David Horowitz, Golden West College, Huntington Beach, CA 92647
The College Mathematics Journal, September 1990, Volume 21,
Number 4, pp. 307–311.
Only a few contemporary calculus textbooks provide even a cursory presentation of tabular integration by parts [see for example, G. B. Thomas and R. L. Finney, Calculus
and Analytic Geometry, Addison-Wesley, Reading, MA 1988]. This is unfortunate because tabular integration by parts is not only a valuable tool for finding integrals but can also be applied to more advanced topics including the derivations of some important theorems in analysis.
The technique of tabular integration allows one to perform successive integrations by parts on integrals of the form
E
F s t d
G s t d
dt (1) without becoming bogged down in tedious algebraic details [V. N. Murty, Integration by parts, Two-Year College Mathematics Journal 11 (1980) 90-94]. There are several ways to illustrate this method, one of which is diagrammed in Table 1. (We assume throughout that F and G are “smooth” enough to allow repeated differentation and integration, respectively.)
Table 1
____________________________________________________
Column #1 Column #2
____________________________________________________
1
F
2
F s
1 d
1
F s
2 d
2
F s
3 d
.
: s 2
1 d n F s n d s 2
1 d n
1
1 F s n
1
1 d
- - - G
G
G
G
G
G s 2
1 d s 2
2 d s 2
3 d
.
: s 2 n d s 2 n
2
1 d
____________________________________________________
In column #1 list F s t d and its successive derivatives. To each of the entries in this column, alternately adjoin plus and minus signs. In column #2 list successive antiderivatives. (The notation G s 2 n d
G s t d and its denotes the nth antiderivative of G. Do not include an additive constant of integration when finding each antiderivative.) Form successive terms by multiplying each entry in column #1 by the entry in column #2 that lies just below it. The integral (1) is equal to the sum of these terms. If F s x d is a

polynomial, then there will be only a finite number of terms to sum. Otherwise the process may be truncated at any level by forming a remainder term defined as the integral of the product of the entry in column #1 and the entry in column #2 that lies directly across from it. Symbolically,
E
F s t d
G s t d
dt
5
FG s 2
1 d 2
F s
1 d
G s 2
2 d 1
F s
2 d
G s 2
3 d 2
.
. .
1 s 2
1 d n F s n d
G s 2 n
2
1 d 1 s 2
1 d n
1
1
E
F s n
1
1 d s t d
G s 2 n
2
1 d s t d
dt
5 k n o
0 s 2
1 d k F s k d
G s 2 k
2
1 d 1 s 2
1 d n
1
1
E
F s n
1
1 d s t d
G s 2 n
2
1 d s t d
dt.
(2)
A proof follows from continued application of the formula for integration by parts
[K. W. Folley, Integration by parts, American Mathematical Monthly 54 (1947)
542-543].
The technique of tabular integration by parts makes an appearance in the hit motion picture Stand and Deliver in which mathematics instructor Jaime Escalante of James A.
Garfield High School in Los Angeles (portrayed by actor Edward James Olmos) works the following example.
Example.
e x 2 sin x dx column #1 column #2
____________________________________________________
1 x 2
2
2x
1
2 sin x
2 cos x
2 sin x
0 cos x
____________________________________________________
Answer:
2 x 2 cos x
1
2x sin x
1
2 cos x
1
C
The following are some areas where this elegant technique of integration can be applied.
Miscellaneous Indefinite Integrals. Most calculus textbooks would treat integrals of the form
E
P s t d s at
1 b d r
dt, where P s t d is a polynomial, by time-consuming algebraic substitutions or tedious partial fraction decompositions. However, tabular integration by parts works particularly well on such integrals (especially if r is not a positive integer).
2

Example.
E
12t 2 1
36
3t
1
2
dt column #1 column #2
____________________________________________________
1
2
12t 2
24t
1
36 s
3t
1
2 d 2
1 y
5
5
12 s
3t
1
2 d
4 y
5
1
24
25
324 s
3t
1
2 d
9 y
5
0
125 s
3t
1
2 d
14 y
5
13608
____________________________________________________
Answer: s
5t 2
1
15 ds
3t
1
2 d
4 y
5
2
50t
27 s
3t
1
2 d
9 y
5
1
125
567 s
3t
1
2 d
14 y
5
1
C
Laplace Transforms. Computations involving the Laplace transform
L
H f s t dJ 5
E
`
0 e
2 st f s t d
dt (3) often require several integrations by parts because of the form of the integrand in (3).
Tabular integration by parts streamlines these integrations and also makes proofs of operational properties more elegant and accessible. (Many introductory differential equations textbooks omit formal proofs of these properties because of the lengthy detail involved in their derivations.) The following example uses this technique to establish the fundamental formula for the Laplace transform of the nth derivative of a function.
Theorem. Let n be a positive integer and suppose that f s n d s t d is piecewise continuous on the interval t
≥
0.
f s t d is a function such that
Furthermore suppose that there exist constants A, b, and M such that for all k
5
0, 1, 2, . . . , n
2
1.
| f s k d s t d | ≤
Ae bt if t
≥
M
Then if s > b,
L
H f s n d s t dJ 5 2 f s n
2
1 d s
0 d 2 sf s n
2
2 d s
0 d 2 ? ? ?
2 s n
2
1 f s
0 d 1 s n L
H f s t dJ
.
(4)
3

