 Miguel Reynolds

2022-01-05

What is the proof of the following:
${\int }_{0}^{1}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt=\frac{{\pi }^{2}}{3}$ ? yotaniwc

Expert

This is by no means a complete solution but a possible route.
Letting $t=\frac{1}{x}$ note that
$I={\int }_{0}^{1}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt$
$={\int }_{1}^{\mathrm{\infty }}{\left(\frac{\mathrm{ln}t}{1-t}\right)}^{2}dt$
$={\int }_{0}^{\mathrm{\infty }}{\left(\frac{\mathrm{ln}\left(1+t\right)}{t}\right)}^{2}dt$
Setting $1+t={e}^{x}$, we get
$I={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{{\left({e}^{x}-1\right)}^{2}}{e}^{x}dx$
$={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}}{{\left({e}^{\frac{x}{2}}-{e}^{-\frac{x}{2}}\right)}^{2}}dx$
$=2{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{{x}^{2}}{{\text{sinh}}^{2}\left(x\right)}dx$
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. encolatgehu

Expert

${\int }_{0}^{1}{\left(\frac{\mathrm{log}t}{1-t}\right)}^{2}={\int }_{0}^{\mathrm{\infty }}\frac{{x}^{2}{e}^{x}}{{\left({e}^{x}-1\right)}^{2}}dx$
Heres star233

Expert

Consider
$\sum _{k=0}^{\mathrm{\infty }}{t}^{k}=\frac{1}{1-t}$ (1)
Differentiating (1) with respect to t yields
$\sum _{k=1}^{\mathrm{\infty }}k{t}^{k-1}=\frac{1}{\left(1-t{\right)}^{2}}$ (2)
Multiplying both sides of (2) by ${\mathrm{ln}}^{2}t$ and integrating from t=0 to t=1 yields
${\int }_{0}^{1}\left(\frac{\mathrm{ln}t}{1-t}{\right)}^{2}dt={\int }_{0}^{1}\sum _{k=1}^{\mathrm{\infty }}k{t}^{k-1}{\mathrm{ln}}^{2}tdt$
$=\sum _{k=1}^{\mathrm{\infty }}k{\int }_{0}^{1}{t}^{k-1}{\mathrm{ln}}^{2}tdt$ (3)
Using formula
(4)
then (3) becomes
${\int }_{0}^{1}\left(\frac{\mathrm{ln}t}{1-t}{\right)}^{2}dt=2\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}}$
$=\frac{{\pi }^{2}}{3}$
where $\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{k}^{2}}=\zeta \left(2\right)=\frac{{\pi }^{2}}{6}$

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