Prove that: \int_0^1\frac{x\ln(1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}

William Cleghorn 2022-01-06 Answered
Prove that:
\(\displaystyle{\int_{{0}}^{{1}}}{\frac{{{x}{\ln{{\left({1}+{x}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}={\frac{{\pi^{{2}}}}{{{96}}}}+{\frac{{{{\ln}^{{2}}{2}}}}{{{8}}}}\)

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Expert Answer

Foreckije
Answered 2022-01-07 Author has 891 answers
Put
\(\displaystyle{f{{\left({s}\right)}}}={\int_{{0}}^{{1}}}{\frac{{{x}{\ln{{\left({s}+{x}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)
We want to determine \(\displaystyle{f{{\left({1}\right)}}}\). After differentiation we have
\(\displaystyle{f}'{\left({s}\right)}={\int_{{0}}^{{1}}}{\frac{{{x}}}{{{\left({s}+{x}\right)}{\left({1}+{x}^{{2}}\right)}}}}{\left.{d}{x}\right.}=-{\int_{{0}}^{{1}}}{\frac{{{s}}}{{{\left({s}^{{2}}+{1}\right)}{\left({s}+{x}\right)}}}}{\left.{d}{x}\right.}+{\int_{{0}}^{{1}}}{\frac{{{s}{x}+{1}}}{{{\left({s}^{{2}}+{1}\right)}{\left({1}+{x}^{{2}}\right)}}}}{\left.{d}{x}\right.}\)
\(\displaystyle=-{\frac{{{s}{\ln{{\left({1}+{s}\right)}}}}}{{{1}+{s}^{{2}}}}}+{\frac{{{\ln{{s}}}}}{{{1}+{s}^{{2}}}}}+{\frac{{{s}{\ln{{2}}}}}{{{2}{\left({1}+{s}^{{2}}\right)}}}}+{\frac{{\pi}}{{{4}{\left({1}+{s}^{{2}}\right)}}}}\)
Now we integrate wrt s between 0 and 1. That yields
\(\displaystyle{f{{\left({1}\right)}}}-{f{{\left({0}\right)}}}=-{f{{\left({1}\right)}}}+{f{{\left({0}\right)}}}+{\frac{{{{\ln}^{{2}}{\left({2}\right)}}}}{{{4}}}}+{\frac{{\pi^{{2}}}}{{{16}}}}\)
Consequently
\(\displaystyle{f{{\left({1}\right)}}}={f{{\left({0}\right)}}}+{\frac{{{{\ln}^{{2}}{\left({2}\right)}}}}{{{8}}}}+{\frac{{\pi^{{2}}}}{{{32}}}}\)
But
\(\displaystyle{f{{\left({0}\right)}}}={\int_{{0}}^{{1}}}{\frac{{{x}{\ln{{\left({x}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}={\sum_{{{k}={0}}}^{\infty}}{\int_{{0}}^{{1}}}{\left(-{1}\right)}^{{k}}{x}^{{{2}{k}+{1}}}{\ln{{x}}}{\left.{d}{x}\right.}\)
\(\displaystyle={\sum_{{{k}={0}}}^{\infty}}{\frac{{{\left(-{1}\right)}^{{{k}+{1}}}}}{{{4}{\left({k}+{1}\right)}^{{2}}}}}=-{\frac{{\pi^{{2}}}}{{{48}}}}\)
Finally we substitute that into (1)
\(\displaystyle{f{{\left({1}\right)}}}={\frac{{{{\ln}^{{2}}{\left({2}\right)}}}}{{{8}}}}+{\frac{{\pi^{{2}}}}{{{96}}}}\)
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Serita Dewitt
Answered 2022-01-08 Author has 5497 answers
This is a solution using complex analysis.
The integral
\(\displaystyle{\int_{{0}}^{{1}}}{\frac{{{\ln{{x}}}}}{{{1}+{x}}}}{\left.{d}{x}\right.}={\sum_{{{k}={0}}}^{\infty}}{\left(-{1}\right)}^{{k}}{\int_{{0}}^{{1}}}{x}^{{k}}{\ln{{x}}}{\left.{d}{x}\right.}={\sum_{{{k}={1}}}^{\infty}}{\left(-{1}\right)}^{{k}}{\frac{{{1}}}{{{k}^{{2}}}}}\) (1)
\(\displaystyle=-{\frac{{\pi^{{2}}}}{{{12}}}}\)
will be useful.
Put
\(\displaystyle{I}={\int_{{0}}^{{1}}}{\frac{{{x}{\ln{{\left({1}+{x}\right)}}}}}{{{1}+{x}^{{2}}}}}{\left.{d}{x}\right.}\)
After integration by parts we have
\(\displaystyle{I}={\frac{{{1}}}{{{2}}}}{{\ln}^{{2}}{2}}-{\frac{{{1}}}{{{2}}}}{I}_{{1}}\) (2)
where
\(\displaystyle{I}_{{1}}={\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}+{x}^{{2}}\right)}}}}}{{{1}+{x}}}}{\left.{d}{x}\right.}\)
To evaluate \(\displaystyle{I}_{{1}}\) we integrate \(\displaystyle{\frac{{{\log{{\left({1}+{z}^{{2}}\right)}}}}}{{{1}+{z}}}}\) over \(\displaystyle\gamma_{{1}}+\gamma_{{2}}\), where \(\displaystyle\gamma_{{1}}\) is the imaginary axis from 0 to i and \(\displaystyle\gamma_{{2}}\) the unit circle from i to 1 clockwise. With \(\displaystyle{\log{{z}}}\) we mean \(\displaystyle{\ln}{\left|{z}\right|}+{i}{a}{r}{g}\ {z}\), where \(\displaystyle-\pi{ < }{a}{r}{g}\ {z}{ < }\pi\)
Put
\(\displaystyle{I}_{{2}}=\int_{{\gamma_{{1}}}}{\frac{{{\log{{\left({1}+{z}^{{2}}\right)}}}}}{{{1}+{z}}}}{\left.{d}{z}\right.}={\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}-{t}^{{2}}\right)}}}}}{{{1}+{i}{t}}}}{i}{\left.{d}{t}\right.}\)
