Integral $I=\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx$ Hi I am stuck on showing that
$$
\int_0^\infty \frac{\ln(1+x)\ln(1+x^{-2})}{x} dx=\pi G-\frac{3\zeta(3)}{8}
$$
where G is the Catalan constant and $\zeta(3)$ is the Riemann zeta function.  Explictly they are given by
$$
G=\beta(2)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2}, \ \zeta(3)=\sum_{n=1}^\infty \frac{1}{n^3}.
$$ 
I have tried using 
$$
\ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1} x^n}{n},
$$ 
but didn't get very far.  
 A: Incomplete answer:
$$
\begin{align*}
I&=\int^{\infty}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx\\
&=\int^{1}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx+\int^{\infty}_{1}\frac{\log(1+x)\log(1+x^{-2})}{x}dx\\
&=\int^{1}_{0}\frac{\log(1+x)\log(1+x^{-2})}{x}dx+\int^{1}_{0}\frac{\log(1+x^{-1})\log(1+x^2)}{x}dx\\
&=2\underbrace{\int^{1}_{0}\frac{\log(1+x)\log(1+x^2)}{x}dx}_{=I_1}-\underbrace{\int^{1}_{0}\frac{\log(x)(2\log(1+x)+\log(1+x^2))}{x}dx}_{=I_2}.\\
\end{align*}
$$
$$
\begin{align*}
I_2&=\int^{1}_{0}\frac{\log(x)(2\log(1+x)+\log(1+x^2))}{x}dx\\
&=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}\int^{1}_{0}\frac{\log(x)(2x^n+x^{2n})}{x}dx\\
&=\sum^{\infty}_{n=1}\frac{(-1)^{n+1}}{n}\left(-\frac{2}{n^2}-\frac{1}{(2n)^2}\right)\\
&=\frac{27}{16}\zeta(3).
\end{align*}
$$
$$
\begin{align*}
I_1&=\int^{1}_{0}\frac{\log(1+x)\log(1+x^2)}{x}dx\\
&=\sum^{\infty}_{m,n=1}\frac{(-1)^{m+n}}{mn}\int^{1}_{0}x^{2m+n-1}dx\\
&=\sum^{\infty}_{m,n=1}\frac{(-1)^{m+n}}{mn(2m+n)}\\
&=\sum^{\infty}_{m=1}\frac{(-1)^{m}}{m}\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n(2m+n)}\\
&=\sum^{\infty}_{m=1}\frac{(-1)^{m}}{2m^2}\sum^{\infty}_{n=1}(-1)^{n}\left(\frac{1}{n}-\frac{1}{2m+n}\right)\\
&=\sum^{\infty}_{m=1}\frac{(-1)^{m}}{2m^2}\left(\sum^{\infty}_{n=1}\frac{(-1)^{n}}{n}-\sum^{\infty}_{n=1}\frac{(-1)^{2m+n}}{2m+n}\right)\\
&=\sum^{\infty}_{m=1}\frac{(-1)^{m}}{2m^2}\sum^{2m}_{n=1}\frac{(-1)^{n}}{n}\\
&=\sum^{\infty}_{m=1}\frac{(-1)^{m-1}(H_{2m}-H_{m})}{2m^2}\\
&=\sum^{\infty}_{m=1}\frac{H_{2m}-H_{m}}{2m^2}-2\sum^{\infty}_{m=1}\frac{H_{4m}-H_{2m}}{2(2m)^2}\\
&=\frac14\sum^{\infty}_{m=1}\frac{-H_{4m}+3H_{2m}-2H_{m}}{m^2}\\
&=-\frac14\sum^{\infty}_{m=1}\frac{H_{4m}}{m^2}+\frac{17}{16}\zeta(3)
\end{align*}$$
Therefore, $I=2I_1-I_2=-\frac12\sum^{\infty}_{m=1}\frac{H_{4m}}{m^2}+\frac{61}{16}\zeta(3)$.
It remains to prove that $\sum^{\infty}_{m=1}\frac{H_{4m}}{m^2}\stackrel?=\frac{67}{8}\zeta(3)-2\pi G$, but I haven't found a way yet.
A: The infinite sum in Chen Wang's answer, that is, $ \displaystyle \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}}$, can be evaluated using contour integration by considering the function $$f(z) = \frac{\pi \cot(\pi z) [\gamma + \psi(-4z)]}{z^{2}}, $$
where $\psi(z)$ is the digamma function and $\gamma$ is the Euler-Mascheroni constant.
The function $f(z)$ has poles of order $2$ at the positive integers, simple poles at the negative integers, simple poles at the positive quarter-integers, and a pole of order $4$ at the origin.
The function $\psi(-4z)$ does have simple poles at the positive half-integers, but they are cancelled by the zeros of $\cot( \pi z)$.
