On the New Evaluation of an Old Integral 

by William Walters and Michael Huber 

ARL-TR-4689 December 2008 

Approved for public release; distribution is unlimited. 
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Army Research Laboratory 

Aberdeen Proving Ground, MD 21005-5069 

ARL-TR-4689 December 2008 

On the New Evaluation of an Old Integral 

William Walters 

Weapons and Materials Research Directorate, ARL 

Michael Huber 

Muhlenberg College 

Approved for public release; distribution is unlimited. 
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4. TITLE AND SUBTITLE 

On the New Evaluation of an Old Integral 

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6. AUTHOR(S) 

William Walters and Michael Huber*

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AH80 

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*Muhlenberg College, Allentown, PA 18104 

14. ABSTRACT 

A simple technique is shown to evaluate   .

)1( m n xx

dx

15. SUBJECT TERMS

calculus, integration, power series 

16. SECURITY CLASSIFICATION OF: 

17. LIMITATION 

OF ABSTRACT 

UL 

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OF PAGES 

12 

19a. NAME OF RESPONSIBLE PERSON

William Walters 

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Standard Form 298 (Rev. 8/98) 

Prescribed by ANSI Std. Z39.18 
iii

Contents 

1. Introduction 1

2. Using Infinite Sums 1

3. A Generalization 3

4. Examples 3

Distribution List 5 
iv

INTENTIONALLY LEFT BLANK.
1 

1. Introduction 

Consider the integral 

  . xx )1(

dx

m (1) 

In the CRC Standard Mathematical Tables,1 equation 1 can require repeated integral evaluations. 

Integral 87 shows 

. )( )(

1

)(  1    1      m pn m pn nm pn bxax

dx

a

b

bxax

dx

bxax a

dx (2) 

Enter this integral into your favorite computer algebra system. Maple returns 

, 1

),1,( LerchPhi)1(

)1(

)1(

)1( m

mx mx

mm

mx

xx

dx m m

m 

     

 

 (3) 

which involves Lerch’s Phi transcendent function, defined as 

. n)(m

x ),,( LerchPhi

0n

a

n





 

max  (4) 

Similarly, Mathematica gives 

, 1

Hypergeome ],2,1,1[ 2F1 tric – )1(

1

m

x xmm

xx

dx m

m 

  



 (5) 

which involves knowing the implementation of Hypergeometric functions. In this report, we 

seek to provide a simpler evaluation for integrals of this form (equation 1). We state up front 

that the exponent m need not be an integer. 

2. Using Infinite Sums 

The integrand of equation 1 can be rewritten as follows: 

. 1 x

x m





 (6) 

By long division, 

 1

 Beyer, W. H., Ed. CRC Standard Mathematical Tables. CRC Press, Inc.: West Palm Beach, FL, 1978. 
2 

, 1

1 2 3   



mm m m

m

xxxx

x

x

 (7) 

which is an infinite sum in the form 

.)1(

0







  i

imi

x (8) 

Therefore, the integral becomes 





 























 



 

0

1

0

0

, 1

)1(

)1(

)1( )1(

i

imi

i

imi

i

imi

m

C

im

x

dxx

dxx

xx

dx

 (9)

for some constant C. Noting that 







  0)1( 1

1

i

ii

x

x

 (10) 

converges for x

2

 < 1 allows us to rewrite equation 9 as 













  

 0

1

; 1

1 )1(

)1( i

ii

m C

mi

x

x x mx

dx

 (11) 

again, m need not be an integer. However, if m = 1, 2, 3, ... , then for i = m–1, x

–m+i

 = x

–1, so 

x

x

dx  ln  (12) 

for the i = m – 1 term, or 

  









 





 



  

2

0

1 1

. 1

)1( ln

1

1 )1(

)1(

m

i m i

ii ii

m m C

mi

x

x

mi

x

xxx

dx (13) 

This allows us to sum past the singularity. 
3 

3. A Generalization 

Generalizing, consider the following: 

    



. 1)1( n

m

m n x

dxx

xx

dx (14) 

From long division, the integrand can be rewritten as x

-m (1 – xn

 + x

2n

 – x

3n

 + ...), 

or 











  0.)1( 1 i

mini

n

m

x

x

x (15) 

Thus, 

      – 1

–

0 0

–1

–1

1 – 1

i in m

i in m

m n i i

dx x

x dx C

x x ni m

  

 

 

     , (16) 

and we sum prior to integration past any singularity. As before, m and n need not be integers, 

and this technique converges for | x | < 1. 

4. Examples 

As a first example, we consider the case with m = 3, an integer. We start with equation 2 (shown 

previously), where m = 3, a = b = n = 1, and p = 0. Continuing, we find that 

3 32 – (1 ) (1 )

dx dx dx

x x xxx     

 

–2

2 –– – 2 (1 )

x dx dx

x x x

         

 

–2

–1 – 1n –1n (1 ) 2

x

    x x xC

 

234

2

1 1 – 1n – – – ...

2 234

xxx

x x C

x x

 

        

 

234

2

1 1 – 1n – – – , .. . 

2 234

xxx

x x C

x x

     (17) 
4 

which is valid for 1 < x < 1. Using our approach and equation 9, 

   )1( 3

xx

dx 





  dxx

i

ii

0

3 ( )1

  dxxxxxx    10123 

 , 432

ln 1

2

1 432

2 C xxx

xx

x x  (18) 

which matches the CRC’s solution. When m is a positive integer, equation 2 is adequate. 

However, as m gets large, the CRC’s proposal will involve evaluating several integrals. Our 

technique is much simpler. 

Now, suppose m = 

5

1 , n = 3, and we keep a = b = 1 and p = 0. Equation 2 becomes 







 )1(

–

)1(

3 5 3 14–

5

1

5

1

xx

dx

x

dx

xx

dx . (19) 

While the first integral on the right-hand side is easy enough to evaluate, the second is not trivial, 

unless the process is repeated or the integrand is expanded in a power series. Turning to 

equation 15, we find 

  







 0

5

1 –3

3 5

1 )1(–

)1( i

i dxx

xx

dx i

[ 5 ] xdxxxx 44

5

29

5

14

5

1

   

 . 49

5

34

5

19

5

4

5 5

49

5

34

5

19

5

4

 Cxxxx (20) 

We do not offer any practical applications for the integral. However, the proposed method 

converges for x

2

 < 1 or when the independent variable is normalized, to guarantee that the 

normalized value of x is less than 1 when nondimensionalized; otherwise, the series diverges. 
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