David Young

Answered

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?

What are the hints on how to compute this integral?

Answer & Explanation

esfloravaou

Expert

2022-01-05Added 43 answers

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

$\int}_{0}^{\mathrm{\infty}}\frac{dy}{y}({e}^{-y}-{e}^{-2y})\sum _{k=0}^{\mathrm{\infty}}{(-1)}^{k}{e}^{-ky$

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

$\sum _{k=0}^{\mathrm{\infty}}{(-1)}^{k}{\int}_{0}^{\mathrm{\infty}}\frac{dy}{y}({e}^{-(k+1)y}-{e}^{-(k+2)y})$

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

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

$=\underset{n\to \mathrm{\infty}}{lim}\mathrm{log}\frac{\frac{12}{\frac{34}{\dots}}\frac{2n-1}{2n}}{\frac{23}{\frac{56}{\dots}}\frac{2n+1}{2n+2}}$

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

Now Taylor expand the denominator and get

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

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

jgardner33v4

Expert

2022-01-06Added 35 answers

I tried with Differentiation under integration sign:

$J\left(\alpha \right)={\int}_{0}^{1}\frac{1-{x}^{\alpha}}{1+x}\frac{dx}{\mathrm{ln}x}\text{}\alpha 0$

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.

Then as usual,

Then, integrating

I used Mathematica to evaluate

nick1337

Expert

2022-01-11Added 573 answers

Consider

Differentiating (1) with respect

Now consider polygamma function

and

Hence by using (4) we obtain

Using (3) and (5) then (2) becomes

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