\(\displaystyle={\int_{{0}}^{{1}}}{\frac{{{\ln{{\left({1}-{t}^{{2}}\right)}}}}}{{{1}+{t}^{{2}}}}}{i}{\left({1}-{i}{t}\right)}{\left.{d}{t}\right.}\)
Since \(\displaystyle{I}_{{1}}\) is real we are only interested in the real part of \(\displaystyle{I}_{{2}}\). After the substitution \(\displaystyle{t}={\frac{{{1}}}{{{s}+{1}}}}\) we get
\(\displaystyle{I}_{{2}}={\int_{{0}}^{{1}}}{\frac{{{t}{\ln{{\left({1}-{t}^{{2}}\right)}}}}}{{{1}+{t}^{{2}}}}}{\left.{d}{t}\right.}={I}+{\int_{{0}}^{{1}}}{\frac{{{t}{\ln{{\left({1}-{t}\right)}}}}}{{{1}+{t}^{{2}}}}}{\left.{d}{t}\right.}\)
\(\displaystyle={I}+{\int_{{0}}^{\infty}}{\frac{{{\ln{{s}}}-{\ln{{\left({s}+{1}\right)}}}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{2}{s}+{2}\right)}}}}{d}{s}\)
\(\displaystyle={I}+{\int_{{0}}^{\infty}}{\frac{{{\ln{{s}}}}}{{{\left({s}+{1}\right)}{\left({s}^{{2}}+{2}{s}+{2}\right)}}}}{d}{s}-{\int_{{0}}^{\infty}}{\frac{{{\ln{{\left({s}+{1}\right)}}}}}{{{\left({s}+{1}\right)}{\left({\left({s}+{1}\right)}^{{2}}+{1}\right)}}}}{d}{s}\)
The first integral can be evaluated by integrating \(\displaystyle{\frac{{{{\log}^{{2}}{z}}}}{{{\left({z}+{1}\right)}{\left({z}^{{2}}+{2}{z}+{2}\right)}}}}\) along a keyhole contour. We get the value \(\displaystyle{\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}-{\frac{{\pi^{{2}}}}{{{32}}}}\) via residue calculus.
In the the second integral we substitute \(\displaystyle{s}+{1}\) to \(\displaystyle{\frac{{{1}}}{{\sqrt{{{u}}}}}}\). If we combine that with (1) we get
\(\displaystyle-{\frac{{{1}}}{{{4}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{u}}}}}{{{1}+{u}}}}{d}{u}={\frac{{\pi^{{2}}}}{{{48}}}}\)
Consequently
\(\displaystyle{I}_{{2}}={I}+{\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}-{\frac{{\pi^{{2}}}}{{{32}}}}-{\frac{{\pi^{{2}}}}{{{48}}}}={I}+{\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}-{\frac{{{5}\pi^{{2}}}}{{{96}}}}\) (3)
Now to the unit circle.
\(\displaystyle{I}_{{3}}=\int_{{\gamma_{{2}}}}{\frac{{{\log{{\left({1}+{z}^{{2}}\right)}}}}}{{{1}+{z}}}}{\left.{d}{z}\right.}\)
\(\displaystyle=-{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\frac{{{\log{{\left({1}+{e}^{{{i}{2}{t}}}\right\rbrace}}}{\left\lbrace{1}+{e}^{{{i}{t}}}\right\rbrace}{i}{e}^{{{i}{t}}}{\left.{d}{t}\right.}}}{}}\)
\(\displaystyle=-{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\frac{{{\ln{{\left({2}{\cos{{t}}}\right)}}}+{i}{t}}}{{{2}{\cos{{\left(\frac{{t}}{{2}}\right)}}}}}}{i}{\left({\cos{{\left(\frac{{t}}{{2}}\right)}}}+{i}{\sin{{\left(\frac{{t}}{{2}}\right)}}}\right)}{\left.{d}{t}\right.}\)
We extract the real part and finally use the substitution \(\displaystyle{s}={\cos{{t}}}\) and (1)
\(\displaystyle{I}_{{3}}={\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left({t}+{\tan{{\left(\frac{{t}}{{2}}\right)}}}{\ln{{\left({2}{\cos{{t}}}\right)}}}\right)}{\left.{d}{t}\right.}\)
\(\displaystyle={\frac{{\pi^{{2}}}}{{{16}}}}+{\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{2}}}}}{{{1}+{s}}}}{d}{s}+{\frac{{{1}}}{{{2}}}}{\int_{{0}}^{{1}}}{\frac{{{\ln{{s}}}}}{{{1}+{s}}}}{d}{s}\)
\(\displaystyle={\frac{{\pi^{{2}}}}{{{16}}}}+{\frac{{{1}}}{{{2}}}}{{\ln}^{{2}}{2}}-{\frac{{\pi^{{2}}}}{{{24}}}}\)
\(\displaystyle={\frac{{{1}}}{{{2}}}}{{\ln}^{{2}}{2}}+{\frac{{\pi^{{2}}}}{{{48}}}}\)
We substitute (3) and (4) in (2)
\(\displaystyle{I}={\frac{{{1}}}{{{2}}}}{{\ln}^{{2}}{2}}-{\frac{{{1}}}{{{2}}}}{\left({I}+{\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}-{\frac{{{5}\pi^{{2}}}}{{{96}}}}+{\frac{{{1}}}{{{2}}}}{{\ln}^{{2}}{2}}+{\frac{{\pi^{{2}}}}{{{48}}}}\right)}\)
\(\displaystyle={\frac{{{3}}}{{{16}}}}{{\ln}^{{2}}{2}}+{\frac{{\pi^{{2}}}}{{{64}}}}-{\frac{{{1}}}{{{2}}}}{I}\)
Finally
\(\displaystyle{I}={\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}+{\frac{{\pi^{{2}}}}{{{96}}}}\)
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nick1337
Answered 2022-01-11 Author has 9672 answers

