A Lecture on Row Integration by Parts (RIP)jarock/RIP_LMU_170314.pdf · A Lecture on Row...

∫u dv = uv −

∫v du

± diff. int. (±∫

)

+ u dv (∫u dv)

↘− du v (−

∫v du)

A Lecture on Row Integration by Parts (RIP)

John A. Rock

[email protected]

Loyola Marymountπ.2017

John A. Rock (Cal Poly Pomona) Row Integration by Parts (RIP) Loyola Marymountπ.2017 1 / 21

Integration by Parts

We can iterate “ultra-violet voodoo”:∫u dv = uv −

∫v du

= u1v1−∫v1 du1

= u1v1 − u2v2 +∫v2 du2

= u1v1 − u2v2 + u3v3−∫v3 du3

...

. . . but when do we stop?

It completely depends on the last integral (−1)n∫vn dun.


Tabular Row Integration by Parts (RIP)

Given∫u dv = uv −

∫v du, define

{u1 = u, v1 =

∫v0 =

∫dv ,

un+1 = u′n, vn+1 =∫vn.


)

+ u1 v0 (The top row is the original∫u dv .)

↘− u2 v1 (−

∫u2v1 = −

∫v du = −

∫v1 du1.)

↘+ u3 v2 (+

∫v2 du2? Not yet. . .)

↘− u4 v3 (−

∫v3 du3? Stop!)

∫u dv = u1v1 − u2v2 + u3v3−

∫v3 du3.




∫v du, define

{u1 = u, v1 =

∫v0 =

∫dv ,

un+1 = u′n, vn+1 =∫vn.


)


↘− u2 v1 (−

∫u2v1 = −

∫v du = −

∫v1 du1.)

↘+ u3 v2

(+∫v2 du2? Not yet. . .)

↘− u4 v3 (−

∫v3 du3? Stop!)

∫u dv = u1v1 − u2v2 + u3v3−

∫v3 du3.




∫v du, define

{u1 = u, v1 =

∫v0 =

∫dv ,

un+1 = u′n, vn+1 =∫vn.


)


↘− u2 v1 (−

∫u2v1 = −

∫v du = −

∫v1 du1.)

↘+ u3 v2 (+

∫v2 du2?

Not yet. . .)↘

− u4 v3 (−∫v3 du3? Stop!)

∫u dv = u1v1 − u2v2 + u3v3−

∫v3 du3.




∫v du, define

{u1 = u, v1 =

∫v0 =

∫dv ,

un+1 = u′n, vn+1 =∫vn.


)


↘− u2 v1 (−

∫u2v1 = −

∫v du = −

∫v1 du1.)

↘+ u3 v2 (+

∫v2 du2? Not yet. . .)

↘− u4 v3 (−

∫v3 du3? Stop!)

∫u dv = u1v1 − u2v2 + u3v3−

∫v3 du3.


Stand and Deliver “Tic-Tac-Toe”

Example 1.∫x2 sin x dx


)

+ x2 sin x↘

− 2x − cos x (∫

2x cos x? Keep going. . .)↘

+ 2 − sin x (−∫

2 sin x = 2 cos x + C . Stop!)

∫x2 sin x dx = −x2 cos x + 2x sin x +2 cos x + C .





)

+ x2 sin x

↘− 2x − cos x (

∫2x cos x? Keep going. . .)

↘+ 2 − sin x (−

∫2 sin x = 2 cos x + C . Stop!)






)

+ x2 sin x↘

− 2x − cos x

(∫


+ 2 − sin x (−∫







)

+ x2 sin x↘

− 2x − cos x (∫

2x cos x?

Keep going. . .)↘

+ 2 − sin x (−∫







)

+ x2 sin x↘

− 2x − cos x (∫

2x cos x? Keep going. . .)

↘+ 2 − sin x (−

∫2 sin x = 2 cos x + C . Stop!)






)

+ x2 sin x↘

− 2x − cos x (∫


+ 2 − sin x

(−∫







)

+ x2 sin x↘

− 2x − cos x (∫


+ 2 − sin x (−∫

2 sin x

= 2 cos x + C . Stop!)






