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

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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Foreckije
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)
$$\displaystyle{f{{\left({1}\right)}}}={\frac{{{{\ln}^{{2}}{\left({2}\right)}}}}{{{8}}}}+{\frac{{\pi^{{2}}}}{{{96}}}}$$
###### Not exactly what you’re looking for?
Serita Dewitt
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
$$\displaystyle{I}={\frac{{{1}}}{{{8}}}}{{\ln}^{{2}}{2}}+{\frac{{\pi^{{2}}}}{{{96}}}}$$
nick1337

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