What is the proof of the following: \int_0^1(\frac{\ln t}{1-t})^2dt=\frac{\pi^2}{3} ?

Miguel Reynolds 2022-01-05 Answered
What is the proof of the following:
\(\displaystyle{\int_{{0}}^{{1}}}{\left({\frac{{{\ln{{t}}}}}{{{1}-{t}}}}\right)}^{{2}}{\left.{d}{t}\right.}={\frac{{\pi^{{2}}}}{{{3}}}}\) ?

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Expert Answer

yotaniwc
Answered 2022-01-06 Author has 4975 answers
This is by no means a complete solution but a possible route.
Letting \(\displaystyle{t}={\frac{{{1}}}{{{x}}}}\) note that
\(\displaystyle{I}={\int_{{0}}^{{1}}}{\left({\frac{{{\ln{{t}}}}}{{{1}-{t}}}}\right)}^{{2}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{1}}^{\infty}}{\left({\frac{{{\ln{{t}}}}}{{{1}-{t}}}}\right)}^{{2}}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{{\infty}}}{\left({\frac{{{\ln{{\left({1}+{t}\right)}}}}}{{{t}}}}\right)}^{{2}}{\left.{d}{t}\right.}\)
Setting \(\displaystyle{1}+{t}={e}^{{x}}\), we get
\(\displaystyle{I}={\int_{{0}}^{\infty}}{\frac{{{x}^{{2}}}}{{{\left({e}^{{x}}-{1}\right)}^{{2}}}}}{e}^{{x}}{\left.{d}{x}\right.}\)
\(\displaystyle={\int_{{0}}^{\infty}}{\frac{{{x}^{{2}}}}{{{\left({e}^{{\frac{{x}}{{2}}}}-{e}^{{-\frac{{x}}{{2}}}}\right)}^{{2}}}}}{\left.{d}{x}\right.}\)
\(\displaystyle={2}{\int_{{-\infty}}^{\infty}}{\frac{{{x}^{{2}}}}{{{{\text{sinh}}^{{2}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}\)
The last integral can be done by the method of residues to get
\(\displaystyle{\int_{{-\infty}}^{\infty}}{\frac{{{x}^{{2}}}}{{{{\text{sinh}}^{{2}}{\left({x}\right)}}}}}{\left.{d}{x}\right.}={\frac{{\pi^{{2}}}}{{{6}}}}\)
I will fill this in once I get back home.
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encolatgehu
Answered 2022-01-07 Author has 1476 answers
\(\displaystyle{\int_{{0}}^{{1}}}{\left({\frac{{{\log{{t}}}}}{{{1}-{t}}}}\right)}^{{2}}={\int_{{0}}^{\infty}}{\frac{{{x}^{{2}}{e}^{{x}}}}{{{\left({e}^{{x}}-{1}\right)}^{{2}}}}}{\left.{d}{x}\right.}\)
Here's a different way to complete his argument, plus a generalization in the comments.
If \(\displaystyle{p}={1}-{e}^{{-{x}}}\), and Y is geometric (p), then
\(\displaystyle{E}{\left[{Y}\right]}={\frac{{{1}}}{{{p}}}}={\frac{{{1}}}{{{1}-{e}^{{-{x}}}}}}={\frac{{{e}^{{x}}}}{{{e}^{{x}}-{1}}}}\)
But, by definition
\(\displaystyle{E}{\left[{Y}\right]}={\sum_{{{k}={0}}}^{\infty}}{k}{\left({1}-{p}\right)}^{{{k}-{1}}}{p}={\sum_{{{k}={1}}}^{\infty}}{k}{\left({e}^{{-{x}}}\right)}^{{{k}-{1}}}{\left({1}-{e}^{{-{x}}}\right)}\)
\(\displaystyle={\sum_{{{k}={1}}}^{\infty}}{k}{\left({e}^{{-{x}}}\right)}^{{k}}{\left({e}^{{x}}-{1}\right)}\)
Thus we have
\(\displaystyle{\int_{{0}}^{\infty}}{\frac{{{x}^{{2}}{e}^{{x}}}}{{{\left({e}^{{x}}-{1}\right)}^{{2}}}}}{\left.{d}{x}\right.}={\int_{{0}}^{\infty}}{x}^{{2}}{\left({\sum_{{{k}={1}}}^{\infty}}{k}{\left({e}^{{-{x}}}\right)}^{{k}}\right)}{\left.{d}{x}\right.}\)
\(\displaystyle={\sum_{{{k}={1}}}^{\infty}}{\left({k}{\int_{{0}}^{\infty}}{x}^{{2}}{e}^{{-{k}{x}}}{\left.{d}{x}\right.}\right)}\)
Finally, if we let \(\displaystyle{u}={k}{x}\), we get
\(\displaystyle{\sum_{{{k}={1}}}^{\infty}}{\left({\frac{{{1}}}{{{k}^{{2}}}}}{\int_{{0}}^{\infty}}{u}^{{2}}{e}^{{-{u}}}{d}{u}\right)}={\left({\int_{{0}}^{\infty}}{u}^{{2}}{e}^{{-{u}}}{d}{u}\right)}{\left({\sum_{{{k}={1}}}^{\infty}}{\frac{{{1}}}{{{k}^{{2}}}}}\right)}=\Gamma{\left({3}\right)}\zeta{\left({2}\right)}={\frac{{\pi^{{2}}}}{{{3}}}}\)
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star233
Answered 2022-01-11 Author has 0 answers

Consider
\(\sum_{k=0}^\infty t^k=\frac{1}{1-t}\) (1)
Differentiating (1) with respect to t yields
\(\sum_{k=1}^\infty kt^{k-1}=\frac{1}{(1-t)^2}\) (2)
Multiplying both sides of (2) by \(\ln^2 t\) and integrating from t=0 to t=1 yields
\(\int_0^1(\frac{\ln t}{1-t})^2dt=\int_0^1\sum_{k=1}^\infty kt^{k-1}\ln^2 tdt\)
\(=\sum_{k=1}^\infty k\int_0^1 t^{k-1}\ln^2tdt\) (3)
Using formula
\(\int_0^1 x^\alpha\ln^n xdx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}},\text{ for }n=0,1,2,...\) (4)
then (3) becomes
\(\int_0^1(\frac{\ln t}{1-t})^2dt=2\sum_{k=1}^\infty\frac{1}{k^2}\)
\(=\frac{\pi^2}{3}\)
where \(\sum_{k=1}^\infty\frac{1}{k^2}=\zeta(2)=\frac{\pi^2}{6}\)

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