)

+ x2 sin x↘

− 2x − cos x (∫


+ 2 − sin x (−∫





Example 1 (as in the film).∫x2 sin x dx


)

+ x2 sin x↘

− 2x − cos x↘

+ 2 − sin x↘

− 0 cos x (−∫

0 = C .)

∫x2 sin x dx = −x2 cos x + 2x sin x + 2 cos x + C .





)

+ x2 sin x↘

− 2x − cos x↘

+ 2 − sin x↘

− 0 cos x

(−∫

0 = C .)






)

+ x2 sin x↘

− 2x − cos x↘

+ 2 − sin x↘

− 0 cos x (−∫

0 = C .)



LIPET, LIATE, etc.

Example 2.∫x ln x dx


)

+ ln x x↘

− 1/x x2/2 (−(1/2)∫x = −x2/4 + C , stop.)

∫x ln x dx = (1/2)x2 ln x −x2/4 + C .


LIPET, LIATE, etc.



)

+ ln x x

↘− 1/x x2/2 (−(1/2)

∫x = −x2/4 + C , stop.)

∫x ln x dx = (1/2)x2 ln x −x2/4 + C .


LIPET, LIATE, etc.



)

+ ln x x↘

− 1/x x2/2

(−(1/2)∫x = −x2/4 + C , stop.)

∫x ln x dx = (1/2)x2 ln x −x2/4 + C .


LIPET, LIATE, etc.



)

+ ln x x↘

− 1/x x2/2 (−(1/2)∫x

= −x2/4 + C , stop.)

∫x ln x dx = (1/2)x2 ln x −x2/4 + C .


LIPET, LIATE, etc.



)

+ ln x x↘

− 1/x x2/2 (−(1/2)∫x = −x2/4 + C , stop.)

∫x ln x dx = (1/2)x2 ln x −x2/4 + C .


An example without polynomials

Example 3.∫e2x cos 3x dx


)

+ e2x cos 3x↘

− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x? Try again.)

↘+ 4e2x (− cos 3x)/9 (−(4/9)

∫e2x cos 3x , stop.)

The last integral is a copy of the original!∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)

∫e2x cos 3x+C .∫

e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.





)

+ e2x cos 3x

↘− 2e2x (sin 3x)/3 (−(2/3)

∫e2x sin 3x? Try again.)

↘+ 4e2x (− cos 3x)/9 (−(4/9)









)

+ e2x cos 3x↘

− 2e2x (sin 3x)/3

(−(2/3)∫e2x sin 3x? Try again.)

↘+ 4e2x (− cos 3x)/9 (−(4/9)









)

+ e2x cos 3x↘

− 2e2x (sin 3x)/3 (−(2/3)∫e2x sin 3x?

Try again.)↘

+ 4e2x (− cos 3x)/9 (−(4/9)∫e2x cos 3x , stop.)








)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9 (−(4/9)









)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9

(−(4/9)∫e2x cos 3x , stop.)








)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9 (−(4/9)









)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9 (−(4/9)


The last integral is a copy of the original!

∫e2x cos 3x dx = (1/3)e2x sin 3x + (2/9)e2x cos 3x−(4/9)







)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9 (−(4/9)



∫e2x cos 3x+C .

∫e2x cos 3x dx = (3/13)e2x sin 3x + (2/13)e2x cos 3x + C0.





)

+ e2x cos 3x↘


↘+ 4e2x (− cos 3x)/9 (−(4/9)






One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫

2x2 ln x dx , same as above.)↘

+ 2/x x3/3 (+∫

2x2/3 = 2x3/9 + C )

∫3x2 ln2 x dx = x3 ln2 x − (2/3)x3 ln x +2x3/9 + C


One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3

(−∫

2x2 ln x dx)

− 2 ln x x2 ← (−∫


+ 2/x x3/3 (+∫

2x2/3 = 2x3/9 + C )



One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫


+ 2/x x3/3 (+∫

2x2/3 = 2x3/9 + C )



One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫

2x2 ln x dx , same as above.)

