also add new results; more than half is new!

    1 Introduction

    We investigate the Mittag-Leffler function,

    ( ) =∞X

    =0Γ ( + )

    (1)

    introduced by Gösta Mittag-Leffler [27, 26] in 1903 with = 1, which he denoted as ( ) =P∞

    =0 Γ(1+ )

    ∆= 1 ( ).

    We consider the broader definition ( ) and not ( ), because the functional relations for

    ( ) are closed and expressed in terms of ( ), whereas confinement to ( ) only, deprives the

    analysis from a complete and more elegant picture. There exist generalizations1 of the Mittag-Leffler

    function ( ), which are beyond the present scope, but discussed by Haubold et al. [22] and also

    covered in the recent book by Gorenflo et al. [16] on Mittag-Leffler functions and their applications.

    The Mittag-Leffler function ( ) is treated by Erdelyi et al. [10, Sec. 18.1 on p. 206-211] and by

    Sansone and Gerretsen [35, Sec. 6.13 on p. 345-348]. Our aim2 here is to deduce the most relevant

    functional properties of the Mittag-Leffler function ( ) defined in (1). Since about half of the

    results have been established before, the manuscript in the form of articles (art.) as in our book [40],is more a review, without detailed historical citations as in [16], but enriched with new results: art.5, 8 10, 12, 14, 15, 16, 17, 18, 19, 23, 24, 26, 27, 28, 36, 37, 39, 56, 57 and part of art. 20, 31, 34, 38,

    ∗Delft University of Technology, Faculty of Electrical Engineering, Mathematics and Computer Science, P.O Box

    5031, 2600 GA Delft, The Netherlands; email : [email protected] generalized Mittag-Leffler function is defined as ( ) = ∞=0

    Γ( + )Γ( ) Γ( + )

    , where 0( ) = ( ).2During my sabattical at Stanford in 2015, I encountered the rather exotic Mittag-Leffler distribution, which I

    intended to include in my Performance Analysis book [41] as another example of a power law-like distribution. However,

    the functional properties of the Mittag-Leffler function ( ) require attention first, before one can turn to probability

    theory.

    1

  • 40, 41, 42. For self-consistency, we have included an appendix A on the Gamma function Γ ( ), whose

    properties are essential for the Mittag-Leffler function ( ).

    The Mittag-Leffler function ( ) naturally arises in fractional calculus and in the solution of

    fractional order integral or differential equations, as illustrated in [22] and [16]; for example, in the

    fractional generalization of the heat equation, random walks, Lévy flights, superdiffusive transport and

    viscoelasticity, and fractional Ohm’s Law. Abel’s integral equation, whose solution involves ( ),

    is treated in [16, Chapter 7]. A “fractional” generalization of the Poisson renewal process, discussed in

    [16, Sec. 9.4], consists of replacing the exponential interarrival time between events by a Mittag-Leffler

    distribution 1 (− ) with real ≥ 0 and 0 ≤ 1. Section 6 presents in art. 40 a different proofof the monotonicity of the Mittag-Leffler function ( ), while art. 41 and art. 42 focus on theMittag-Leffler random variable. The last section 7, also new, explores the related integral

    ( ) =

    Z ∞

    0 Γ ( + )(2)

    that naturally appears in the Euler-Maclaurin summation for ( ) in art. 27.

    2 Complex function theory

    Since the Mittag-Leffler function ( ) is defined in (1) as a power series in , a first concern is its

    validity range in . The radius of convergence of the power series ( ) =P∞

    =0 of a function

    satisfies 1 = lim sup →∞ | |1 or1 = lim →∞

    ¯̄¯ +1

    ¯̄¯ when the latter exists [37]. Using Γ( + )Γ( + ) ∼

    in [1, 6.1.47] when →∞, the radius of convergence of the power series in (1) is

    1= lim

    →∞

    ¯̄¯̄ Γ ( + )Γ ( + + )

    ¯̄¯̄ = lim

    →∞| |− = 0

    for Re ( ) 0 and all complex . An entire function has a power series with infinite radius of

    convergence and is, thus, analytic in the entire complex plane with an essential singularity at infinity.

    Associated to entire complex functions is the concept of the order , which is defined [37, p. 248] for

    any 0 as ( ) =³| | +

    ´when | |→∞. A necessary and sufficient condition [37, p. 253] that

    ( ) =P∞

    =0 should be an entire function of order is that

    1= lim

    →∞

    − log | |log

    Applied to the power series in (1), after invoking Stirling’s asymptotic formula [1, 6.1.39] that follows

    from (132) in Appendix A,

    Γ ( + ) ∼√2 − ( ) + −

    12

    yields1= lim

    →∞

    log |Γ ( + )|log

    = lim→∞

    log√2 − +

    ¡+ − 12

    ¢log ( )

    log=

    Hence, we find one of the important properties that the Mittag-Leffler function ( ) is an entire

    complex function in of order = 1 for Re ( ) 0 and any . For real and , the Mittag-

    Leffler function ( ) is real on the real axis. Moreover, for real positive and , the Mittag-Leffler

    2

  • function ( ) attains its maximum on the real positive axis, because¯̄ ¡ ¢¯̄

    ≤P∞

    =0| |

    |Γ( + )| =P∞=0 Γ( + ) = ( ).

    The number ( ) of zeros 1 2 of an entire function ( ) for which | | ≤ is non-decreasingin and, if ( ) is of order , then ( ) = ( + ). Roughly stated [37, p. 249], the higher the order

    of an entire function, the more zeros it may have in a given region of the complex plane. Moreover,

    if the modulus of a zero is = | |, then

    ∞X

    =1

    1

    ( )converges if

    and the lower bound of is the exponent 1 of convergence; thus 1 ≤ . IfP∞

    =1

    ³ ´ +1converges

    for an integer , then + 1 1 and the smallest integer is called the genus of ( ). In any case,

    ≤ 1 ≤ . We may have 1 , for example for ( ) = , whose order is = 1, but the exponentof convergence 1 = 0, because does not have zeros. Applied to the Mittag-Leffler function ( )

    of order = 1 , the theory indicates that more zeros are expected for small than for large , which

    seems contradictory to the monotonicity of ( ) for 0 1 on the negative real axis in art. 40.The determination of the zeros of ( ) is generally difficult [16, Sec. 4.6],[32] and omitted here.

    For some special values of the parameter , the Taylor series (1) reduces to known functions, such

    as 0 ( ) = 1Γ( )11− for | | 1, 1 1 ( ) = and 2 1 ( ) = cosh (

    √). From the incomplete Gamma

    function [1, 6.5.29], we have

    1 ( ) =∗ ( − 1 ) =

    Γ ( − 1)

    ∞X

    =0

    (− )! ( − 1 + )

    = 1−µ1−

    1

    Γ ( − 1)

    Z ∞− −2

    ¶(3)

    which can also be written in terms of Kummer’s confluent hypergeometric function [1, 13.1.2]

    ( ; ) =Γ ( )

    Γ ( )

    ∞X

    =0

    Γ ( + )

    Γ ( + ) !

    as

    1 ( ) =∞X

    =0Γ ( + )

    =(1 )

    Γ ( )

    Generalizations of (3) to 1 ( ) of fractional = 1 , where is an integer, are derived in (22) in art.7 below. Many analytic functions can be expressed in terms of the hypergeometric function, defined

    by Gauss’s series [1, 15.1.1]

    ( ; ; ) =Γ( )

    Γ( )Γ( )

    ∞X

    =0

    Γ( + )Γ( + )

    Γ( + ) !convergent for | | 1 (4)

    where the argument of the Gamma functions is of the form + with = 1, in contrast to the

    Mittag-Leffler function in (1) where = is real positive. Just the fact that is real and not an

    integer colors the theory of the Mittag-Leffler function ( ) and causes its main challenges.

    3

  • 3 Deductions from the definition (1) of ( )

    1. Special values of and . If = 0, then, since lim →0 1Γ( ) = 0, we have

    0 ( ) =∞X

    =1Γ ( )

    =∞X

    =0

    +1

    Γ ( + )

    and, hence,

    0 ( ) = ( ) (5)

    If = 0, then

    0 ( ) =1

    Γ ( )

    1

    1−for | | 1

    From ( ) = 1Γ( ) +P∞

    =1 Γ( + ) , we observe that lim →∞ ( ) =1

    Γ( ) .

    2. After splitting odd and even indices in the -sum of (1), we obtain

    (− ) =∞X

    =0

    (−1)Γ ( + )

    =∞X

    =0

    2

    Γ ( + 2 )−

    ∞X

    =0

    2 +1

    Γ ( + + 2 )

    and

    (− ) = 2¡2¢− 2 +

    ¡2¢

    (6)

    Property (6) cannot be expressed for (− ) in terms of itself and motivates our viewpoint that thecomplex function theory of the Mittag-Leffler function should focus on ( ), rather than on ( ).

    The differentiation rule in art. 6 below is the more fundamental motivation.Adding ( ) = 2

    ¡2¢+ 2 +

    ¡2¢to (6) leads to

    2

    ¡2¢=

    ( ) + (− )2

    (7)

    and, similarly,

    2 +

    ¡2¢=

    ( )− (− )2

    (8)

    Examples From 1 1 ( ) = 1 ( ) = , the relation (7) indicates that 2 ( ) = cosh (√) and

    next

    4 ( ) =1

    2

    ncosh

    ³14

    ´+ cos

    ³14

    ´o

    The odd variant (8) gives 2 2¡2¢= sinh .

    3. Cyclotomic property. When introducing the identityP −1

    =0

    2

    = 1−2

    1−2 = 1 | into (1),

    −1X

    =0

    ³2

    ´=

    ∞X

    =0

    P −1=0

    2

    Γ ( + )=

    ∞X

    =0

    1 |

    Γ ( + )=

    ∞X

    =0Γ ( + )

    we obtain

    ( ) =1

    −1X

    =0

    ³2´

    (9)

    4

  • For = 2 in (9),

    2 2¡2¢= ( ) + (− )

    we retrieve the relation equivalent to (6), because (9) essentially follows by multisectioning of a power

    series [34, Section 4.3] of which splitting in odd and even terms is the obvious case in = 2 sections.

    Example The case = = 1 in (9),

    1 ( ) = ( ) =1

    −1X

    =0

    1 2

    (10)

    can be extended to certain integer values of . Indeed, using

    1 ( ) =∞X

    =0( + − 1)!

    =∞X

    = −1

    −( −1)

    != −( −1)

    ⎝ −−2X

    =0!

    in (9) yields

    ( ) =1

    −1X

    =0

    1

    ³1 2

    ´=

    − −1 −1X

    =0

    − 2 ( −1)

    ⎝1 2

    −−2X

    =0

    2

    !

    The last double sum

    −1X

    =0

    −2X

    =0

    − 2 ( −1− )!=

    −1X

    =1

    −1−

    ( − 1− )!

    −1X

    =0

    − 2 =−1X

    =1

    −1−1 |

    ( − 1− )!

    vanishes for all integers ≤ . Thus, for 0 ≤ ≤ , we arrive at

    ( ) =− −1 −1X

    =0

    − 2 ( −1)1 2

    (11)

    4. Mittag-Leffler function with = + where ∈ Z. Another rewriting of the definition (1) of( ),

    ( ) =∞X

    =0Γ ( + )

    =∞X

    =1

    −1

    Γ ( − + )=1Ã ∞X

    =0Γ ( − + )

    −1

    Γ ( − )

    !

    leads to “the shift down of by ” formula

    ( ) =1µ

    − ( )−1

    Γ ( − )

    ¶(12)

    or, similarly after → + ,

    ( ) =1

    Γ ( )+ + ( )

    from which, for = 0, we find again (5). If = + in (12), then we obtain a recursion in

    + ( ) =1µ

    +( −1) ( )−1

    Γ ( + ( − 1) )

    ¶(13)

    5

  • After iteration of (13), we find

    + ( ) = ( )−−1X

    =0Γ ( + )

    (14)

    The case for = 1 in (14) is again (12). Relation (14) is directly retrieved from the definition (1) of

    ( ),

    + ( ) =∞X

    =0Γ ( + ( + ))

    =∞X

    =

    Γ ( + )= −

    Ã ∞X

    =0Γ ( + )

    −−1X

    =0Γ ( + )

    !