An ''elementary'' solution.
\(J=\int_0^1\frac{x\ln(1+x)}{1+x^2}dx,\ K=\int_0^1\frac{x\ln(1-x)}{1+x^2}dx \\J+K=\int_0^1\frac{x\ln(1-x^2)}{1+x^2}dx \\=\frac{1}{2}\int_0^1\frac{\ln(\frac{2y}{1+y})}{1+y}dy \\=\frac{1}{2}\int_0^1\frac{\ln 2}{1+y}dy+\frac{1}{2}\int_0^1\frac{\ln y}{1+y}dy-\frac{1}{2}\int_0^1\frac{\ln(1+y)}{1+y} \\=\frac{1}{4}\ln^2 2-\frac{1}{24}\pi^2 \\K-J=\int_0^1\frac{x\ln(\frac{1-x}{1+x})}{1+x^2}dx \\=\int_0^1\frac{(1-y)\ln y}{(1+y)(1+y^2)}dy \\=\int_0^1(\frac{\ln y}{1+y}-\frac{y\ln y}{1+y^2})dy \\=-\frac{\pi^2}{12}-\int_0^1\frac{y\ln y}{1+y^2}dy \\=-\frac{\pi^2}{12}-\frac{1}{4}\int_0^1\frac{\ln z}{1+z}dz \\=-\frac{1}{16}\pi^2 \\\text{Therefore,} \\J=\frac{1}{2}((J+K)-(K-J)) \\=\frac{1}{2}(\frac{1}{4}\ln^2 2-\frac{1}{24}\pi^2+\frac{1}{16}\pi^2) \\=\frac{1}{8}\ln^2 2+\frac{1}{96}\pi^2\)

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