↘+ 2/x x3/3 (+

∫2x2/3 = 2x3/9 + C )



One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫


+ 2/x x3/3

(+∫

2x2/3 = 2x3/9 + C )



One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫


+ 2/x x3/3 (+∫

2x2/3

= 2x3/9 + C )



One more example

Example 4.∫

3x2 ln2 x dx


)

+ ln2 x 3x2

↘− (2 ln x)/x x3 (−

∫2x2 ln x dx)

− 2 ln x x2 ← (−∫


+ 2/x x3/3 (+∫

2x2/3 = 2x3/9 + C )



Your turn!

Exercises. Find the following antiderivatives.

1∫

(x3 − 2x + 7)ex dx

2∫

sin 2t sin 5t dt

3∫

cos x ln (sin x) dx

4∫

ln (x2 + 1) dx

5∫

(arcsin x)2 dx

(For fun, let’s see what WolframAlpha or Symbolab gives us.)


Solutions

Exercise 1.∫

(x3 − 2x + 7)ex dx


)

+ x3 − 2x + 7 ex

↘− 3x2 − 2 ex

↘+ 6x ex

↘− 6 ex

↘+ 0 ex (

∫0 = C )

∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C


Solutions

Exercise 1.∫

(x3 − 2x + 7)ex dx


)

+ x3 − 2x + 7 ex

↘− 3x2 − 2 ex

↘+ 6x ex

↘− 6 ex

↘+ 0 ex

(∫

0 = C )

∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C


Solutions

Exercise 1.∫

(x3 − 2x + 7)ex dx


)

+ x3 − 2x + 7 ex

↘− 3x2 − 2 ex

↘+ 6x ex

↘− 6 ex

↘+ 0 ex (

∫0 = C )

∫(x3 − 2x + 7)ex dx = (x3 − 3x2 + 4x + 3)ex + C


Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5↘

+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫

sin 2t sin 5t dt)

The last integral is a copy of the original!∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t

+(4/25)∫

sin 2t sin 5t dt + C

∫sin 2t sin 5t dt = (−5/21) sin 2t cos 5t + (2/21) cos 2t sin 5t + C0.


Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t

↘− 2 cos 2t (− cos 5t)/5

↘+ −4 sin 2t (− sin 5t)/25 (+(4/25)

∫sin 2t sin 5t dt)


+(4/25)∫




Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5

↘+ −4 sin 2t (− sin 5t)/25 (+(4/25)

∫sin 2t sin 5t dt)


+(4/25)∫




Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5↘

+ −4 sin 2t (− sin 5t)/25

(+(4/25)∫

sin 2t sin 5t dt)


+(4/25)∫




Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5↘

+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫

sin 2t sin 5t dt)


+(4/25)∫




Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5↘

+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫

sin 2t sin 5t dt)

The last integral is a copy of the original!

∫sin 2t sin 5t dt = (−1/5) sin 2t cos 5t + (2/25) cos 2t sin 5t

+(4/25)∫




Solutions

Exercise 2.∫

sin 2t sin 5t dt


)

+ sin 2t sin 5t↘

− 2 cos 2t (− cos 5t)/5↘

+ −4 sin 2t (− sin 5t)/25 (+(4/25)∫

sin 2t sin 5t dt)


+(4/25)∫




Solutions

Exercise 3.∫

cos x ln (sin x) dx


)

+ ln (sin x) cos x↘

− cos x

sin xsin x (−

∫cos x = − sin x + C )

∫cos x ln sin x dx = sin x ln (sin x)− sin x + C


Solutions

Exercise 3.∫

cos x ln (sin x) dx


)

+ ln (sin x) cos x

↘− cos x

sin xsin x (−




Solutions

Exercise 3.∫

cos x ln (sin x) dx


)


− cos x

sin xsin x

(−∫

cos x = − sin x + C )



Solutions

Exercise 3.∫

cos x ln (sin x) dx


)