    Similarly,

    − ( ) =∞X

    =0Γ ( + ( − ))

    =∞X

    =−

    +

    Γ ( + )=

    Ã ∞X

    =0Γ ( + )

    +X

    =1

    Γ ( − )

    !

    and, hence,

    −− ( ) = ( ) +

    X

    =1

    ¡1¢

    Γ ( − )(15)

    Subtracting (15) from (14) yields3

    + ( )− − − ( ) = −−1X

    =−Γ ( + )

    illustrating for real 0, ≥ 0 and positive that 2 + ( ) − ( ).

    5. Differentiation with respect to . The derivative of the definition (1) with respect to is

    ( ) =∞X

    =1

    −1

    Γ ( + )=

    ∞X

    =0

    ( + 1)

    Γ ( + + )

    =1∞X

    =0

    ( + + − 1)Γ ( + + )

    −− 1 ∞X

    =0Γ ( + + )

    =1∞X

    =0Γ ( + − 1 + )

    −− 1 ∞X

    =0Γ ( + + )

    and, with the definition (1),

    ( ) =1

    + −1 ( )−− 1

    + ( )

    3which we can write as

    −−1

    =−Γ ( + )

    =−

    { + ( )} =−

    =0

    +

    Γ ( + ( + ))

    = log−

    + ( ) −−

    =0

    ( + ( + ))

    Γ ( + ( + ))

    where ( ) = log Γ( ) is the digamma function [1, 6.3.1].

    6

  • Using (12), + ( ) = 1³

    ( )− 1Γ( )´, simplifies to

    ( ) = −1 ( )− ( − 1) ( ) (16)

    The -derivative

    ( ) =−1

    −1 + −1 ( )− ( − 1)−1

    −1 + ( )

    has a recursive structure, when denoting ( ) = ( ) and 0 ( ) = ( ),

    ( ) = −1 ( + − 1)− ( − 1) −1 ( + )

    which can be iterated resulting in

    ( ) =X

    =0

    ( ) + − ( )

    where the coefficients ( ) = 1, −1 ( ) = −³

    + ( −1)2 −( +1)2

    ´and 0 ( ) =

    Q −1=0 ( − 1 + ). In general, ( ) are polynomials in and of order − . Unfortunately

    4,

    it is not easy to write all coefficients ( ) in closed form. With (14), we have

    ( ) ( ) =X

    =0

    ( )

    Ã

    − ( )−−1X

    =0Γ ( − + )

    !

    (17)

    For = 0, we obtain from (17) with ( )¯̄=0= 1Γ( + ) ,

    1

    Γ ( + )=

    1 X

    =0

    ( )

    Γ ( − + )

    and, with Γ( + )Γ( − + ) =Q

    =1 ( + − ), the polynomial nature of ( ) is apparent:

    X

    =1

    Y

    =1

    ( + − ) ( ) = −−1Y

    =0

    ( − 1 + )

    Art. 10 below will present a closed form for ( ).

    4For example,

    22 ( )

    2= +2 −2 ( )− (2 + − 3) +2 −1 ( ) + ( − 1) ( + − 1) +2 ( )

    33 ( )

    3= +3 −3 ( )− (3 + 3 − 6) +3 −2 ( )

    + {(2 + − 3) ( + 2 − 2) + ( − 1) ( + − 1)} +3 −1 ( )− ( − 1) ( − 1 + ) ( − 1 + 2 ) +3 ( )

    44 ( )

    4= +4 −4 ( )− (4 + 6 − 10) +4 −3 ( )

    + {(2 + − 3) ( + 2 − 2) + ( − 1) ( + − 1) + 3 ( + − 2) ( + 3 − 3)} +4 −2 ( )

    − {( + 3 − 2) (2 + − 3) ( + 2 − 2) + ( − 1) ( + − 1) ( + 3 − 2) + ( − 1) ( − 1 + ) ( − 1 + 2 )} +4 −1 ( )

    + ( − 1) ( − 1 + ) ( − 1 + 2 ) ( − 1 + 3 ) +4 ( )

    7

  • 6. Differentiation recursion. Using the functional equation of the Gamma function, Γ ( + 1) = Γ ( ),we write

    ( ) =∞X

    =0Γ ( + )

    =∞X

    =0( − 1 + )Γ ( − 1 + )

    Thus,

    −1 ( ) =∞X

    =0

    −1+

    ( − 1 + )Γ ( − 1 + )

    Differentiation with respect to gives us

    n−1 ( )

    o=

    ∞X

    =0

    −2+

    Γ ( − 1 + )= −2

    ∞X

    =0

    ( )

    Γ ( − 1 + )

    With the definition (1), we arrive for any at the differentiation recursion in ,

    n−1 ( )

    o= −2 −1 ( ) (18)

    Differentiating (18) again -times and using the recursion (18) yields

    n−1 ( )

    o= −1− − ( ) (19)

    7. Fractional values of . Let = where ≤ are integers, then the differentiation formula (19)becomes, for = 1,

    n−1

    ³ ´o= −1

    ∞X

    =0

    Γ¡− +

    ¢ = −1∞X

    =0

    ( − )

    Γ¡+ ( − )

    ¢

    = −1∞X

    =−Γ¡+

    ¢ = −1X

    =1

    Γ¡−

    ¢ + −1∞X

    =0Γ¡+

    ¢

    and we find that

    n−1

    ³ ´o= −1

    X

    =1

    Γ¡−

    ¢ + −1³ ´

    (20)

    For = 1, (20) is

    n−1 ( )

    o=

    −1−

    Γ ( − )+ −1 ( )

    In particular, when = 1, then it holds for ≥ 1 that ( ) = ( ), which illustrates that= ( ), explicitly given in (10), is a solution of the differential equation = .

    For = 1, (20) reduces to

    n−1

    1

    ³1´o

    =X

    =1

    −1− 1

    Γ¡− 1

    ¢ + −1 1³

    from whichn− −1

    1

    ³1´o

    = −X

    =1

    −1− 1

    Γ¡− 1

    ¢

    8

  • Integrating both sides from 0 to yields, for ≥ 1,

    1

    ³1´= 1−

    (

    1{ =1} +X

    =1

    1

    Γ¡− 1

    ¢Z

    0

    −1− 1 −

    )

    (21)

    where the indicator function 1 equals one if the condition is true, else 1 = 0. After letting =1

    in (21) and replacing = − in the summation, we obtain5

    1 ( ) = (1− )

    ⎧⎨

    ⎩1{ =1} +−1X

    =0

    1

    Γ³− 1 +

    ´Z

    0

    ( −1+ )−1 −

    ⎫⎬

    ⎭ (22)

    that reduces, for = 1, to Wiman’s form in [46]

    1 ( ) =

    ⎧⎨

    ⎩1 +Z

    0

    −−1X

    =1

    −1

    Γ³ ´

    ⎫⎬

    For positive real , it holds thatR0

    ( −1+ )−1 − R∞0

    ( −1+ )−1 − = Γ³− 1 +

    ´and (22)

    shows that 1 ( ) (1− )©1{ =1} + − 1{ =1}

    ªand

    1 ( ) (1− )

    which illustrates for = 1 that the entire function 1 ( ) has order = 1 = . This bound also

    reappears in art. 33. Moreover, it is interesting to compare (11) formally, written for an integer,

    ( ) =

    1− (1

    +−1X

    =1

    − 2 ( −1)1 2

    )

    with Bieberbach’s integral (105) for ( ) and with the bound, where 1 is replaced by ,

    ( )

    1−1

    We return to the relation between ( ) and 1 ( ) later in art. 33.Examples If = 1 in (21), we retrieve (3) and when = 2 in (21), we find for = 1

    12

    ³12

    ´=

    (

    1 +1

    Γ¡12

    ¢Z

    0

    − 12−

    )

    =

    (

    1 +2√

    Z √

    0

    − 2)

    5Using the recursion ( ) = ( + 1 ) +Γ( +1)

    − in [1, 6.5.21] for the incomplete Gamma function ( ) =1

    Γ( ) 0−1 − in (22), 1 ( ) = (1− ) 1{ =1} +

    −1=0 − 1 + , shows that

    1 ( ) =∞

    =0Γ +

    = (1− ) 1{ =1} +−1

    =0

    + +−1

    =0Γ +

    which agrees with the Wiman-recursion (22) when → + 1,

    =Γ +

    = 1 +1 ( ) =(1−( +1)) 1{ =0} +

    −1

    =0

    +

    9

  • so that, with the definition [1, 7.1.1] of the error function erf ( ) = 2√R0

    − 2 ,

    12( ) =

    2

    ½1 +

    2√

    Z

    0

    − 2¾=

    2{1 + erf ( )} (23)

    8. Hadamard’s seriesP∞

    =0 ( !) and ( ). Sharp bounds of the Gamma function for + 0,

    that follow from (132) in Appendix A, are

    √2 ( + ) + −

    12 −( + ) ≤ Γ ( + ) ≤

    √2 ( + ) + −

    12 −( + )

    112( + )

    Now,

    √2 ( + ) + −

    12 −( + ) =

    √2 ( + )−

    12

    µ+

    ¶ −12

    õ+

    ¶( + − 12) −( + )!

    ≤√2 ( + )−

    12

    µ+

    ¶ −12

    ÃΓ¡+

    ¢√2

    !

    and replacing the inequality, we approximatively have

    Γ ( + ) ≈¡√2¢1−

    √+

    Ã r

    + Γ

    µ+

    ¶!

    If = 1, then for + 1 0, the above reduces to

    Γ ( + 1) ≈¡√2¢1−

    √+ 1

    Ã r+1Γ

    µ+1¶!

    √ ¡√2¢1−

    √µΓ

    µ+1+ 1

    ¶¶√ ³√

    2´1− µ

    Γ

    µ+1+ 1

    ¶¶

    For large , we approximate as

    Γ ( + 1) ≈√ ¡√

    2¢1−

    √³Γ ( + 1)

    ´(24)

    In art. 51, Gauss’s multiplication formula is written as Γ ( + 1) = (2 )−−12 +

    12Q

    =1 Γ¡+

    ¢

    and indicates (assuming = is an integer) that

    Γ ( + 1) =√ ³√

    2´1− Y

    =1

    Γ

    µ+

    from which, for ≥ 2,√ ¡√

    2¢1− ³

    Γ ( + 1)´

    Γ ( + 1)√ ³√

    2´1− ³

    Γ ( + 1)´

    After introducing (24) in the definition (1) of the Mittag-Leffler function for real, positive , we

    approximately obtain

    ( ) =∞X

    =0Γ ( + 1)

    ≈¡√2¢ −1√

    ∞X

    =0

    √ ¡ ¢

    ( !)

    10

  • but Gauss’s multiplication formula produces a lower bound

    ( ) =∞X

    =0Γ ( + 1)

    ¡√2¢ −1√

    ∞X

    =0

    ¡ ¢

    ( !)(25)

    About 10 years before Mittag-Leffler has introduced his function ( ), Hadamard [20, p. 180]

    suggests in his study of entire functions that ( ) =P∞

    =0 Γ( +1) ∼(√2 )

    −1

    √P∞

    =0( )( !)

    for large .

    Hadamard derives an exact -fold integral for the last series, from which he deduces thatP∞

    =0 ( !)1

    . Art. 7 shows that ( ) 11

    . Combined with (25) leads to

    ∞X

    =0( !)

    ¡√2¢1−√

    1 1

    which is considerably sharper for 1 than Hadamard’s [20, p. 180] bound.

    We also give Hadamard’s [20, p. 180] nice argument, starting from

    1

    =∞X

    =0Γ ( + 1)

    and letting 0 = , which runs over fractions for 1, so that1

    =P∞

    0=0

    0

    Γ( 0 +1) . Comparing

    terms with ( ) =P∞

    =0 Γ( +1) , then shows that ( )1

    . For 1, Hadamard states that

    ( )£1¤ 1 1

    . However, the bounds presented here based on the theory of the Mittag-Leffler

    function are sharper than Hadamard’s estimates.

