I cannot prove that:
$$ \int_0^1 \! \frac{x}{\operatorname{artanh} x} \, {\rm d}x = \frac{7 \, \zeta(3)}{\pi^2} $$
where $\operatorname{artanh}$ is the inverse hyperbolic tangent function, and $\zeta$ is the Rieman zeta function. This integral is equivalent to the following integrals:
$$ \int_1^\infty \! \frac{{\rm d}x}{x^3 \operatorname{arcosh}x} ,\quad \int_0^1 \! \frac{x}{\operatorname{arsech}x} \, {\rm d}x $$
Also, increasing the order of the integrand yields:
$$ \begin{align} \int_0^1 \! \frac{x^2}{\operatorname{artanh}^2 x} \, {\rm d}x &= \frac{124 \, \zeta(5)}{\pi^4} – \frac{14 \, \zeta(3)}{3\pi^2} \\[3pt] \int_0^1 \! \frac{x^3}{\operatorname{artanh}^3 x} \, {\rm d}x &= \frac{1905 \, \zeta(7)}{\pi^6} – \frac{124 \, \zeta(5)}{\pi^4} \\[3pt] \int_0^1 \! \frac{x^4}{\operatorname{artanh}^4 x} \, {\rm d}x &= \frac{28616 \, \zeta(9)}{\pi^8} – \frac{2540 \, \zeta(7)}{\pi^6} + \frac{124 \, \zeta(5)}{5\pi^4} \end{align} $$
Sorry for my bad English.
Best Answer
$\newcommand{\d}{\,\mathrm{d}}\newcommand{\artanh}{\operatorname{artanh}}$We have: $$I=\int_0^1\frac{x}{\artanh x}\d x=\int_0^\infty\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{1}{2}\int_\Bbb R\frac{\sinh t}{t\cdot\cosh^3t}\d t$$
It suffices to show: $$\int_{\Bbb R}\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{14\zeta(3)}{\pi^2}$$
Which is amenable to contour integration. By multiplying with a regularising term $e^{i\cdot\delta z}$ and integrating around a box contour, the box part vanishes and we just collect the residues. By taking $\delta\to0^+$ we recover the original integral.
What are the poles? These will be whenever $\cosh z=0$ so they occur at $z=\frac{\pi i}{2}(2n+1)$ for $n\ge0$. The poles are all of third order; that means we have our work cut out for us. This method will also generalise to the second, third, fourth, etc. order integrands, it will just get extremely tedious.
The Taylor expansion of $\cosh$ at $\zeta_n:=\frac{\pi i}{2}(2n+1)$ will be: $i\cdot(-1)^n((z-\zeta_n)+(1/3!)(z-\zeta_n)^3+\cdots)$ so the Taylor expansion of $\cosh^3$ at $\zeta_n$ is $i\cdot(-1)^{n+1}((z-\zeta_n)^3+(1/2)(z-\zeta_n)^5+\cdots)$. Taking reciprocals: $$\frac{1}{\cosh^3z}=\frac{i(-1)^n}{(z-\zeta_n)^3}+\frac{i(-1)^{n+1}}{2(z-\zeta_n)}+o(1)$$In a neighbourhood of $\zeta_n$. The Taylor expansion of $\sinh$ at $\zeta_n$ will be $i(-1)^n(1+(1/2)(z-\zeta_n)^2+(1/4!)(z-\zeta_n)^4+\cdots)$. The Taylor expansion of $\frac{1}{z}$ at $\zeta_n$ is $\zeta_n^{-1}-\zeta_n^{-2}(z-\zeta_n)+\zeta_n^{-3}(z-\zeta_n)^2+\cdots$.
By multiplying series it follows the residue of the integrand at $\zeta_n$ is: $$i(-1)^n\cdot i(-1)^n(1/2)\cdot\zeta_n^{-1}+i(-1)^n\cdot i(-1)^n\cdot\zeta_n^{-3}+(i(-1)^{n+1}/2)\cdot i(-1)^n\cdot\zeta_n^{-1}\\=-\zeta_n^{-3}=\frac{8}{i\cdot\pi^3(2n+1)^3}$$So that: $$I=2\pi i\sum_{n=0}^\infty\frac{8}{i\cdot\pi^3(2n+1)^3}=\frac{16}{\pi^2}\sum_{n=0}^\infty\frac{1}{(2n+1)^3}$$The sum over the reciprocal of all odd cubes is $\zeta(3)-2^{-3}\zeta(3)=\frac{7}{8}\zeta(3)$. So we should have: $$\int_\Bbb R\frac{\sinh t}{t\cdot\cosh^3t}\d t=\frac{16\cdot7}{8\pi^2}\zeta(3)=\frac{14}{\pi^2}\zeta(3)$$As expected.
Rigorously, when the modifier $e^{i\cdot\delta z}$ is present, this has Taylor series $e^{i\cdot\delta\zeta_n}(1+i\delta(z-\zeta_n)+\cdots)$ and contributes to the residues a factor which $\to1$ as $\delta\to0^+$. As long as one is happy exchanging the limits with the series, we can ignore this in the residue calculation.
About the box contour. For $N\in\Bbb N$ large consider the box contour with vertices $\pm2\pi N,\pm2\pi N+2\pi N\cdot i$. On the lowest horizontal strip we get an approximation to the desired integral. On the upper horizontal strip the imaginary parts are fixed to $2\pi N\cdot i$; for $z=x+2\pi N\cdot i$, $\cosh(z)=\cosh(x)\cosh(2\pi N\cdot i)+\sinh(x)\sinh(2\pi N\cdot i)=\cosh(x)$ and $\sinh(z)=\sinh(x)\cosh(2\pi N\cdot i)+\cosh(x)\sinh(2\pi N\cdot i)=\sinh(x)$. Therefore, with the modified integrand $f(z)=\frac{\sinh z}{z\cdot\cosh^3z}e^{i\cdot\delta z}$, $z\cdot f(z)$ is bounded in modulus by $e^{-2\pi N\cdot\delta}$ on the upper horizontal. Thus the integral on the upper horizontal vanishes.
On the vertical segments $\cosh(z)$ is comparable in magnitude to $\cosh(2\pi N)$ and similarly for the $\sinh$ term; it follows the integral on the verticals vanishes.