− cos x

sin xsin x (−

∫cos x =

− sin x + C )



Solutions

Exercise 3.∫

cos x ln (sin x) dx


)


− cos x

sin xsin x (−




Solutions

Exercise 4.∫

ln (x2 + 1) dx


)

+ ln (x2 + 1) 1↘

− 2x/(x2 + 1) x (−∫

2x2/(x2 + 1))

∫ln (x2 + 1) dx = x ln (x2 + 1)−

∫2x2/(x2 + 1)

= x ln (x2 + 1)−∫

2(x2 + 1− 1)/(x2 + 1)

= x ln (x2 + 1)−∫

2 +∫

2/(x2 + 1)

= x ln (x2 + 1)−2x + 2 arctan x + C .


Solutions

Exercise 4.∫

ln (x2 + 1) dx


)

+ ln (x2 + 1) 1

↘− 2x/(x2 + 1) x (−

∫2x2/(x2 + 1))

∫ln (x2 + 1) dx = x ln (x2 + 1)−

∫2x2/(x2 + 1)

= x ln (x2 + 1)−∫

2(x2 + 1− 1)/(x2 + 1)

= x ln (x2 + 1)−∫

2 +∫

2/(x2 + 1)

= x ln (x2 + 1)−2x + 2 arctan x + C .


Solutions

Exercise 4.∫

ln (x2 + 1) dx


)

+ ln (x2 + 1) 1↘

− 2x/(x2 + 1) x

(−∫

2x2/(x2 + 1))

∫ln (x2 + 1) dx = x ln (x2 + 1)−

∫2x2/(x2 + 1)

= x ln (x2 + 1)−∫

2(x2 + 1− 1)/(x2 + 1)

= x ln (x2 + 1)−∫

2 +∫

2/(x2 + 1)

= x ln (x2 + 1)−2x + 2 arctan x + C .


Solutions

Exercise 4.∫

ln (x2 + 1) dx


)

+ ln (x2 + 1) 1↘

− 2x/(x2 + 1) x (−∫

2x2/(x2 + 1))

∫ln (x2 + 1) dx = x ln (x2 + 1)−

∫2x2/(x2 + 1)

= x ln (x2 + 1)−∫

2(x2 + 1− 1)/(x2 + 1)

= x ln (x2 + 1)−∫

2 +∫

2/(x2 + 1)

= x ln (x2 + 1)−2x + 2 arctan x + C .


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘

− 2 arcsin x√1−x2 x (−

∫2x arcsin x√

1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2 = −2x + C )

∫(arcsin x)2 dx = x(arcsin x)2 + 2(arcsin x)

√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1

↘− 2 arcsin x√

1−x2 x (−∫

2x arcsin x√1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2 = −2x + C )


√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘

− 2 arcsin x√1−x2 x

(−∫

2x arcsin x√1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2 = −2x + C )


√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘


∫2x arcsin x√

1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2 = −2x + C )


√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘


∫2x arcsin x√

1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2

(−∫

2 = −2x + C )


√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘


∫2x arcsin x√

1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2

= −2x + C )


√1− x2−2x + C


Solutions

Exercise 5.∫

(arcsin x)2 dx


)

+ (arcsin x)2 1↘


∫2x arcsin x√

1−x2 )

− arcsin x 2x√1−x2 ← (−

∫2x arcsin x√

1−x2 )

↘+ 1√

1−x2 −2√

1− x2 (−∫

2 = −2x + C )


√1− x2−2x + C


Taylor’s Formula with Integral Remainder

Theorem 1 (Horowitz). Under suitable conditions,

f (b) = f (a) +f ′(a)

1!(b − a) +

f ′′(a)

2!(b − a)2 + · · ·+ f (n)(a)

n!(b − a)n

+

∫ b

a

f (n+1)(t)

n!(b − t)n dt.

To derive this formula using RIP, there are two tricks:

1. Start with the Fundamental Theorem of Calculus:

f (b)− f (a) =∫ ba f ′(t) dt =

∫ ba (−1)(−f ′(t)) dt.