    9. Logarithmic derivative. The logarithmic derivative log ( ) = ( )( ) follows directly from(16) as

    log ( )=1µ

    −1 ( )

    ( )− ( − 1)

    ¶(26)

    Since − 1 + − 1 for ≥ 1 because 0, we have for positive real and 1,

    ( ) =∞X

    =0( − 1 + )Γ ( − 1 + )

    1

    − 1

    ∞X

    =0Γ ( − 1 + )

    =−1 ( )

    − 1

    and that ( − 1) −1( )( ) . The logarithmic derivative (26) becomeslog ( )

    0, illustrating that

    log ( ) is increasing for real ≥ 0 and 1. A little more precisely with − 1 + − 1 +for ≥ 1, we have for real positive and for 1− (because then Γ ( − 1 + ) 0 for ≥ 1),

    ( ) =1

    Γ ( )+

    ∞X

    =1( − 1 + )Γ ( − 1 + )

    1

    Γ ( )+

    1

    − 1 +

    ̰X

    =0Γ ( − 1 + )

    −1

    Γ ( − 1)

    !

    and

    ( )1

    ( − 1 + )

    µ

    Γ ( )+ −1 ( )

    11

  • Thus, µ1−

    1

    Γ ( ) ( )

    ¶−1 ( )

    ( )− ( − 1)

    and (26) becomeslog ( ) 1

    µ1−

    1

    Γ ( ) ( )

    Example If = 1 and = 1, then 1 1 ( ) = and the above inequality results in the well-knownbound (see e.g. [41, p. 103]) that − 1− for all real .

    10. Taylor expansion around 0. Since the Mittag-Leffler function ( ) is an entire function, anyTaylor expansion around an arbitrary (finite) point 0 has infinite radius of convergence,

    ( ) =∞X

    =0

    1

    !

    ( )¯̄¯̄= 0

    ( − 0)

    With the -th derivative (17), we find

    ( ) =∞X

    =0

    ⎧⎨

    ⎩X

    =0

    ( )

    Ã

    − ( 0)−−1X

    =0

    0

    Γ ( − + )

    !⎫⎬

    ³− 00

    ´

    !

    but, unfortunately (see art. 5), the coefficients ( ) are not available in closed form. However,the closed form (19), which is a rather fundamental property of ( ), opens a new avenue. The

    Taylor expansion around 0 of

    −1 ( ) =∞X

    =0

    1

    !

    n−1 ( )

    o¯̄¯̄= 0

    ( − 0)

    becomes with the differentiation recursion (19)

    −1 ( ) = −10

    ∞X

    =0

    − ( 0)

    !−0 ( − 0)

    The function −1 ( ) has a branch cut at the negative real axis for real and , implying that

    the radius of convergence equals | 0|. Using its definition in Section 2,

    1= lim

    →∞

    ¯̄¯̄¯

    − −10 − −1 ( 0) !−0 − ( 0) ( + 1)!

    ¯̄¯̄¯= −10 lim→∞

    ¯̄¯̄ − −1 ( 0)

    − ( 0) ( + 1)

    ¯̄¯̄

    then indicates for any finite 6= 0, and that¯̄¯̄ − −1 ( )

    − ( )

    ¯̄¯̄ ∼ ( + 1) if →∞ (27)

    Introducing (26) for 6= 0 that

    1 = lim→∞

    ¯̄¯̄ − −1 ( )

    − ( ) ( + 1)

    ¯̄¯̄ = lim

    →∞

    ¯̄¯̄ log − ( ) −

    µ1−

    ¶¯̄¯̄

    shows that

    lim→∞

    ¯̄¯̄ 1 log − ( )

    ¯̄¯̄ = 0

    12

  • Thus, the logarithmic derivative for 6= 0 and finite and is of order log − ( ) =¡

    1− ¢ forany positive 0.

    We proceed by removing real powers of = log that destroy analyticity in the complex plane

    and introduce formally the Taylor series³

    0

    ´1−=³

    0− 1 + 1

    ´1−=P∞

    =0

    ¡1− ¢³

    0− 1´, valid

    for any and | | | 0|, in

    ( ) =

    µ

    0

    ¶1− ∞X

    =0

    − ( 0)

    !

    µ

    0− 1¶

    yielding, after executing the Cauchy product,

    ( ) =∞X

    =0

    (X

    =0

    µ1−−

    ¶− ( 0)

    !

    0− 1¶

    for | | | 0|

    Next, letting = and 0 = 0 , we first expand in a Taylor series around 0õ

    0

    ¶ 1− 1

    !

    =∞X

    =0

    1

    !

    õ

    0

    ¶ 1− 1

    ! ¯̄¯̄¯= 0

    ( − 0)

    asõ

    0

    ¶ 1− 1

    !

    =X

    =0

    µ ¶µ

    0

    ¶(−1) − =

    X

    =0

    µ ¶µ− 00

    + 1

    ¶(−1) −

    =∞X

    =0

    ⎧⎨

    ⎩X

    =0

    µ ¶µ ¶(−1) −

    ⎫⎬

    µ− 00

    Substitution and reversing the - and -sum yields

    ( ) =∞X

    =0

    ⎣∞X

    =0

    X

    =0

    µ1−−

    ¶− ( 0)

    !

    ⎧⎨

    ⎩X

    =0

    µ ¶µ ¶(−1) −

    ⎫⎬

    ⎦µ− 00

    =∞X

    =0

    ⎣∞X

    =0

    − ( 0)

    !

    ∞X

    =

    µ1−−

    ¶X

    =0

    µ ¶µ ¶(−1) −

    ⎦µ− 00

    The characteristic coefficients [39, Appendix] of a complex function ( ) with Taylor series ( ) =P∞

    =0 ( 0) ( − 0) , defined by [ ]| ( 0) =1!

    ³( )− ( 0)

    ´¯̄¯= 0, possesses a general

    form

    [ ]| ( 0) =X

    =1 = ; 0

    Y

    =1

    ( 0) (28)

    and obeys [ ]| ( 0) = 0 if 0 and . Also, [1 ]| ( 0) = ( 0), while (28) indicatesthat [ ]| ( 0) = 1 ( 0). We can show [38] that

    [ ]|(1+ ) =X

    =1

    (−1) +µ ¶µ ¶

    =!

    !

    X

    =

    ( )S( ) (29)

    13

  • where ( ) and S( ) are the Stirling numbers of the first and second kind [1, Sec. 24.1.3 and 24.1.4],respectively. We apply these properties to the Taylor series

    ( ) =∞X

    =0

    "∞X

    =0

    − ( 0)

    !

    ∞X

    =

    µ1−−

    ¶[ ]|

    (1+ )1

    #µ− 00

    =∞X

    =0

    "X

    =0

    − ( 0)

    !

    X

    =

    µ1−−

    ¶[ ]|

    (1+ )1

    #µ− 00

    Finally, we arrive with (29) at the Taylor series of ( ) around 0,

    ( ) =∞X

    =0

    "−0

    !

    X

    =0

    − ( 0)X

    =

    µ ¶Γ (2− )

    Γ (2− − + )

    X

    =

    ( )S( )1#

    ( − 0) (30)

    from which the closed form of the -th derivative of the Mittag-Leffer function, evaluated at 0 = ,

    follows as

    ( )¯̄¯̄=

    = −X

    =0

    − ( )X

    =

    µ ¶Γ (2− )

    Γ (2− − + )

    X

    =

    ( )S( )1

    (31)

    Since the Mittag-Leffer function ( ) is an entire function, the -th Taylor coefficient around

    decreases faster than any power of ( − ) for large and any , we deduce that ( )¯̄¯=

    =¡!¢. Thus, the -th Taylor coefficient 1!

    ( )¯̄¯=

    increases at most as a polynomial in of

    order . In contrast to the relatively simple Taylor series of ( ) in (1) around the origin, the

    general form (31) emphasizes the complicated nature of the Mittag-Leffer function ( ) elsewhere

    in the complex plane.

    As a check of (31), for = = 1, we have with the orthogonality condition of the Stirling numbers

    X

    =

    ( )S( ) = (32)

    that

    ( ) =∞X

    =0

    "X

    =0

    1 1− ( 0)X

    =

    µ ¶Γ (2− )

    Γ (2− − + )

    # ³ − 00

    ´

    !

    Applying (15) yields − 1 1− ( ) = 1 1 ( ) +P

    =1( 1 )Γ(1− ) and 1 1− ( ) = so that

    ( ) = 0∞X

    =0

    "X

    =0

    −0

    µ ¶1

    Γ (1− + )

    #( − 0)

    !

    = 0∞X

    =0

    ( − 0)!

    = 0 − 0 =

    11. Second order logarithmic derivative. Similarly as in art. 9, we can directly differentiate (26)again,

    2 log ( )2

    = −12

    µ−1 ( )

    ( )− ( − 1)

    ¶+1 −1 ( )

    ( )

    14

  • using (16), resulting in

    −1 ( )

    ( )=1(

    −2 ( )

    ( )+

    −1 ( )

    ( )−µ

    −1 ( )

    ( )

    ¶2)

    After tedious manipulations, we arrive at the second order logarithmic derivative

    2 log ( )2

    =1

    ( )2

    (−2 ( )

    ( )+ ( − 1)− ( − 1) −1

    ( )

    ( )−µ

    −1 ( )

    ( )

    ¶2)

    (33)

    12. The Taylor series of log ( ) around = 0. From the power series definition (1) of ( ),we have

    lim→0

    log ( )= lim

    →0

    1P∞

    =0 Γ( + )

    lim→0

    1̰X

    =0Γ ( − 1 + )

    −∞X

    =0

    ( − 1)Γ ( + )

    !

    = Γ ( ) lim→0

    ∞X

    =1

    −1

    Γ ( + )

    and find

    lim→0

    log ( )=

    Γ ( )

    Γ ( + )

    Proceeding in this way to higher-order derivatives becomes cumbersome. The general theory of

    characteristic coefficients [39, Appendix] provides us with

    log ( ) = log 0 ( 0) +∞X

    =1

    ÃX

    =1

    (−1) −1

    0 ( 0)[ ] ( 0)

    !

    ( − 0)

    Confining to 0 = 0, the characteristic coefficients (28) of the Mittag-Leffler function ( ) are

    [ ]| ( ) =X

    =1 = ; 0

    Y

    =1

    1

    Γ ( + )

    and the Taylor series of log ( ) around 0 = 0 follows as

    log ( ) =∞X

    =0

    where we define 0 = log (−Γ ( )) and the coefficients for 0

    = −X

    =1

    (−Γ ( ))[ ]| ( )

    =Γ ( )

    Γ ( + )+

    −1X

    =2

    (−1) −1 X

    =1 = ; 0

    Y

    =1

    Γ ( )

    Γ ( + )+(−1) −1

    µΓ ( )

    Γ ( + )

    15

  • The list of the coefficients for = 1 up to = 5 is

    1 =Γ ( )

    Γ ( + )

    2 =Γ ( )

    Γ ( + 2 )−1

    2

    µΓ ( )

    Γ ( + )

    ¶2

    3 =Γ ( )

    Γ ( + 3 )+

    Γ2 ( )

    Γ ( + 2 )Γ ( + )+1

    3

    µΓ ( )

    Γ ( + )

    ¶3

    4 =Γ ( )

    Γ ( + 4 )−

    Γ2 ( )

    Γ ( + 3 )Γ ( + )−1

    2

    µΓ ( )

    Γ ( + 2 )

    ¶2+

    Γ3 ( )

    Γ ( + 2 )Γ2 ( + )−1

    4

    µΓ ( )

    Γ ( + )

    ¶4

    5 =Γ ( )

    Γ ( + 5 )−

    Γ2 ( )

    Γ ( + 4 )Γ ( + )−

    Γ2 ( )

    Γ ( + 3 )Γ ( + 2 )+

    Γ3 ( )

    Γ ( + 3 )Γ2 ( + )+

    Γ3 ( )

    Γ ( + )Γ2 ( + 2 )

    −Γ4 ( )

    Γ3 ( + )Γ ( + 2 )+1

    5

    µΓ ( )

    Γ ( + )

    ¶5

    Unfortunately, it seems difficult to further sum the terms in . The closest zero to 0 = 0 lies at a

    distance from 0 equal to the radius of convergence of log ( ), which is

    1= lim

    →∞

    ¯̄¯̄ +1

    ¯̄¯̄ = lim

    →∞

    ¯̄¯P +1

    =1(−Γ( )) [ + 1]| ( )

    ¯̄¯

    ¯̄¯P

    =1(−Γ( )) [ ]| ( )

    ¯̄¯

    The generalization towards the closest zero to 0 requires the Taylor coefficients of (30).

    13. Derivation of ( ) with respect to the parameters and . From the definition (1), partialdifferentiating yields

    ( ) =

    Ã1

    Γ ( )+

    ∞X

    =1Γ ( + )

    !

