real analysis - 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.

Answer

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.