# Evaluating $\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\ln x}$

Some time ago i came across to the following integral:

$$I=\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\ln x}$$ What are the hints on how to compute this integral?

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What happens if you substitute $u=\ln x$? (On second thought, it doesn't look too useful.) –  Harald Hanche-Olsen Jan 16 '13 at 15:30
@HaraldHanche-Olsen: Isn't it an improper integral? –  B. S. Jan 16 '13 at 15:39
Fwiw Wolfram alpha couldn't find an indefinite integral, although it can find the definite integral, presumably by approximation. –  JSchlather Jan 16 '13 at 15:40
@JacobSchlather: But it converges. –  B. S. Jan 16 '13 at 15:40
Maple says the value is $-\ln(\pi/2)$. –  Harald Hanche-Olsen Jan 16 '13 at 15:42

Make the substitution $x=e^{-y}$ and do a little algebra to get the value of the integral to be

$$\int_0^{\infty} \frac{dy}{y} \frac{e^{-y} - e^{-2 y}}{1+e^{-y}}$$

Now Taylor expand the denominator and get

$$\int_0^{\infty} \frac{dy}{y} (e^{-y} - e^{-2 y}) \sum_{k=0}^{\infty} (-1)^k e^{-k y}$$

If we can reverse the order of sum and integral, we get

$$\sum_{k=0}^{\infty} (-1)^k \int_0^{\infty} \frac{dy}{y} (e^{-(k+1) y} - e^{-(k+2) y})$$

The integral inside may be evaluated exactly, and the result is the sum

$$\sum_{k=0}^{\infty} (-1)^k \log{\frac{k+1}{k+2}}$$

$$= \lim_{n \rightarrow \infty} \; \log{\frac{\frac{1}{2} \frac{3}{4} \ldots \frac{2 n-1}{2 n}}{\frac{2}{3} \frac{5}{6} \ldots \frac{2 n+1}{2 n+2}}}$$

$$= \log{\left ( \frac{2}{\pi} \right )}$$

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Seems reasonable. Thank you! –  Martin Gales Jan 16 '13 at 16:27
Thank you for asking it - this was fun! –  Ron Gordon Jan 16 '13 at 16:40
Yeah, math is about having fun! :-) (+1) –  Chris's sis Jan 16 '13 at 16:54
To reverse the order of sum and integra, I understand that you appeal to the absolute convergence of sum and integral, that is, to the convergence of the positive series $\sum\limits_{k\geqslant0}\int\limits_0^{+\infty}(e^{-(k+1)y}-e^{-(k+2)y})dy/y$. But this series diverges. Maybe I did not understand the result you are applying, could you explain? –  Did Jan 21 '13 at 9:17
Maybe we are speaking across each other. Look, do you agree that the sum in Line 2 absolutely converges (in this case, to $(1-e^{-y})^{-1})$? And that the integral absolutely converges (in this case, to $\log{\frac{k+1}{k+2}}$). Because each integral is also continuous, then we get a convergent result in exchanging the order of the sum and integral in Line 2. As I said, this does not apply to Line 3 because, as you point out, that the sum does not converge absolutely. But I am not seeking to reverse the order of integration here, just in Line 2. –  Ron Gordon Jan 21 '13 at 14:38

I tried with "Differentiation under integration sign":

$$J(\alpha)=\int_0^1\frac{1-x^\alpha}{1+x}\frac{dx}{\ln x}\quad \alpha>0$$

Then, as usual, $$\frac{dJ}{d\alpha}=-\int_0^1\frac{x^\alpha}{1+x}dx=-f(\alpha),\text{(say)}$$

Then, integrating $$J(\alpha)=-\int f(\alpha)d\alpha+c$$ I used Mathematica to evaluate $f(\alpha)$ (it gives difference of two Harmonic numbers) and its integral. The integral is just $\ln\frac{\Gamma(\frac{2m+1}{2})}{\Gamma(\frac{m+1}{2})}$. Note that $J(0)=0$ and putting $\alpha=1$, the result follows.

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Very interesting approach! Thanks! –  Martin Gales Jan 17 '13 at 16:05
Thanks, Martin! –  Tapu Jan 17 '13 at 17:13
Division? And what did you do with the $\ln x$ term in the denominator? –  L. F. Jan 26 '13 at 15:34