    =∞X

    =1

    1

    Γ ( )

    ¯̄¯̄= +

    = −∞X

    =1

    ( )

    Γ ( )

    ¯̄¯̄= +

    while, similarly but containing index = 0,

    ( ) = −∞X

    =0

    ( )

    Γ ( )

    ¯̄¯̄= +

    Hence, we observe that

    ( ) =2

    ( )

    Partial differentiating -times gives

    ( ) =∞X

    =0

    1

    Γ ( )

    ¯̄¯̄= +

    which suggest to let = so that

    ( ) =∞X

    =0

    1

    Γ ( )

    ¯̄¯̄= +

    =∞X

    =0

    1

    Γ ( )

    ¯̄¯̄= +

    16

  • leading to the partial differentiation equation for any integer ≥ 0 and any (independent ofand ),

    ( ) =2

    ( ) (34)

    14. A Taylor series approach with Fermi-Dirac integrals. We introduce the Taylor series of the entirefunction 1Γ( + ) around into the definition (1) of ( ),

    ( ) =1

    Γ ( )+

    ∞X

    =1Γ ( + )

    =1

    Γ ( )+

    ∞X

    =1

    ∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    ( + − )

    and

    ( ) =1

    Γ ( )+

    1

    Γ ( ) 1−+

    ∞X

    =1

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    ∞X

    =1

    µ+−

    The reversal in the - and -sum leads to a confinement of | | 1. We will now choose = andevaluate the series

    P∞=1 , that converges for | | 1.

    The Fermi-Dirac integral of order is defined as

    ( ) =1

    Γ( + 1)

    Z ∞

    0 1 +− (35)

    The value of the zero argument in is immediately written in terms of the Eta function,

    (0) = ( + 1) (36)

    where the Eta function ( ) is related to the Riemann Zeta function ( ) as

    ( ) = (1− 21− ) ( ) (37)

    By expanding 11+ − =− +

    1+ − + =P∞

    =1(−1)−1 − ( − ) for Re ( ) 0 in (35), the Dirichlet series

    for all complex is readily deduced as

    ( ) =∞X

    =1

    (−1) −1( )+1

    (38)

    In particular, −1 ( ) = 11+ − . Hence, we can writeP∞

    =1 (− ) = − − −1(log ) for | | 1 and

    (− ) =1

    Γ ( )−

    ∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    − −1(log ) (39)

    For integer negative order and 0, it can be shown [38] that

    − ( ) =−1

    −1

    µ1

    1 + −

    ¶=X

    =1

    ( − 1)!(−1) −1S( )µ

    1

    1 + −

    where S( ) is the Stirling Number of the Second Kind [1, 24.1.4]. Since 11+ −

    = 1 − 11+ , which isequivalent to −1 ( ) = 1− −1 (− ), the -th derivative shows that, for 1,

    − ( ) = (−1) − (− ) (40)

    17

  • and, thus extending the above for | | 1 to,

    (− ) =∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    − −1(− log ) (− ) (41)

    Stretching the convergence constraint in (41) to = 1 and using (36) results in

    (−1) =∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    (− ) (− )

    Further by (37), it holds that (− ) = (1− 21+ ) (− ) = (1− 21+ ) (−1)+1 +1, because (− ) =(−1)+1 +1 and (−2 ) = 0 for 0. Taking into account that the odd Bernoulli numbers 2 +1 = 0

    for 0, we find

    (−1) = −1

    2Γ ( )−

    ∞X

    =1

    2

    (2 )!

    2 −1

    2 −11

    Γ ( )

    ¯̄¯̄=

    (22 − 1) 2 −1

    which converges fast for small . Since (−1) in (1) is an alternating series with decreasing coeffi-cients for 0 and 1 462, it holds that 1Γ( ) (−1)

    1Γ( + ) .

    The major interest of the expansion (41) lies in its fast convergence for small , whereas the

    definition (1) is converging slower for small . Moreover, rewriting (9) as

    ( ) =1

    −1X

    =0

    ³1 2

    ´

    illustrates that any real can be transformed to a value smaller than 1 by choosing = [ ]+1, where

    [ ] is the largest integer smaller or equal to . Indeed, for | | 1,

    ( ) =∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    ⎧⎨

    P[ ]=0 − −1

    ³− log[ ]+1 −

    ³2[ ]+1 + 1

    ´´

    [ ] + 1

    ⎫⎬

    µ−[ ] + 1

    15. Möbius inversion. The first Möbius inversion pair is

    ( ) =∞X

    =1

    ( ) ⇐⇒ ( ) =∞X

    =1

    ( ) ( ) (42)

    where ( ) is the Möbius function. Let ( ) = Γ( + ) in (42), then

    ( ) =∞X

    =1

    ( ) =∞X

    =1Γ ( + )

    = ( )−1

    Γ ( )

    and Möbius inversion ( ) =P∞

    =1 ( ) ( ) in (42) yields, for 6= 0,

    Γ ( + )=

    ∞X

    =1

    ( )

    µ( )−

    1

    Γ ( )

    ¶(43)

    With = , (43) simplifies to Γ( + ) =P∞

    =1 ( )³

    ( )− 1Γ( )´.

    18

  • Invoking 2¡2¢=

    ( )+ (− )2 in art. 2 in (43)

    Γ ( + )=

    ∞X

    =1

    ( )

    µ2 2

    ¡2¢− (− )−

    1

    Γ ( )

    = 2∞X

    =1

    ( )

    µ

    2

    ¡2¢−

    1

    Γ ( )

    ¶−

    ∞X

    =1

    ( )

    µ(− )−

    1

    Γ ( )

    and using (43) leads to

    ∞X

    =1

    ( )

    µ(− )−

    1

    Γ ( )

    ¶=

    2 2

    Γ ( + 2 )−Γ ( + )

    (44)

    which is an instance of the general Möbius function identity, proved in [38],

    X

    =1

    ( )X

    =1

    (−1) ( ) = 2 (2)− (1) (45)

    holds for any function and any 1. With = , (44) simplifies to 22

    Γ( +2 ) − Γ( + ) =P∞

    =1 ( )³

    (− )− 1Γ( )´.

    16. Mertens function. Applying Abel summation using the Mertens6 function −1 ( ) =P

    =1 ( ),

    we obtain

    ( ) =∞X

    =1

    −1 ( ) ( ( )− (( + 1) ))) + lim→∞

    ( ) −1 ( )

    Hence,

    Γ ( + )=

    ∞X

    =1

    −1 ( )³ ³ ´

    − +³

    +´´+ lim

    →∞

    µ ¡ ¢−

    1

    Γ ( )

    ¶−1 ( )

    and the limit vanishes if 0. WithR ( +1)

    ( ) = +¡

    +¢−

    ¡ ¢, the

    corresponding integral representation is

    Γ ( + )= −

    ∞X

    =1

    −1 ( )

    Z ( +1)( )

    = −∞X

    =1

    Z ( +1)−1

    ³h i´( )

    and

    Γ ( + )= −

    Z ∞−1

    ³h i´( ) for 0 (46)

    Similarly for (44), it holds that

    2 2

    Γ ( + 2 )−Γ ( + )

    = −Z ∞

    −1

    ³h i´(− ) for 0 (47)

    Although (46) and (47) are remarkable, further progress to estimate the order of −1 ([ ]) requires a

    study of ( ).

    6A sufficient condition to prove the Riemann Hypothesis is to demonstate that the Mertens function behaves as

    −1 ( ) =12+ for large .

    19

  • 4 Integrals containing ( )

    17. Integral duplication formula for (− ). Using the duplication formula of the Gamma function,Γ (2 ) = 1√ 22 −1Γ ( )Γ

    ¡+ 12

    ¢, the definition (1) is rewritten as

    ( ) =∞X

    =0Γ ( + )

    =22 −1√

    ∞X

    =0

    Γ¡12 + +

    ¢ ¡22

    ¢

    Γ (2 + 2 )

    Invoking the Euler integral Γ ( ) =R∞0

    −1 − for Re ( ) 0,

    (− ) =22 −1√

    ∞X

    =0

    ¡−22

    ¢

    Γ (2 + 2 )

    Z ∞

    0

    − 12

    =22 −1√

    Z ∞

    0

    − 12−

    ∞X

    =0

    (− (4 ) )Γ (2 + 2 )

    and the definition (1), we find an integral duplication formula for Re ( ) ≥ 0 and Re ( ) ≥ −12 ,

    (− ) =22 −1√

    Z ∞

    0

    − 12−

    2 2 (− (4 ) ) (48)

    After substituting the Gamma duplication formula in the slightly rewritten power series

    ( ) =∞X

    =0Γ ( + )

    =∞X

    =0( − 1 + )Γ ( − 1 + )

    =22 −2√

    ∞X

    =0

    (4 )

    Γ (2 − 1 + 2 )Γ

    µ+ −

    1

    2

    we find alternatively,

    (− ) =22 −2√

    ∞X

    =0

    (−4 )Γ (2 − 1 + 2 )

    Z ∞

    0

    − 32

    =22 −2√

    Z ∞

    0

    − 32−

    ∞X

    =0

    (− (4 ) )Γ (2 − 1 + 2 )

    for Re ( ) ≥ 12 and Re ( ),

    (− ) =22( −1)√

    Z ∞

    0

    −32−

    2 2 −1 (− (4 ) ) (49)

    which also applies to ( ) = 1 ( ) after choosing = 1.

    Let = (4 ) or = 41

    , then (49) is, for real, nonnegative ,

    (− ) =22 −2√

    µ1

    4− 1¶ −1

    2Z ∞

    0

    −32− 1

    4− 1

    2 2 −1 (− )

    Let = 14− 1 or = (4 )− , then we arrive at the Laplace transform

    Z ∞

    0

    − − 32 2 2 −1 (− ) =

    22 −2 −12

    ¡− (4 )−

    ¢(50)

    20

  • Example For = 1 and = 12 , the Laplace transform (50) becomes√√ 1

    2

    µ−

    1

    2æ=

    Z ∞

    0

    − 12− −

    √= 2

    Z ∞

    0

    − 2−

    With 2 + =¡+ 12

    ¢2 − 14 , we have√√ 1

    2

    µ−

    1

    2æ= 2

    14

    Z ∞

    0

    − ( + 12 )2

    = 214

    Z ∞

    12

    − 2 =142√

    Z ∞

    12√

    − 2

    Simplified with erfc( ) = 2√R∞ − 2 = 1− erf ( ) and erf (− ) = 2√

    R −0

    − 2 = − erf ( )

    12(− ) =

    2 2√

    Z ∞− 2 =

    2erfc ( )

    is again (23), because erf (− ) = 2√R −0

    − 2 = − erf ( ).

    18. Integral multiplication formula for ( ). The method of art. 17 is readily generalized.Invoking Gauss’s multiplication formula (123) into the definition (1) yields

    ( ) =∞X

    =0Γ ( + )

    = (2 )12(1− ) − 1

    2

    ∞X

    =0

    −1Y

    =1

    Γ³+ +

    ´

    Γ ( + )( )

    We introduce the Mellin transform (143) of a product of Gamma functions for Re ( ) 0,

    −1Y

    =1

    Γ

    µ+

    ¶=

    Z ∞

    0

    −1 ( )

    where the inverse function ( ) is specified in (146) in art. 57 as a Taylor series in , and we obtain

    ( ) = (2 )12(1− ) − 1

    2

    Z ∞

    0( ) −1

    ∞X

    =0

    ( )

    Γ ( + )

    Thus, for any integer , we arrive at an integral multiplication formula for the Mittag-Leffler functions,

    ( ) = (2 )12(1− ) − 1

    2

    Z ∞

    0( ) −1 ( ) (51)

    The companion of (51) follows similarly from ( ) =P∞

    =0 ( −1+ )Γ( −1+ ) as

    ( ) = (2 )12(1− ) ( −1)+ 1

    2

    Z ∞

    0( ) −2 ( −1)+1 ( ) (52)

    which directly reduces for = 1 to the Mittag-Leffler function ( ) = 1 ( ),

    ( ) = (2 )12(1− ) 1

    2

    Z ∞

    0( ) −1 ( )

    19. Special cases of the integral multiplication formula for ( ). The case = 3 in (51) with 3 ( )in (150) expressed in terms of the modified Bessel function ( ), defined in (148), becomes

    ( ) =33 −

    12Z ∞

    013

    ¡2√ ¢ −1

    2 3 3

    ¡33

    ¢(53)

    21

  • The companion of (53) follows from (52) as

    ( ) =33( −1)+

    12Z ∞

    013

    ¡2√ ¢ −3

    2 3 3 −2¡33

    ¢(54)

    In [2, eq. (25), (28) and (31)], Apelblat has recently presented three remarkable integral functional

    relations,

    ( ) =1√

    Z ∞

    0

    −2

    4 2

    ¡2¢

    (55)

    ( ) =1Z ∞

    0

    r13

    Ã2

    32

    √27

    !