Now consider a square on the complex plane (call it $C_{N}$) with vertices at $\pm (N + \frac{1}{2}) \pm i (N +\frac{1}{2})$. 
On the sides of the square, $\cot (\pi z)$ is uniformly bounded.
And when $z$ is large in magnitude and not on the positive real axis, $\psi(-4z) \sim \ln(-4z)$.
So $ \displaystyle \int_{C_{N}} f(z) \ dz $ vanishes as $N \to \infty$ through the positive integers.
Therefore, 
$$\sum_{n=1}^{\infty} \text{Res} [f(z), n] + \sum_{n=1}^{\infty} \text{Res}[f(z),-n] + \text{Res}[f(z),0] + \sum_{n=0}^{\infty} \text{Res}\Big[f(z), \frac{2n+1}{4} \Big] =0 .$$
To determine the residues, we need the following Laurent expansions.
At the positive integers,
$$ \gamma + \psi (-4z) = \frac{1}{4} \frac{1}{z-n} + H_{4n} + \mathcal{O}(z-n) $$
and 
$$ \pi \cot (\pi z) = \frac{1}{z-n} + \mathcal{O}(z-n) .$$
At the origin,
$$ \gamma+ \psi(-4z) = \frac{1}{4z} -4 \zeta(2) z -16 \zeta(3) z^{2} + \mathcal{O}(z^{3})$$
and 
$$ \pi \cot (\pi z) = \frac{1}{z} - 2 \zeta(2) z + \mathcal{O}(z^{3}) .$$
And at the positive quarter-integers,
$$ \gamma + \psi(-4z) = \frac{1}{4} \frac{1}{z-\frac{2n+1}{4}} + \mathcal{O}(1)$$
and
$$ \pi \cot (\pi z) = (-1)^{n} \pi + \mathcal{O}\Big(z- \frac{2n+1}{4} \Big) .$$
Then at the positive integers,
$$f(z) = \frac{1}{z^{2}} \Big( \frac{1}{4} \frac{1}{(z-n)^{2}} + \frac{H_{4n}}{z-n} + \mathcal{O}(1) \Big), $$
which implies
$$\begin{align} \text{Res} [f(z),n] &= \text{Res} \Big[ \frac{1}{4z^{2}} \frac{1}{(z-n)^{2}} , n \Big] + \text{Res} \Big[ \frac{1}{z^{2}} \frac{H_{4n}}{z-n}, n \Big] \\ &= - \frac{1}{2n^{3}} + \frac{H_{4n}}{n^{2}} .\end{align}$$
At the negative integers,
$$ \text{Res}[f(z),-n] = \frac{\gamma + \psi(4n)}{n^{2}} = \frac{H_{4n-1}}{n^{2}} = \frac{H_{4n}}{n^{2}} - \frac{1}{4n^{3}} . $$
At the origin,
$$ f(z) = \frac{1}{z^{2}} \Big( \frac{1}{4z^{2}} - \frac{\zeta(2)}{2} - 4 \zeta(2) - 16 \zeta(3) z + \mathcal{O}(z^{2}) \Big),$$
which implies 
$$\text{Res}[f(z),0] = -16 \zeta(3) .$$
And at the positive quarter-integers,
$$ f(z) = \frac{\pi}{4z^{2}} \frac{(-1)^{n}}{z- \frac{2n+1}{4}} + \mathcal{O}(1),$$
which implies
$$ \begin{align} \text{Res} \Big[ f(z),\frac{2n+1}{4} \Big] &= \text{Res} \Big[\frac{\pi}{4z^{2}} \frac{(-1)^n}{z- \frac{2n+1}{4}}, \frac{2n+1}{4} \Big] \\ &= 4 \pi \ \frac{(-1)^{n}}{(2n+1)^{2}} . \end{align} $$
Putting everything together, we have
$$ - \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^{3}} + 2 \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} - \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{3}} - 16 \zeta(3) + 4 \pi \sum_{n=0}^{\infty} \frac{(-1)^{n}}{(2n+1)^{2}} $$
$$ = - \frac{1}{2} \zeta(3) + 2 \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} - \frac{1}{4} \zeta(3) - 16 \zeta(3) + 4 \pi G = 0 .$$
Therefore,
$$ \sum_{n=1}^{\infty} \frac{H_{4n}}{n^{2}} = \frac{67}{8} \zeta(3) - 2 \pi G .$$
EDIT:
I found the Laurent expansion of $\psi(-4z)$ at the positive integers by using the functional equation of the digamma function to express $\psi(4z)$ as 
$$ \psi(4z) = \psi(4z+4n+1) - \frac{1}{4z+4n} - \frac{1}{4z+4n-1} - \ldots - \frac{1}{4z} .$$ 
Then I evaluated the limit $$\lim_{z \to -n} (z+n) \psi(4z) = - \frac{1}{4}$$ and the limit $$\lim_{z \to -n} \Big(\psi(4z) + \frac{1}{4} \frac{1}{z+n} \Big) = - \gamma +H_{4n} .$$ 
This leads to the expansion $$\gamma + \psi (-4z) = \frac{1}{4} \frac{1}{z-n} + H_{4n} + \mathcal{O}(z-n) .$$
I did something similar to find the expansion at the positive quarter-integers.
