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I have an interest in contour integration. I am not that good at it, but I enjoy learning what I can about it.

Here is a version of a rational log integral rarely encountered.

$\displaystyle \int_{0}^{1}\frac{\ln(x^{2}+1)}{x+1}dx=\frac{3}{4}\ln^{2}(2)-\frac{{\pi}^{2}}{48}$

I can do this using real methods (via double integral and substitution). I can post my workings if anyone would be interested.

My question is, can this be evaluated using contour integration due to the limits being $[0,1]$ instead of $[0,\infty)$?. Contours may not be the most efficient way to go about it, but what is the course of action when the limits are 0 to 1 instead of 0 to infinity?.

Start wearing purple
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Cody
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    $x=1$ is not a zero/pole/branch point of the integrand. I would therefore be surprised if there is a way to compute this using contour integrals unless I overlooked some symmetry. – Start wearing purple May 30 '13 at 13:38
  • Problem is that you would want to extend integration region to $[0,\infty)$, but you can't because the resulting integral does not converge. – Ron Gordon May 30 '13 at 13:50
  • Yes, I noticed the 0 to infinity is divergent. Sorry, I thought perhaps there was a clever way you all would know that I do not. Thanks Ron and OL. – Cody May 30 '13 at 14:29
  • @O.L. indeed, neither is $0$ a special point. – not all wrong May 30 '13 at 16:02
  • If anyone wants to post their own method of evaluation, please feel free. It does not have to be contour integration. It's always fun to see clever methods of going about problems. I derived a digamma sum as one way to approach it. Maybe overkill, but it is interesting. $\displaystyle \frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi(n+1)}{n}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^{n}\psi(n+1/2)}{n}$ – Cody May 31 '13 at 14:27

1 Answers1

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Here is my favorite method:

Consider a parameter $a$ dependent integral:

$$I(a)=\int_{0}^{1}\frac{\ln(ax^{2}+1)}{x+1}dx$$

After differentiating with respect to $a$ we get:

$$\frac{dI}{da}=\int_{0}^{1}\frac{x^2dx}{(1+x)(1+ax^2)}=$$

$$=\frac{1}{2}\frac{\ln(1+a)}{a(1+a)}+\frac{\ln 2}{1+a}-\frac{\arctan\sqrt{a}}{\sqrt{a}(1+a)}$$

Integration of $I'(a)$ yields $$I(a)=\ln2\ln(1+a)-\frac{1}{4}\ln^2(1+a)-\arctan^2\sqrt{a}+\frac{1}{2}\int_{1}^{a}\frac{\ln(1+z)}{z}dz+\text{const}$$

In order to find out the constant of integration , we note that $I(0)=0$, implying that

$$\text{const}=\frac{1}{2}\int_{0}^{1}\frac{\ln(1+z)}{z}dz=\frac{1}{2}\frac{\pi^2}{12}=\frac{\pi^2}{24}$$

The last integral can be easily calculated by expanding the log into Taylor series and integrating term by term. Finally the original integral $I$ yields $I=I(1)$

Martin Gales
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  • +1 for reminding me that parametrizing and differentiating often solves many problems. – Sangchul Lee Jun 10 '13 at 15:54
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    The last integral becomes: $$ \int_{1}^{a}{\ln(1 + z) \over z},{\rm d}z =\int_{-1}^{-a}{\ln(1 - z) \over z},{\rm d}z =-\int_{-1}^{-a}{\rm Li}{2}'(z),{\rm d}z ={\rm Li}{2}(-1) - {\rm Li}{2}(-a) =-,{\pi^{2} \over 12} - {\rm Li}{2}(-a) $$ – Felix Marin Sep 04 '14 at 02:59