    3

    ¡3¢

    (56)

    ( )− 1√ =

    Z ∞

    0

    1

    ¡2√ ¢√ ( ) (57)

    which he has skilfully derived by manipulations of Laplace transforms. Apart from the last relation

    (57), where ( ) is the Bessel function of order , the first two are special cases of the multiplication

    formula for ( ) in art. 18. Indeed, after substitution of = 142 in the first integral (55), we

    obtain

    1 ( ) =

    12

    √Z ∞

    0

    − 12−

    2 1 ((4 ) )

    which is a special case of (50), more easily noticed from its generalization (60) below, for → 2 ,= 1 and = (4 ) . With = and after substituting =

    3

    27 in (54), we obtain

    ( ) =

    12− Z ∞

    013

    Ã

    2

    r3

    27

    !3 − 5

    2 3 3 −2¡3¢

    which leads to (56) by choosing = 1.

    20. Laplace transform of¡ ¢

    . The Laplace transform of −1¡ ¢

    is

    Z ∞

    0

    − −1³ ´

    =

    Z ∞

    0

    −∞X

    =0

    −1+

    Γ ( + )=

    ∞X

    =0Γ ( + )

    Z ∞

    0

    − + −1

    Fubini’s theorem states that the summation and integration can be reversed provided the integralsR∞0

    − ¡ ¢ andR∞0

    − + −1 = Γ( + )+ exist and the resulting series converges, leading

    to Z ∞

    0

    − −1³ ´

    =∞X

    =0

    Γ ( + )

    Γ ( + ) +(58)

    valid for | | ≤ | |.A first choice is = and = in (58), which restricts so that | | | |,

    Z ∞

    0

    − −1 ( ) =1

    ∞X

    =0

    ³ ´=

    −(59)

    The Laplace transform (59) plays a key role in the theory of the Mittag-Leffler function.

    Two other choices in (58) follow after the introduction of the duplication formula of the Gamma

    functionΓ ( + )

    Γ ( + )=

    2 −1Γ ( + )

    2 Γ¡2 + 2

    ¢Γ¡+12 + 2

    ¢

    22

  • as = 2 and = 2 and =+12 and = 2 , respectively. If = 2 and = 2 , then we find

    Z ∞

    0

    −2−1

    ³2

    ´=

    2 −1 2

    ∞X

    =0

    1

    Γ¡+12 + 2

    ¢µ

    2 2

    resulting, with the definition (1), in

    Z ∞

    0

    −2−1

    ³2

    ´=

    2 −1 2 2+12

    Ã

    (4 ) 2

    !

    (60)

    while the third choice = +12 and = 2 leads to

    Z ∞

    0

    − +12−1

    ³2

    ´=

    2 −1+12 2 2

    Ã

    (4 ) 2

    !

    (61)

    Both (60) and (61) are slightly more general than and reduce to (50) and (48), respectively, for = −1.

    21. Generalized integration. By using a variation of the integral of the Beta-function [1, 6.2],R0

    −1 ( − ) −1 = + −1 Γ( )Γ( )Γ( + ) for real 0, Re ( ) 0 and Re ( ) 0, we obtain ageneralized integral variant of the Mittag-Leffler function ( ) in (1),

    1

    Γ ( )

    Z

    0( − ) −1 −1

    ³ ´=

    1

    Γ ( )

    ∞X

    =0Γ ( + )

    Z

    0

    + −1 ( − ) −1

    = + −1∞X

    =0Γ ( + )

    Γ ( + )

    Γ ( + + )

    which reduces for = and = to

    1

    Γ ( )

    Z

    0( − ) −1 −1 ( ) = −1+ + ( ) (62)

    The -fold integral7 (62) for = and = 1 possesses the same form as the -fold differentiation

    in (19) n−1 ( )

    o= −1− − ( )

    that is better recognized with Cauchy’s integral for the -th derivative of an analytic function,( )¯̄¯= 0

    = !2R( 0)

    ( )( − 0) +1

    , where ( 0) is a closed contour around 0,

    (−1) +1 !2

    Z

    ( )( − )− −1 −1 ( ) = −1− − ( )

    Hence, (62) written with the reflection formula (122), suggests that

    sin Γ (1− )2

    Z

    ( )( − ) −1 −1 ( ) = −1− − ( )

    7The -fold integral is

    ( ) = 11

    2

    −1

    ( ) =1

    ( − 1)!( − ) −1 ( )

    The generalization towards fractional calculus, where the integer is extended to a real number, is treated in [17].

    23

  • holds for any negative real , leading to fractional derivatives. The Laplace transform (59) easier

    connects to fractional derivatives, avoiding the contour integral. Applications of the Mittag-Leffler

    function to fractional calculus are amply illustrated in [16].

    Example For = = 1 in (62) and, next, = 1, = and + 1→ , we obtain, for Re ( ) 1,

    1 ( ) =1

    Γ ( − 1)

    Z 1

    0(1− ) −2

    Let = 1− , then the incomplete Gamma function appears

    1 ( ) =Γ ( − 1)

    Z 1

    0

    −2 − =1−

    Γ ( − 1)

    Z

    0

    −2 −

    which is again equal to (3).

    22. Integration of a product of Mittag-Leffler functions. We extend the idea in art. 21 and consider,for ≥ 0,

    =1

    Γ ( )

    Z

    0( − ) −1 −1 ( ) ( ( − ) )

    =1

    Γ ( )

    ∞X

    =0

    ∞X

    =0Γ ( + ) Γ ( + )

    Z

    0( − ) + −1 + −1

    Introducing the integral of the Beta-function results in

    =+ −1

    Γ ( )

    ∞X

    =0

    ∞X

    =0

    ( )

    Γ ( + )

    ( )

    Γ ( + )

    Γ ( + )Γ ( + )

    Γ ( + + + )

    After the choice = , = , = and = , the double sum simplifies to

    =+ −1

    Γ ( )

    ∞X

    =0

    ∞X

    =0

    ( ) ( )

    Γ ( + + + )

    Further computations require the choice = ,

    =+ −1

    Γ ( )

    ∞X

    =0

    ∞X

    =0

    ( ) ( )

    Γ ( + + ( + ))

    Let = + , then 0 ≤ and = − ≥ 0, while ≥ 0, thus

    =+ −1

    Γ ( )

    ∞X

    =0

    X

    =0

    Γ ( + + )=

    + −1

    Γ ( )

    ∞X

    =0Γ ( + + )

    X

    =0

    µ ¶

    Executing the finite geometric seriesP

    =0

    ³ ´=

    +1−1

    −1leads to

    =+ −1

    Γ ( ) ( − )

    ⎧⎨

    ∞X

    =0

    ( )

    Γ ( + + )−

    ∞X

    =0

    ( )

    Γ ( + + )

    ⎫⎬

    Finally, we arrive for 0 and 0 atZ

    0( − ) −1 −1 ( ) ( ( − ) ) = + −1 +

    ( )− + ( )−

    (63)

    24

  • Clearly8, since lim →0 ( ( − ) ) = 1Γ( ) , then (63) reduces to (62).

    23. An asymptotic result. We invoke the device9, used by Gauss in his grand treatise [15, p. 146] onthe hypergeometric function [23, p. 74] to deduce the Euler integral for the Gamma function from the

    Beta-integral based on

    lim→∞

    µ1−

    ¶= −

    and start from (62)Z

    0

    ³1−

    ´ −1 −1 ( ) = Γ ( ) + ( )

    Let = 1 + with Re ( ) 0 and is real, thenZ

    0

    ³1−

    ´−1 ( ) = Γ ( + 1) + +1 ( )

    and, after taking the limit for →∞Z ∞

    0

    − −1 ( ) = lim→∞Γ ( + 1) + +1 ( )

    Comparison with the Laplace transform (59) for | | | | shows that

    lim→∞Γ ( + 1) + +1 ( ) =

    and, for large real ,

    + ( ) ∼1

    1−1

    Γ ( ) ( )

    Let = , obeying | | | |, then for real = | |→∞

    + ( ) ∼− −

    1− − −1

    Γ ( )∼

    1√2

    12− −

    1− − −− (ln( )−1) cos + sin

    µ1 +

    µ1¶¶

    where in the last step (136) is used. Thus for 0 and for − 2 2 , we evidently find that( )→ 0 for Re ( )→∞ and for 2

    32 that ( )→∞ for Re ( )→ −∞. For = ± 2 , it

    holds that + ( ) ∼ 1√212− − 2

    1− − − 22 , from which | ± ( )| ∼

    12√2for real →∞.

    8 In case → , then, after using de l’Hospital’s rule,

    lim→

    + ( )− + ( )−

    = + ( ) ++ ( )

    =

    and the differential rule (16), formula (63) becomes

    0

    ( − ) −1 −1 ( ) ( ( − ) ) =+ −1

    { + −1 ( ) + ( + 1− ( + )) + ( )}

    9Starting from the above Beta-integral, it holds that

    Γ ( ) lim→∞

    Γ ( + 1)

    Γ ( + + 1)= lim

    →∞ 0

    −1 1− =∞

    0

    −1 − = Γ ( )

    from whichΓ ( + )

    Γ ( )∼

    25

  • 24. Apelblat series. Inspired by series in [2], we call

    ( ) =∞X

    =0

    + −1+ ( )

    an Apelblat series, where the Taylor series of the function ( ) =P∞

    =0 around 0 = 0 converges

    for | | ≤ . Evidently, if = 0 for , then ( ) is a polynomial of order in and ( ) isan entire function. We generalize the method of Apelblat [2]. We take the Laplace transform L [ ] ofboth sides and use (59),

    L [ ( )] =∞X

    =0

    Lh

    + −1+ ( )

    i=

    ∞X

    =0

    − −

    −=

    − +

    ∞X

    =0

    − −

    and obtain a product of Laplace transformed functions

    L [ ( )] = Lh− −1

    − ( )i 1

    µ1¶

    (64)

    The convolution theorem for the Laplace transform suggests us to find the inverse Laplace transform

    L−1£1¡1¢¤of 1

    ¡1¢. Apelblat [2] observes and demonstrates that elegant series follow if a closed

    form for L−1£1¡1¢¤exist, else we can proceed with (152) and Hankel’s integral (141)

    L−1∙1

    µ1¶¸

    =1

    2

    Z + ∞

    − ∞

    1µ1¶

    =1

    2

    ∞X

    =0

    Z + ∞

    − ∞

    − −

    = −1∞X

    =0Γ ( + )

    After taking the inverse Laplace transform of both sides in (64), we formally arrive at the Apelblat

    series, for the free parameter Re ( ) 0,

    ∞X

    =0

    + −1+ ( ) =

    Z

    0( − ) − −1 − ( ( − ) ) −1

    ∞X

    =0Γ ( + )

    (65)

    which directly follows from (62) for = + and → − . Hence, the property (65) of the Apelblatseries is a consequence of the generalized integration property in art. 21.

    Examples 1. The Taylor series of (1 + ) =P∞

    =0

    ¡ ¢converges for all complex provided

    | | 1. The right-hand series in (65) becomes with =¡ ¢

    = Γ( +1)!Γ( − +1) =Γ(− + )!Γ(− ) (− )

    ∞X

    =0Γ ( + )

    =1

    Γ (− )

    ∞X

    =0

    Γ (− + )!Γ ( + )

    (− )

    which reduces if we choose = − toP∞

    =0 Γ( + ) =1

    Γ(− )− . The Apelblat series (65) then

    becomes, for = − and 0 Re ( ) Re ( ),

    ∞X

    =0

    µ−¶

    + −1+ ( ) =

    1

    Γ ( )

    Z

    0

    − −1− ( ) ( − ) −1 − ( − )

    For = 1, = 2, = = 1, we retrieve the series [2, eq. (64)] and for = 2, = 2, = = 1, the

    series [2, eq. (68)].