A: I have a simple solution for this problem. Let
$$ I(\alpha,\beta)=\int_0^\infty\frac{\ln(1+\alpha x)\ln(1+\beta x^{-2})}{x}dx. $$
Then $I(0,0)=I(\alpha,0)=I(0,\beta)=0$ and $I(1,1)=I$. It is easy to check
\begin{eqnarray*}
\frac{\partial^2 I}{\partial \alpha\partial\beta}&=&\int_0^\infty\frac{1}{(1+\alpha x)(x^2+\beta)}dx\\
&=&\frac{1}{2}\left(\frac{\pi}{\sqrt{\beta}}+2\alpha\ln\alpha+\alpha\ln\beta\right)\frac{1}{1+\alpha^2\beta}\\
&=&\frac{1}{2}\left(\frac{\pi}{\sqrt{\beta}}+2\alpha\ln\alpha+\alpha\ln\beta\right)\sum_{n=0}^\infty(-1)^n\alpha^{2n}\beta^n\\
&=&\frac{1}{2}\left(\pi\sum_{n=0}^\infty(-1)^n\alpha^{2n}\beta^{n-\frac{1}{2}}+2\sum_{n=0}^\infty(-1)^n(\alpha^{2n+1}\ln\alpha)\beta^n+\sum_{n=0}^\infty(-1)^n\alpha^{2n+1}\beta^n\ln\beta\right).
\end{eqnarray*}
Thus
\begin{eqnarray*}
I&=&\int_0^1\int_0^1\frac{\partial^2 I}{\partial \alpha\partial\beta}d\beta d\alpha\\
&=&\frac{1}{2}\int_0^1\left(\pi\sum_{n=0}^\infty\frac{(-1)^n}{n+\frac{1}{2}}\alpha^{2n}+2\sum_{n=0}^\infty\frac{(-1)^n}{n+1}\alpha^{2n+1}\ln\alpha-\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^2}\alpha^{2n+1}\right)d\alpha\\
&=&\pi\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}-\sum_{n=0}^\infty\frac{(-1)^n}{4(n+1)^3}-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n}{2(n+1)^3}\\
&=&\pi G-\frac{1}{2}\sum_{n=0}^\infty\frac{(-1)^n}{(n+1)^3}\\
&=&\pi G-\frac{3\zeta(3)}{8}.
\end{eqnarray*}
Here we use
$$\int_0^1 x^n\ln x dx=-\frac{1}{(n+1)^2}. $$
A: $$I=\int_0^\infty\frac{\ln(1+x)\ln(1+x^{-2})}{x}\ dx=\int_0^\infty f(x)\ dx=\int_0^1 f(x)\ dx+\underbrace{\int_1^\infty f(x)\ dx}_{x\to1/x}$$
$$=2\underbrace{\int_0^1\frac{\ln(1+x)\ln(1+x^2)}{x}\ dx}_{\large I_1}-\frac94\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{x}\ dx}_{-3/4\zeta(3)}$$
$$I_1=\sum_{m=1}^\infty\frac{(-1)^{m-1}}{m}\int_0^1 x^{2m-1}\ln(1+x)\ dx$$
$$=\sum^{\infty}_{m=1}\frac{(-1)^{m-1}(H_{2m}-H_{m})}{2m^2}=\frac12\sum_{m=1}^\infty \frac{(-1)^mH_m}{m^2}-\frac12\sum_{m=1}^\infty \frac{(-1)^mH_{2m}}{m^2}$$
The first sum $\sum_{m=1}^\infty \frac{(-1)^mH_m}{m^2}=-\frac58\zeta(3)$ is well known and the second one :
$$\sum_{m=1}^\infty \frac{(-1)^mH_{2m}}{m^2}=4\sum_{m=1}^\infty \frac{(-1)^mH_{2m}}{(2m)^2}=4\Re\sum_{m=1}^\infty \frac{(i)^mH_{m}}{m^2}$$
By using the generating function
$$\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)$$ and setting $x=i$ we get 
$$\sum_{m=1}^\infty \frac{(-1)^mH_{2m}}{m^2}=4\Re\left\{\operatorname{Li}_3(i)-\operatorname{Li}_3(1-i)+\ln(1-i)\operatorname{Li}_2(1-i)+\frac12\ln(i)\ln^2(1-i)+\zeta(3)\right\}\\=\frac{23}{16}\zeta(3)-\pi\ G $$
Collect all results we get $\displaystyle  I=\pi G-\frac{3\zeta(3)}{8}$
