David Young

2022-01-04

Some time ago I came across to the following integral:
$I={\int }_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\mathrm{ln}x}$
What are the hints on how to compute this integral?

esfloravaou

Expert

Make the substitution $x={e}^{-y}$ and do a little algebra to get the value of the integral to be
${\int }_{0}^{\mathrm{\infty }}\frac{dy}{y}\frac{{e}^{-y}-{e}^{-2y}}{1+{e}^{-y}}$
Now Taylor expand the denominator and get
${\int }_{0}^{\mathrm{\infty }}\frac{dy}{y}\left({e}^{-y}-{e}^{-2y}\right)\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}{e}^{-ky}$
If we can reverse the order of sum and integral, we get
$\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}{\int }_{0}^{\mathrm{\infty }}\frac{dy}{y}\left({e}^{-\left(k+1\right)y}-{e}^{-\left(k+2\right)y}\right)$
The integral inside may be evaluated exactly, and the result is the sum
$\sum _{k=0}^{\mathrm{\infty }}{\left(-1\right)}^{k}\mathrm{log}\frac{k+1}{k+2}$
$=\underset{n\to \mathrm{\infty }}{lim}\mathrm{log}\frac{\frac{12}{\frac{34}{\dots }}\frac{2n-1}{2n}}{\frac{23}{\frac{56}{\dots }}\frac{2n+1}{2n+2}}$
$=\mathrm{log}\left(\frac{2}{\pi }\right)$

jgardner33v4

Expert

I tried with Differentiation under integration sign:

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

nick1337

Expert

Consider
$I\left(\alpha \right)={\int }_{0}^{1}\frac{1-{x}^{\alpha }}{\left(1+x\right)\mathrm{ln}x}dx$ (1)
Differentiating (1) with respect $\alpha$ yields
$\frac{dI}{d\alpha }={\int }_{0}^{1}\frac{d}{d\alpha }\left[\frac{1-{x}^{\alpha }}{\left(1+x\right)\mathrm{ln}x}\right]dx\phantom{\rule{0ex}{0ex}}=-{\int }_{0}^{1}\frac{{x}^{\alpha }}{1+x}dx\phantom{\rule{0ex}{0ex}}=-{\int }_{0}^{1}\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}{x}^{\alpha +k}dx\phantom{\rule{0ex}{0ex}}=-\sum _{k=0}^{\mathrm{\infty }}\left(-1{\right)}^{k}{\int }_{0}^{1}{x}^{\alpha +k}dx\phantom{\rule{0ex}{0ex}}=-\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{\alpha +k+1}$ (2)
Now consider polygamma function
${\psi }_{n}\left(z\right)=\frac{{d}^{n+1}}{d{z}^{n+1}}\mathrm{ln}\mathrm{\Gamma }\left(z\right)$ (3)
and
${\psi }_{n}\left(z\right)=\left(-1{\right)}^{n+1}n!\sum _{k=0}^{\mathrm{\infty }}\frac{1}{\left(z+k{\right)}^{n+1}}$ (4)
Hence by using (4) we obtain
$\sum _{k=0}^{\mathrm{\infty }}\frac{\left(-1{\right)}^{k}}{\left(z+k{\right)}^{n+1}}=\frac{1}{\left(-2{\right)}^{n+1}n!}\left[{\psi }_{n}\left(\frac{z}{2}\right)-{\psi }_{n}\left(\frac{z+1}{2}\right)\right]$ (5)
Using (3) and (5) then (2) becomes

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