    26

  • 2. The Taylor series of the Bessel function¡2

    ¢−( ) =

    P∞=0

    (−1)!Γ( +1+ )

    ¡2

    ¢2 in [1, 9.1.10] has

    only even Taylor coefficients 2 =(−14)!Γ( +1+ ) and the odd 2 +1 = 0. If 2 +1 = 0, then

    ∞X

    =0Γ ( + )

    =∞X

    =0

    2

    2

    Γ ( + 2 )=

    2 −1

    ∞X

    =0

    2

    ¡2

    ¢2

    Γ³2 +

    ´Γ³

    +12 +

    ´

    where the duplication formula of Gamma function is used. If 2 =¡ ¢

    2 = Γ(− + )!Γ(− )¡− 2

    ¢then

    ∞X

    =0Γ ( + )

    =

    2 −1Γ (− )

    ∞X

    =0

    Γ (− + ) (−1)¡2

    ¢2

    !Γ³2 +

    ´Γ³

    +12 +

    ´

    Choosing − = 2 yields

    ∞X

    =0Γ ( + )

    =

    2 −1Γ³2

    ´∞X

    =0

    (−1)¡2

    ¢2

    !Γ³−12 + 1 +

    ´ =√ ¡

    2

    ¢− −12

    2 −1Γ³2

    ´ −12( )

    and

    ∞X

    =0

    µ− 2¶

    2 +2 −1+2 ( ) =

    √ ¡2¢ −12

    2 −1Γ³2

    ´Z

    0( − ) − −1 − ( ( − ) )

    −12 −1

    2( )

    while choosing − = +12 yields

    ∞X

    =0Γ ( + )

    =

    2 −1Γ³

    +12

    ´∞X

    =0

    (−1)¡2

    ¢2

    !Γ³2 +

    ´ =√ ¡

    2

    ¢− −22

    2 −1Γ³

    +12

    ´ −22( )

    and the Apelblat series (65) then becomes

    ∞X

    =0

    µ− +12

    ¶2 +2 −1

    +2 ( ) =

    √ ¡2

    ¢1−2

    2 −1Γ³

    +12

    ´Z

    0( − ) − −1 − ( ( − ) ) 2

    2−1 ( )

    For = 0, = 1, = = 1, we retrieve the series [2, eq. (74)],

    ∞X

    =0

    µ−12¶2

    1+2 ( ) = −Z

    0( ) 1 ( − )

    3. Let us now consider the hypergeometric function [1, 15.1] with Taylor series around the origin,

    ( ; ; ) =Γ ( )

    Γ ( )Γ ( )

    ∞X

    =0

    Γ ( + )Γ ( + )

    Γ ( + ) !(66)

    With (66), the Apelblat series (65) at the right-hand side becomes

    ∞X

    =0Γ ( + )

    =Γ ( )

    Γ ( )Γ ( )

    ∞X

    =0

    Γ ( + )Γ ( + )

    Γ ( + )Γ ( + ) !

    27

  • and, if we choose equal to , then

    ∞X

    =0Γ ( + )

    =Γ ( )

    Γ ( )Γ ( )

    ∞X

    =0

    Γ ( + )

    Γ ( + )

    ( )

    !=

    1

    Γ ( )( )

    where ( ) = Γ( )Γ( )P∞

    =0Γ( + )Γ( + ) ! is Kummer’s confluent hypergeometric function [1, 13.1.2].

    The Apelblat series (65) thus becomes

    Γ ( )

    Γ ( )

    ∞X

    =0

    Γ ( + )Γ ( + )

    Γ ( + ) !+ −1

    + ( ) =

    Z

    0( − ) − −1 − ( ( − ) ) −1 ( )

    In order to use the property [1, 13.3.2] of the Kummer function,

    lim→∞

    ³−

    ´= Γ ( )

    1−2 −1

    ¡2√ ¢

    we first choose = −1 and then take the limit →∞ of both sides becomes, with lim →∞Γ( + )Γ( )

    = 1

    (see [1, 6.1.47]),

    ∞X

    =0

    Γ ( + )

    Γ ( + ) !(−1) + −1 + ( ) =

    Z

    0( − ) − −1 − ( ( − ) ) +

    1−2−1

    −1¡2√ ¢

    Let = , then = − 1, we have

    ∞X

    =0

    (−1)+ −1

    !+ ( ) =

    Z

    0( ( − ) )

    −22 −2

    ¡2√ ¢

    which simplified for = 2 to

    ∞X

    =1

    (−1) −1

    ( − 1)! +1( ) =

    Z

    0( ) 0

    ¡2√−

    ¢

    25. Berberan-Santos’ integral for the double argument. Berberan-Santos [4] applied a simplified form(154) of the inverse Laplace contour integral (152), which he deduced in [3], to the Mittag-Leffler

    function ( ). First, let the Laplace transform of a real function ( ) be equal to (− ), hence,R∞0

    − ( ) = (− ). From the Laplace transform with = +

    | (− )| =¯̄¯̄Z ∞

    0

    − − ( )

    ¯̄¯̄ ≤

    Z ∞

    0

    − ( )

    it follows that lim →∞R∞0

    − ( ) = 0 for any direction in which with Re ( ) 0 tends to

    infinity. Since ( ) is entire function of order 1 , ( ) =³ 1 ´

    , we have that (− ) =µ

    1 +¶=

    µ1cos +

    ¶and | (− )| → 0 provided that cos + 0 or 2 −

    32 − . Since = with − 2 ≤ ≤ 2 because Re ( ) 0, the limit lim →∞ | (− )| = 0requires that 0 ≤ ≤ 1. Incidentally, the definition of in Appendix C indicates that = 0 andBerberan-Santos’ inverse Laplace transform (154) then yields

    ( ) =2Z ∞

    0Re ( (− )) cos for 0

    28

  • Next, Berberan-Santos observed from (6) that (− ) = 2¡− 2

    ¢− 2 +2

    ¡− 2

    ¢so that,

    for real ,

    Re ( (− )) = 2¡− 2

    ¢

    Hence, the inverse Laplace transform becomes

    ( ) =2Z ∞

    02

    ¡− 2

    ¢cos for 0

    Finally, taking the Laplace transform of both sides and reversing the integrals

    (− ) =2Z ∞

    02

    ¡− 2

    ¢µZ ∞

    0

    − cos

    results in Berberan-Santos’ remarkable integral for the double argument in (not )

    (− ) =2Z ∞

    0

    2

    ¡− 2

    ¢

    2 + 2for 0 ≤ ≤ 1, Re ( ) 0 (67)

    Example For = 12 , = 1 and with 1 ( ) = , Berberan-Santos’ integral (67) yields again (23),

    12(− ) =

    2Z ∞

    0

    − 2

    2 + 2=

    2erfc ( )

    26. Another proof of Berberan-Santos’ integral (67). Application of Theorem 1 in Appendix B to( ) = ( − ), only valid for 0 ≤ ≤ 1 because then lim →∞

    ( − )2 = 0 (see art. 25), yields

    (− ) =Z ∞

    −∞

    (− )2 + 2

    Using (6), (− ) = 2¡− 2

    ¢− 2 +2

    ¡− 2

    ¢and the fact that 2 +2

    ¡− 2

    ¢is odd

    and 2¡− 2

    ¢is even, again leads to (67).

    UsingR∞0

    − ( 2+ 2) in (67) and reversing the integrals, justified by absolute convergence, gives

    (− ) =2Z ∞

    0

    − 2½Z ∞

    0

    − 22

    ¡− 2

    ¢ ¾=

    Z ∞

    0

    − 2½Z ∞

    0

    − − 12 2 (− )

    ¾

    Thus, Berberan-Santos’ integral (67) is equivalent to the statement that the Laplace transform of the

    Laplace transform of −12 2 (− ) equals

    (−√)√ . In fact, the latter property follows from the

    Stieltjes transform, which is an iteration of the Laplace transform and which is treated in the book

    by Widder [45, Chapter VIII]. Appendix C discusses Gross’ Laplace transform pair [19], in which the

    inverse Laplace transform (152) is of the same form as the Laplace transform (151) itself.

    27. Euler-Maclaurin summation. The Euler-Maclaurin summation formula [33, p. 14] is

    X

    = +1

    ( ) =

    Z( ) +

    X

    =1

    (−1)!

    h( −1)( )− ( −1)( )

    i+ (68)

    with remainder term

    =(−1) −1

    !

    Z( − [ ]) ( )( )

    29

  • where and ( ) are the Bernoulli numbers and the Bernoulli polynomials defined in [1, Chapter

    23].

    The right-hand side summation in the Euler-Maclaurin summation formula (68) requires higher

    order derivatives for the function ( ) = Γ( + ) . We invoke Leibniz’ rule

    ( )( ) =Γ ( + )

    ¯̄¯̄=

    =X

    =0

    µ ¶(log ) −

    1

    Γ ( + )

    ¯̄¯̄=

    which simplifies, with 1Γ( + ) =1

    Γ( )

    ¯̄¯= +

    , to

    ( )( )

    !=

    X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    (log ) −

    ( − )!

    Since 1Γ( ) is an entire function, the -sum converges when → ∞. Applying the Euler-Maclaurin

    summation formula (68) to the definition (1) of ( ) =P∞

    =0 Γ( + ) shows that →∞, but that we

    can choose the integer = . The asymptotic form (136) indicates that lim →∞ 1Γ( + )

    ¯̄¯== 0

    for all and (68) becomes

    ∞X

    = +1Γ ( + )

    =

    Z ∞

    Γ ( + )+

    −1X

    =0

    (−1) +1+ 1

    X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    (log ) −

    ( − )!+

    (69)

    The integral is written in terms of the integral ( ), defined in (2), as

    Z ∞

    Γ ( + )=

    Z ∞

    0 Γ ( + + )= + ( )

    In summary, the Euler-Maclaurin expansion (69) is, withP∞

    = +1 Γ( + ) =P∞

    =0

    + +1

    Γ( + ( +1)+ ) ,

    +1+( +1) ( ) = + ( ) + + (70)

    where

    =−1X

    =0

    (−1) +1+ 1

    X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    (log ) −

    ( − )!

    28. Euler-Maclaurin sum . We concentrate on and reverse the summations,

    =−1X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    −1X

    =

    (−1) +1+ 1

    (log ) −

    ( − )!

    Let = log and use = !( − )!− , then

    −1X

    =

    (−1) +1+ 1

    ( − )!=

    −1X

    =0

    (−1) +1+ 1

    1

    !

    30

  • and

    lim→∞

    −1X

    =

    (−1) +1+ 1

    ( − )!=

    ∞X

    =0

    +1

    ( + 1)!(− )

    =1∞X

    =1!(− ) =

    1Ã ∞X

    =0!(− ) − 1

    !

    The generating function (129) of the Bernoulli numbers ,

    − 1=

    ∞X

    =0!

    convergent for | | 2

    leads, for | | 2 , to∞X

    =

    (−1) +1+ 1

    ( − )!=

    1µ−− − 1

    − 1¶=

    µ1

    1− −− −1

    For →∞ and |log | 2 , we obtain

    ∞ =∞X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    ∞X

    =

    (−1) +1+ 1

    (log ) −

    ( − )!

    =∞X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    µ1

    1− −− −1

    ¶¯̄¯̄=log

    Using the Fermi-Dirac integrals in art. 14, − −1( ) =³

    11+ −

    ´and −1 = (−1) ! −1− , we

    have

    ∞ =∞X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    − −1(log + )−1

    log

    ∞X

    =0

    1

    Γ ( )

    ¯̄¯̄= +

    µ−log

    We recognize from (39) that, for | | 1,

    ¡−

    ¢= ( ) =

    1

    Γ ( )−

    ∞X

    =0!

    1

    Γ ( )

    ¯̄¯̄=

    − −1(log + )

    and from the series (107) for ( ), we obtain, for¯̄¯ log

    ¯̄¯ 1,

    ∞ = − + ( ) +1

    Γ ( + )+ + ( ) (71)

    which is thus valid for | | 1 and¯̄¯ log

    ¯̄¯ 1.

    The Euler-Maclaurin expansion (70) becomes, for → ∞ and with (71) and assuming that→ 0,

    +1+( +1) ( ) + + ( ) = 2 + ( ) + Γ ( + )

    With (14), we arrive, for | | 1,¯̄¯ log

    ¯̄¯ 1 but all , at

    ( ) = + ( ) +X

    =0Γ ( + )

    (72)

    31

  • Numerical computations for ≤ 20 show that (72) is increasingly accurate for increasing aslong as

    ¯̄¯ log

    ¯̄¯ 1, irrespective of . On the other hand, when

    ¯̄¯ log

    ¯̄¯ 1, increasing (up to 20) show

    a decreasing accuracy. Since lim →∞ + ( ) = 0, (72) reduces to an identity when → ∞.Anticipating (104), the Euler-Maclaurin sum (69) becomes

    ∞X

    = +1Γ ( + )

    =

    1−⎧⎨

    ⎩1

    +

    Z ∞

    0

    −1

    +

    Ãsin( + ) ln + cos ( + )

    2 + (ln )2

    ! ⎫⎬

    +−1X

    =0

    (−1) +1+ 1

    X

    =0!