2. Use (b − t) as the antiderivative of (−1) (treating b as a constant).




)

+ −f (1)(t) −1↘

− −f (2)(t) (b − t)↘

+ −f (3)(t) −(b − t)2

2!...

......

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)




)

+ −f (1)(t) −1

↘− −f (2)(t) (b − t)

↘

+ −f (3)(t) −(b − t)2

2!...

......

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)




)

+ −f (1)(t) −1↘

− −f (2)(t) (b − t)

↘

+ −f (3)(t) −(b − t)2

2!...

......

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)




)

+ −f (1)(t) −1↘

− −f (2)(t) (b − t)↘

+ −f (3)(t) −(b − t)2

2!

......

...

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)




)

+ −f (1)(t) −1↘

− −f (2)(t) (b − t)↘

+ −f (3)(t) −(b − t)2

2!...

......

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!

↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)




)

+ −f (1)(t) −1↘

− −f (2)(t) (b − t)↘

+ −f (3)(t) −(b − t)2

2!...

......

(−1)n−1 −f (n)(t) (−1)n(b − t)n−1

(n − 1)!↘

(−1)n −f (n+1)(t) → (−1)n+1 (b − t)n

n!

(+∫ f (n+1)(t)

n! (b − t)n dt)



Therefore,

f (b)−f (a) =

∫ b

a−f (1)(t)(−1) dt

=

[−f (1)(t)(b − t)− f (2)(t)

2!(b − t)2 − · · · − f (n)(t)

n!(b − t)n

]ba

+

∫ b

a

f (n+1)(t)

n!(b − t)n dt

=f ′(a)

1!(b − a) +

f ′′(a)

2!(b − a)2 + · · ·+ f (n)(a)

n!(b − a)n

+

∫ b

a

f (n+1)(t)

n!(b − t)n dt.


Laplace Transform of f (n)

Theorem 2 (Horowitz). Under suitable conditions,

L{f (n)(t)} =

∫ ∞0

e−st f (n)(t) dt

=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0) + snL{f (t)}.

This formula follows immediately from RIP!




)

+ e−st f (n)(t)↘

− −se−st f (n−1)(t)↘

+ s2e−st f (n−2)(t)...

......

(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘

(−1)n (−1)nsne−st → f (t)(+sn

∫e−st f (t) dt

)




)

+ e−st f (n)(t)

↘− −se−st f (n−1)(t)

↘+ s2e−st f (n−2)(t)...

......

(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘

(−1)n (−1)nsne−st → f (t)(+sn

∫e−st f (t) dt

)




)

+ e−st f (n)(t)↘

− −se−st f (n−1)(t)↘

+ s2e−st f (n−2)(t)

......

...

(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘

(−1)n (−1)nsne−st → f (t)(+sn

∫e−st f (t) dt

)




)

+ e−st f (n)(t)↘

− −se−st f (n−1)(t)↘

+ s2e−st f (n−2)(t)...

......

(−1)n−1 (−1)n−1sn−1e−st f (1)(t)

↘(−1)n (−1)nsne−st → f (t)

(+sn

∫e−st f (t) dt

)




)

+ e−st f (n)(t)↘

− −se−st f (n−1)(t)↘

+ s2e−st f (n−2)(t)...

......

(−1)n−1 (−1)n−1sn−1e−st f (1)(t)↘

(−1)n (−1)nsne−st → f (t)(+sn

∫e−st f (t) dt

)



Therefore,

L{f (n)(t)} =

∫ ∞0

e−st f (n)(t) dt

=[e−st f (n−1)(t) + se−st f (n−2)(t) + · · ·+ sn−1e−st f (t)

]∞0

+ sn∫ ∞0

e−st f (t) dt

=−f (n−1)(0)− sf (n−2)(0)− · · · − sn−1f (0)

+ snL{f (t)}.


References

Horowitz, D. (1990). Tabular integration by parts. College Math. J.21, no. 4, 307-311.

Rock, J. A. (2017). A lecture on integration by parts. MathematicalScientist 42, no. 1.


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