Proof.
column #1 column #2
____________________________________________________
1
2
1 e
2 st
2 se
2 st s 2 e
2 st
.
: s 2
1 d n
2
1 s 2
1 d n
2
1 s n
2
1 e
2 st s 2
1 d n s 2
1 d n s n e
2 st f f f f f s n d s n
2
1 d s t d s n
2
2 d s t d s
1 d s t d s
.
: t s t d d
____________________________________________________
L
H f s n d s t dJ 5
1
3 e
2 st f s n
2
1 d s t d 1 se
2 st f s n
2
2 d s t d 1 ? ? ?
1 s n
2
1 e
2 st f s t d
4 t
5 t
5
0
`
E
`
0 s n e
2 st f s t d
dt
5 2 f s n
2
1 d s
0 d 2 sf s n
2
2 d s
0 d 2 ? ? ?
2 s n
2
1 f s
0 d 1 s n L
H f s t dJ
.
Taylor’s Formula. Tabular integration by parts provides a straightforward proof of
Taylor’s formula with integral remainder term.
Theorem. Suppose f s t d has n
1
1 continuous derivatives throughout an interval containing a. If x is any number in the interval then f s x d 5 f s a d 1 f s
1 d
(a ds x
2 a d 1 f s
2 d s a d
2!
s x
2 a d
2 1 ? ? ?
1 f s n d s a d
n!
s x
2 a d n
1
E x a f s n
1
1 d s t d
n!
s x
2 t d n dt.
(5)
4

Proof.
column #1 column #2
____________________________________________________ s s
2
1
2
1
2
1
1 d n d
2 f s
1 d s t d
2 f s
2 d s t d
2 f s
3 d s t d n
1
1
1
2
:
.
2 f
2 f s n s n d s t d
1
1 d s t d
2
1 s x
2 t d
2 s x
2 t d
2
2!
.
: s 2
1 d n s x
2 t d n
2
1 s n
2
1 d
!
s 2
1 d n
1
1 s x
2 t d n
n!
____________________________________________________
E x a f 2 f s
1 d s t dgf 2
1 g
dt
5
3
2 f s
1 d s t ds x
2 t d 2
1 f s
2 d s t d
2!
s x
2 t d
2 2
E x a f s n
1
1 d s t d
n!
s x
2 t d n dt
? ? ?
2 f s n d s t d
n!
s x
2 t d n
4 t
5 x t
5 a
Equation (5) follows immediately.
Residue Theorem for Meromorphic Functions. Tabular integration by parts also applied to complex line integrals and can us used to prove the following form of the residue theorem.
Theorem. Suppose f s z d is analytic in
$ order m at z
0
.
Then if 0 < r < R
5 H
z: 0 <
| z
2 z
0
|
< R
J and has a pole of
| z
2 z r
0
^ 1 | 5 r f s z d
dz
5
2 p i s m
2
1 d
!
lim z
→ z
0
3 d m
2
1 dz m
2
1 s z
2 z
0 d m f s z d
4
.
(6)
5

Proof.
column #1 column #2
____________________________________________________ s 2
1 d m
1
2
1 s z
2 z
0 d m f s z d d dz s z
2 z
0 d m f s z d d 2 dz 2 s z
2 z
0
2
1 d m dz m
.
:
2
1
2
1 d m f s z d s z
2 z
0 d m f s z d s z
2 z
0 d 2 m
2 s z
2 z
0 d 2 m
1
1 m
2
1 s z
2 z
0 d 2 m
1
2 s m
2
1 ds m
2
2 d
.
: s 2
1 d m
2
1 s z
2 z
0 d 2
1 m
2
1!
____________________________________________________
Thus,
| z
2 z r
0
| 5 r s z
2 z
0 d m f s z ds z
2 z
0 d 2 m dz
5
3
2 m
2
2 s m
2 k
2
2 d
!
k o
0 d k dz k s z
2 z s m
2
1 d
!
s z
2 z
0 d m
0 d m f s z d
2 k
2
1
4
| z
2 z
0
| 5 r
1
1 s m
2
1 d
!
| z
2 r z
0
| 5 r d m
2
1 dz m
2
1 s z
2 z
0 z
2 z
0 d m f s z d
dz, (7) where the summation term in (7) must be evaluated for the closed circle
| z
2 z
0
| 5
r.
However, each of the functions in column #1 has a removable singularity at z
0 and is therefore single-valued in
$
. Furthermore, each of the functions in column #2 is also single-valued in
$
. Thus, the summation term in (7) must vanish. The integral term on the right side of (7) is evaluated by the Cauchy integral formula and the result (6) follows directly. (The right-hand side of (6) can be recognized as the formula for the residue of f s z d at the pole z
0
.
) See [Ruel V. Churchill and James W. Brown, Complex
Variables and Applications, McGraw-Hill, New York, 1984].
6