    1

    Γ ( )

    ¯̄¯̄= +

    (log ) −

    ( − )!+ (73)

    Numerical computations (for = −1) indicate that the -summation converges for small and small, with fast convergence when = 1. For = 1, (73) simplifies to

    ∞X

    = +1

    1

    Γ ( + )=1(

    +

    Z ∞

    0

    +

    Ãsin( + ) ln + cos ( + )

    2 + (ln )2

    ! )

    +∞X

    =0

    +1(− )( + 1)!

    1

    Γ ( )

    ¯̄¯̄= +

    where the latter sum converges for | | 1.

    5 Complex integrals for ( )

    29. Basic complex integral for ( ). Inverse Laplace transformation (152) of (59) yields, with0 | |

    1

    for Re ( ) 0,

    −1 ( ) =1

    2

    Z 0+ ∞

    0− ∞

    where the integrand is analytic for Re ( ) . For real ≥ 0, we can move the line of integration to| | = ( 0) and perform an ordinary substitution = . Let = and we arrive at the

    basic complex integral

    ( ) =1

    2

    Z + ∞

    − ∞

    −| | (74)

    The basic complex integral (74) of ( ) can also be deduced from Hankel’s deformed integral

    (139) in the power series of the Mittag-Leffler function (1)

    ( ) =∞X

    =0Γ ( + )

    =1

    2

    ∞X

    =0

    Z ¡ − ¢ − =1

    2

    Z ∞X

    =0

    ¡ − ¢ −

    Only if | − | 1, implying | | | | for any along the contour described in art. 55, then weobtain

    ( ) =1

    2

    Z −

    1− −(75)

    where the radius of the circle at = 0 in the contour (see art. 55) can be appropriately chosento satisfy | | Re or | |

    1Re for any . We obtain again (74) by choosing = 2 and = .

    32

  • 30. Mittag-Leffler function ( ) for negative real . Although ( ), defined by the power series(1), is not valid for negative Re ( ) values, the Taylor series (39) of ( ) in around 0 = 0,

    (− ) =1

    Γ ( )−

    ∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    − −1(log )

    and its companion (41)

    (− ) =∞X

    =0

    1

    !

    1

    Γ ( )

    ¯̄¯̄=

    − −1(− log ) (− )

    indicate that µ−1¶=

    1

    Γ ( )− − (− )

    which may be used, by analytic continuation, as a definition of ( ) for Re ( ) 0. Thus, for

    Re ( ) 0, the Mittag-Leffler function ( ) possesses an essential singularity at = 0 and is not

    an entire function.

    The same result is also deduced in [16, Sec. 4.8, pp. 80-82] from the basic complex integral (75)

    in art. 29

    ( ) =1

    2

    Z −

    is valid for any complex and . Introducing the identity

    −=

    1

    − −=1−

    1

    − 1 +

    leads to

    ( ) =1

    2

    Z−

    1

    2

    Z − −− − 1

    Invoking Hankel’s integral (140) to the first integral and the basic complex integral (75) to the last

    integral leads again to

    ( ) =1

    Γ ( )− −

    µ1¶

    (76)

    The Taylor series (1) of ( ) then shows that

    − ( ) =1

    Γ ( )−

    µ1¶= −

    ∞X

    =1

    Γ ( + )

    31. Evaluation of the basic complex integral for ( ). We will now evaluate the integral (74) byclosing the contour over the negative Re ( )-plane. Consider the closed loop integral

    1

    2

    Z −

    where the contour consists of the vertical line at = + , the circle segment at infinity turning

    from 2 towards − , the line above the negative real axis, the small circle with radius turningaround the origin from − to − + , the line from the origin just below the negative real axis,over the circle segment with infinite radius from − + to − 2 and ending at the negative side of the

    33

  • vertical line. The integrand has only singularities of ( − )−1, which are simple poles because thezeros of − = 0 are all simple and lie at = | |

    1 arg 2

    for each ∈ Z. If is irrational,there are infinitely many zeros. The contour only encloses the poles with argument = arg + 2

    between − ≤ ≤ . Thus, the integer ranges from

    −j

    2+arg

    2

    k≤ ≤

    j

    2−arg

    2

    k

    and there are precisely b c enclosed poles, where b c is the integer smaller than or equal to . Hence,must be at least equal to 1, else no singularities are enclosed. By Cauchy’s residue theorem, we

    obtain

    1

    2

    Z −

    −=

    X

    ∈[−b 2+ arg2 c b 2−arg2 c]lim→

    −−

    =1 X

    ∈[−b 2+arg2 c b 2− arg2 c]

    1− with = | |1 arg 2

    We can always choose 0 small enough, provided that Re ( − ) −1. Evaluation of thecontour yields

    1

    2

    Z −

    −=

    1

    2

    Z + ∞

    − ∞

    −+

    1

    2

    (Z 0

    ¡ ¢ − −

    ( ) −+

    Z ∞

    0

    ¡ − ¢ − −

    ( − ) −+

    )

    = ( ) +1

    2

    (Z ∞

    0

    Ã− ( − )

    − −−

    ( − )

    !− −

    )

    Hence, we have

    ( ) = −1

    2

    (Z ∞

    0

    Ã−( − )

    − −−

    ( − )

    !− −

    )

    +1 X

    ∈[−b 2+ arg2 c b 2− arg2 c]

    1−

    Finally for Re ( ) Re ( )− 1, we arrive at

    ( ) =sin ( − )

    Z ∞

    0

    − −

    2 − 2 cos + 2+sin

    Z ∞

    0

    2 − −

    2 − 2 cos + 2

    +1 X

    ∈[−b 2+ arg2 c b 2− arg2 c]

    1− with = | |1 arg 2

    (77)

    The last residue sum illustrates again that ( ) is an entire function of order 1 . The sign of is

    determined by cos³arg +2

    ´. Consequently for large | |, in the sectors − 2 −2 arg 2 −2 ,

    the function ( ) tends to infinity, while in the complementary sectors 2 −2 arg32 −2 ,

    ( )→ 0.For real = − = in (77) and =

    (2 +1)

    , we find

    (− ) = −sin ( − )

    Z ∞

    0

    − −

    2 + 2 cos + 2+sin

    Z ∞

    0

    2 − −

    2 + 2 cos + 2

    +1− (1− ) b cX

    =0

    cos (2 −1) − sin(2 −1) − 2 (1− )

    34

  • For real , = 1 and 0 1, the condition to enclose a pole is − ≤ arg + 2 ≤ . Sincearg = ∈ [− ] for 0 1, the above sum disappears and we obtain

    (− ) =sin

    Z ∞

    0

    −1 −

    2 + 2 cos + 2for 0 1 (78)

    from which the asymptotic behavior for large , with 2 + 2 cos + 2 ∼ 2 , are

    (− ) ∼sin Γ ( )

    For small, real , the series (1) gives

    (− ) =∞X

    =0

    (−1)Γ (1 + )

    = 1−Γ (1 + )

    +¡2¢= exp

    µ−Γ (1 + )

    ¶+

    ¡2¢

    Hence, (− ) for 0 1 is said to “interpolate” between an exponential (for small ) and apower law (for large ). These two regimes have been studied by Mainardi [24], who illustrated their

    accuracy with several plots for = 0 25 0 5 0 75 0 9 and 0 99. Art. 36 presents another derivationin (86) that is equivalent to (78).

    32. Mittag-Leffler’s contour integral. In a rather long article of 1905, Mittag-Leffler [28, at p. 133-135] proceeds one step further and substitutes =

    1

    in the basic complex integral (74). The map

    →1

    is multi-valued. For = | | with − ≤ ≤ , the inverse map → shows that= | | = | | +2 , from which the argument = +2 for any integer . The contourin the -plane requires that 2 | | , because → 0 for large at the contour . Similarly,

    the transformed contour in the -plane requires that 2 |arg | in order that1

    → 0along the straight lines towards infinity. Thus, the map →

    1

    changes the angles from to of

    the straight lines in the contour (Fig. 43 in Bieberbach’s book [7, p. 273]). Moreover, we must choose

    the branch (i.e. the appropriate integer in = + 2 ) of the function1

    that is positive for

    positive , i.e. along the positive real -axis, because the same holds for along the positive real -axis.

    Performing the substitution =1

    in (74) leads, for | | | | and |arg | 2 , to Mittag-Leffler’scontour integral

    ( ) =1

    2

    Z 1− 1

    −with

    2| | (79)

    Mittag-Leffler’s integral (79) for = 1 is actually more elegant than the basic complex integral (74)

    at the expense of a more complicated contour .

    33. Deductions from Mittag-Leffler’s contour integral (79) for ( ). Based on (79), we followBieberbach [7, p. 273], who deduces two bounds for 0 2 and = 1. First, using

    1

    −= −

    1+

    ( − )

    in (79), we have

    ( ) = −1 1

    2

    Z1− 1

    +1

    2

    Z 1− + 1

    ( − )

    = −1 1

    2

    Z− +

    1

    2

    Z − −1 1

    ( − )

    35

  • With Hankel’s contour integral (139),

    ( ) = −1 1

    Γ ( − )+

    1

    2

    Z − +1 1

    ( − )

    The remaining integral is upper bounded by

    ¯̄¯̄¯̄1

    2

    Z − +1 1

    ( − )

    ¯̄¯̄¯̄

    1

    2 | |

    Z | |− +1

    ¯̄¯

    1 ¯̄¯

    | |¯̄1−

    ¯̄ | | =| |2

    because 12R | |

    − −1 1

    |1− | | | converges for → ∞ if¯̄¯

    1 ¯̄¯ = | |

    1cos arg → 0 for large , which

    requires that arg ≥ 2 . In addition, we must prevent that for large can tend arbitrarily close to 1,which is guaranteed if |arg | ∈

    ¡2

    ¢, because |arg | ∈

    ¡2

    ¢. Hence, for |arg | ∈

    ¡2

    ¢, we

    arrive at ¯̄¯̄ ( ) +

    1

    Γ ( − )

    ¯̄¯̄

    | |2

    For the second bound, Bieberbach [7, p. 275] cleverly observes that a similar integration path 0

    as in Mittag-Leffler’s integral (79) can be followed with the only difference that the circular part of

    the path now has a radius smaller than | |. In other words, while | | | | in Mittag-Leffler’s integral(79), the path 0 now turns over an angle − to with the radius smaller than | |. The closedcontour (see also [35, p. 346, Fig. 6.13-2]), that first follows the Mittag-Leffler path and returns

    via the path 0 , encloses the point = as the only singularity, provided − arg . Hence,by Cauchy’s residue theorem, it holds that

    1

    2

    Z 1− 1

    −−

    1

    2

    Z

    0

    1− 1

    −=1 1− 1

    and

    ( ) =1

    2

    Z

    0

    1− 1

    −+1 1− 1

    from which¯̄¯̄ ( )−

    1 1− 1¯̄¯̄ ≤

    1

    2 | |

    Z

    0

    ¯̄¯1−¯̄¯¯̄¯

    1 ¯̄¯

    ¯̄− 1¯̄ =

    0

    | |

    by the same convergence argument as above, where now on the circular segment | | | | can be chosensmall enough. In summary, the second bound for |arg | ∈

    ¡2

    ¢is

    ¯̄¯̄ ( )−

    1 1− 1¯̄¯̄

    0

    | |

    The derivation illustrates why Bieberbach considers 0 2, because |arg | ∈ (0 2 ).The second bound shows that ( ) ≈ 1− is independent of so that ( ) ≈

    11

    ³1´. Hence, we are led for non-negative real and | | to

    1 ( ) ≈ 2³

    36

  • whose exact corresponding relation (100) for the associated integral ( ) =R∞0 Γ( + ) , explored

    in Section 7, is 1 ( ) = 2³

    2´.

