Miguel Reynolds

Answered

2022-01-05

What is the proof of the following:

$\int}_{0}^{1}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt=\frac{{\pi}^{2}}{3$ ?

Answer & Explanation

yotaniwc

Expert

2022-01-06Added 34 answers

This is by no means a complete solution but a possible route.

Letting$t=\frac{1}{x}$ note that

$I={\int}_{0}^{1}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt$

$={\int}_{1}^{\mathrm{\infty}}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt$

$={\int}_{0}^{\mathrm{\infty}}{\left(\frac{\mathrm{ln}(1+t)}{t}\right)}^{2}dt$

Setting$1+t={e}^{x}$ , we get

$I={\int}_{0}^{\mathrm{\infty}}\frac{{x}^{2}}{{({e}^{x}-1)}^{2}}{e}^{x}dx$

$={\int}_{0}^{\mathrm{\infty}}\frac{{x}^{2}}{{({e}^{\frac{x}{2}}-{e}^{-\frac{x}{2}})}^{2}}dx$

$=2{\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{{x}^{2}}{{\text{sinh}}^{2}\left(x\right)}dx$

The last integral can be done by the method of residues to get

$\int}_{-\mathrm{\infty}}^{\mathrm{\infty}}\frac{{x}^{2}}{{\text{sinh}}^{2}\left(x\right)}dx=\frac{{\pi}^{2}}{6$

I will fill this in once I get back home.

Letting

Setting

The last integral can be done by the method of residues to get

I will fill this in once I get back home.

encolatgehu

Expert

2022-01-07Added 27 answers

Heres

star233

Expert

2022-01-11Added 238 answers

Consider

Differentiating (1) with respect to t yields

Multiplying both sides of (2) by

Using formula

then (3) becomes

where

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