    34. Deductions from Cauchy’s residue theorem. If ( ) is an entire function and lim →∞

    ¯̄¯̄ ( )sin

    ¯̄¯̄→ 0

    in a semicircle with either − 2 2 or 232 , then it follows directly from Cauchy’s residue

    theorem [37] that

    1

    2

    Z + ∞

    − ∞ sin( ) = −1{− 2 2 }

    ∞X

    =0

    (−1) ( ) for 0 1

    The definition (1) of ( ) =P∞

    =0 Γ( + ) contains ( ) = Γ( + ) , which is an entire function in

    . From the asymptotic behavior (136) of 1|Γ( + )| in art. 54, it follows that the above contour

    can be closed over the positive Re ( )-plane, resulting in10

    ( ) = −1

    2

    Z +∞

    +∞ − sin

    (− )Γ ( + )

    for − 1 0 (81)

    where the path above and below the real axis follow the lines + ± , where 0 2 . Thus, the

    line of integration cannot be parallel with the imaginary axis, unless 2. If = = 1, the reflection

    formula (122) leads to Mellin’s integral 12R +∞+∞ Γ ( ) (− ) = .

    Let us consider the contour C, consisting of the line + with 0 ≤ ≤ , the line parallel tothe real axis from + to the left at − + , the line back to the real axis at − (and thereflection of this parallelogram around the real axis). The path parallel to the real axis

    Z − +

    + − sin

    (− )Γ ( + )

    =

    Z −

    sin ( + − )

    (− ) +−

    Γ ( + + − )

    vanishes by (136) for →∞ provided 0 2 . The contour C encloses the poles sin at = −from = 1 to with residue (−1) . Hence, by shifting the lines + ± to − + ± , maintainingthe angle 0 2 , we deform the integral (81) into

    ( ) =X

    =1

    1

    Γ ( − )1−

    1

    2

    Z 0+∞

    0+∞ − sin

    (− )Γ ( + )

    for − 1− 0

    For complex , it is generally for any complicated to bound the integral for large | | to deduce anasymptotic expansion.

    35. We evaluate (81) along the line = + − below the real -axis to the line = + abovethe real -axis for 0 2 ,

    (− ) =−

    2

    Z ∞

    0

    + −

    Γ ( + + − ) sin ( + − )−2

    Z ∞

    0

    +

    Γ ( + + ) sin ( + )

    =1

    2

    Z ∞

    0

    + − − ( 2+ )

    Γ ( + + − ) sin ( + − )+1

    2

    Z ∞

    0

    + ( 2+ )

    Γ ( + + ) sin ( + )

    10The integral (81) is rewritten with the reflection formula of the Gamma function as a Barnes-Mellin type integral

    (− ) = −1

    2

    +∞

    +∞ −

    Γ (1− )Γ ( )Γ ( + )

    for − 1 0 and 02

    (80)

    37

  • Since the integrand sin(− )

    Γ( + ) is real on the real axis, the reflection principle [37, p. 155] leads, for

    real = , to

    (− ) =Z ∞

    0Re

    (( 2+ )

    Γ ( + + ) sin ( + )

    )

    which slightly simplifies, after choosing = −12 , to

    (− ) = −1√

    Z ∞

    0Re

    (( 2+ ) ( log )

    Γ¡− 2 +

    ¢cos

    )

    Choosing = 2 , and thus restricting 0 2, yields

    (− ) =1√

    Z ∞

    0Re

    (log

    Γ¡− 2 +

    ¢

    )

    cosh

    Rather than continuing with this integral, we present another approach.

    36. If we choose = 2 in (81), then we must restrict 0 2. In that case, we evaluate the contouralong the line = + ,

    (− ) = −2

    Z ∞

    −∞

    1

    sin ( + ) Γ ( + + )for − 1 0

    We limit ourselves to real = . Since 0, the reflection formula (122) yields

    1

    Γ ( − | |+ )= −Γ (1− + | |− )

    sin ( | |− − )

    We change the sign of and obtain

    (− ) =1

    2

    Z ∞

    −∞

    sin ( ( − )− )sin ( − )

    Γ ( + 1− − )− for 0 1

    In order to invoke Euler’s integral (109), we must require that Re ( + 1− ) 0

    (− ) =1−

    2

    Z ∞

    −∞

    sin ( ( − )− )sin ( − )

    Z ∞

    0

    −1

    − −

    We can interchange the integrals by absolute convergence requiring that 0 1,

    (− ) =1−

    2

    Z ∞

    0

    −1 Z ∞

    −∞

    sin ( ( − )− )sin ( − )

    ( − ) (82)

    In passing, we check for → 1 and = 1 that, by using the Dirac impulse function ( ),

    1 (− ) =1

    2

    Z ∞

    0

    − −1Z ∞

    −∞

    − log =

    Z ∞

    0

    − −1 (log ) = −

    The latter integral in (82) can be recasted as

    ( ) = −1

    2

    Z ∞

    −∞

    sin ( ( − )− )sin ( − )

    ( − ) =1

    2

    Z + ∞

    − ∞

    sin ( − )sin

    ( ) for 0 1

    (83)

    38

  • In art. 37 below, we prove for real and 0 1 that

    ( ) =1

    2

    Z + ∞

    − ∞

    sin ( − )sin

    ( ) =1µ

    sin ( − )− 2 sin1 + 2 cos + 2

    ¶(84)

    Substitution of (84) into (82) yields, for 0 ≤ 1 and + 1 Re ( ),

    (− ) = −1−sin ( − )

    Z ∞

    0

    1 + 2 cos + 2−

    1

    +

    1−sin

    Z ∞

    0

    2 −

    1 + 2 cos + 2−

    1

    (85)

    which equals (77) without residue sum (after the substitution =1

    ). If = 1, then (85) simplifies

    to

    (− ) =sin

    Z ∞

    0

    −1

    1 + 2 cos + 2−

    1

    (86)

    which is listed in [4, eq. (34)] and deducible from (78).

    37. The integral ( ). The contour of the integral ( ) in (83) for real can be closed overeither half-plane. If | | 1, then we close the contour over Re ( ) 0-plane and obtain

    1

    2

    Z + ∞

    − ∞

    sin ( − )sin

    ( ) = −1∞X

    =1

    (−1) sin ( − )

    =1

    2

    Ã( − )

    ∞X

    =0

    ¡−

    ¢− − ( − )

    ∞X

    =0

    ¡− −

    ¢!

    =2

    Ã( − )

    1 +−

    − ( − )

    1 + −

    !

    =1 sin ( − )− 2 sin

    1 + 2 cos + 2

    If | | 1, then we close the contour of the Re ( ) 0 plane,

    1

    2

    Z + ∞

    − ∞

    sin ( − )sin

    ( ) =1∞X

    =0

    (−1) sin (− − ) −

    =−12

    µ

    1 + −−

    1 + − −

    =1µ

    sin ( − )− 2 sin1 + 2 cos + 2

    Both the case | | 1 and | | 1 lead to the same result! In summary, for any real and for 0 1,we have proved (84).

    The contour in (83) can be rewritten as

    ( ) =2½sin

    ³

    2−´Z ∞

    0

    cosh

    coshcos ( log ) − cos

    ³

    2−´Z ∞

    0

    sinh

    coshsin ( log )

    ¾

    Since 1 ( ) = 1¡1¢, as follows from (84), it holds that

    39

  • Z ∞

    0

    sinh

    coshsin ( log ) =

    − 1+ 1

    tan³

    2

    ´Z ∞

    0

    cosh

    coshcos ( log )

    so that

    ( ) =2

    sin³

    2−´½

    1−− 1+ 1

    cot³

    2−´tan

    ³

    2

    ´¾Z ∞

    0

    cosh

    coshcos ( log )

    It follows then from (84) that

    Z ∞

    0

    cosh

    coshcos ( log ) =

    2 (sin ( − )− sin )

    (1 + 2 cos + 2 )nsin

    ¡2 −

    ¢+ 1−1+ cos

    ¡2 −

    ¢tan

    ¡2

    ¢o

    whose right-hand is independent of , so that can be chosen at will. The simplest choice11 is = 2 ,

    then Z ∞

    0

    cosh

    coshcos ( log ) =

    2 (1 + ) cos 21 + 2 cos + 2

    Suppose that we ignore the restriction that

    1

    2

    Z + ∞

    − ∞

    sin ( − )sin

    ( ) =1∞X

    =1

    (−1) −1 sin ( − )

    only converges for 1 and that we substitute the series formally back in (82) and change the order

    of integration and summation. Then, we find

    (− ) =1∞X

    =1

    sin ( − )Γ ( − + 1)(− )

    = −∞X

    =1

    1

    (− ) Γ ( − )=

    1

    Γ ( )− −

    µ−1¶

    which is precisely equal to (76), in spite of the divergence of the series!

    38. Deductions from the integral (86). Berberan-Santos [4, eq. (35)] mentions12 that partial integra-tion of (86) results in

    (− ) = 1−1

    2+

    1 Z ∞

    0arctan

    µ+ cos ( )

    sin ( )

    ¶−

    1

    (87)

    Indeed, let = in (86), then

    (− ) =sin

    Z ∞

    0

    −1 1

    1 + 2 cos + 2

    With 11+2 cos + 2 =

    12 sin

    ³1

    − +− 1

    +

    ´, we have

    (− ) =1

    2

    Z ∞

    0

    µ1

    − +−

    1

    +

    ¶−

    1 1

    (88)

    11 It is readily verified that

    sin ( − )− sinsin

    2− + 1−

    1+cos

    2− tan

    2

    = cos2

    (1 + )

    12The integral in Berberan-Santos [4, eq. (35)], (− ) = 1 − 12+

    1∞0arctan +cos( )

    sin( )−

    1

    , misses a

    factor of 1 before the integral.

    40

  • as well asZ

    1 + 2 cos + 2=

    1

    2 sin

    µlog

    − +

    +

    ¶=

    1

    2 sin

    µlog

    + cos − sin+ cos + sin

    =1

    2 sin

    Ã

    log1− sin+cos1 + sin+cos

    !

    = −1

    sinarctan

    µsin

    + cos

    because arctan = 2 log1−1+ . Hence, we find the indefinite integral with a constant ,

    Z −1

    1 + 2 cos + 2= −

    1

    sinarctan

    µsin

    + cos

    ¶+ (89)

    With this preparation, partial integration of (86) yields

    (− ) =1arctan

    µsin

    cos

    ¶−

    1 Z ∞

    0arctan

    µsin

    + cos

    ¶−

    1

    = 1−1 Z ∞

    0arctan

    µsin

    + cos

    ¶−

    1

    After invoking arctan = 2 − arctan1 for 0, we arrive at (87).

    Further, using the integral for arctan =R0 1+ 2 ,

    1− (− ) =1 Z ∞

    0

    Z sin+cos

    0

    −1

    1 + 2

    and reverse the integrals, provided that 0 ≤ 12 ,

    1− (− ) =1 Z tan

    0

    R ( sin −cos )1

    0−

    1

    1 + 2

    Hence, we obtain, for 0 ≤ 12 ,

    (− ) =1Z tan

    0

    −( ( sin −cos ))1

    1 + 2(90)

    After letting = sin − cos in (90), we retrieve (86) again.

    39. Bounds from the integral (86). We split the integration interval in (86) into two parts,

    (− ) =sin

    Z ∞

    1

    −1³−

    1 −1+ −

    1 ´

    1 + 2 cos + 2

    First,

    Z ∞

    1

    −1 −1 −1

    1 + 2 cos + 2= −

    1Z ∞

    1

    −11(1− −1)

    1 + 2 cos + 2

    ≥ −1Z ∞

    1

    −1

    1 + 2 cos + 2

    41

  • Using (89) and sin1+cos = tan 2 , we arrive at the lower bound,

    sinZ ∞

    1

    −1 −1 −1

    1 + 2 cos + 2≥1 −

    1

    arctan³tan

    2

    ´=1

    2−

    1

    Second, after partial integration and again using (89), we obtain

    sinZ ∞

    1

    −1 −1

    1 + 2 cos + 2=1

    2−

    1

    −1 Z ∞

    1

    −1

    arctan

    µsin

    + cos

    =1

    2−

    1

    −1 −

    1 Z ∞

    1

    −1( −1) arctan

    µsin

    + cos

    so that

    (− ) ≥ −1

    −1 −

    1 Z ∞

    1

    −1( −1) arctan

    µsin

    + cos

    Further, we may bound the latter integral,Z ∞

    1

    −1( −1) arctan

    µsin

    + cos

    ¶ Z ∞

    1

    −1( −1) arctan

    µsin

    1