# How do I find a closed form of ${\pi^{2n}\over \zeta(2n)}\int_{-1}^{1}{x^{2n-2}\over \pi^2+(2\tanh^{-1}{x})^2}dx$?

How do I evaluate the closed form for $g(n)$?

Where n is an integer, $n\ge 1$

$${\pi^{2n}\over \zeta(2n)}\int_{-1}^{1}{x^{2n-2}\over \pi^2+(2\tanh^{-1}{x})^2}dx=g(n)$$

Make a subsititution $u=\tanh^{-1}{x}\rightarrow dx=sech^2{u}du$

$${\pi^{2n}\over \zeta(2n)}\int_{-\infty}^{\infty}{1\over sinh^2{u}}\cdot{\tanh^{2n}{u}\over \pi^2+4u^2}du=g(n)$$

$${\pi^{2n}\over \zeta(2n)}\int_{-\infty}^{\infty}{1\over sinh^2{u}}\cdot{\tanh^{2n}{u}\over \pi^2[1+\left({2u\over \pi}\right)^2]}du=g(n)$$

$${\pi^{2n-2}\over \zeta(2n)}\int_{-\infty}^{\infty}{1\over sinh^2{u}}\cdot{\tanh^{2n}{u}\over 1+\left({2u\over \pi}\right)^2}du=g(n)$$

Apply geometric series ${1\over 1+x}=\sum_{k=0}^{\infty}(-1)^kx^k$

$${\pi^{2n-2}\over \zeta(2n)}\cdot{\left({2\over \pi}\right)^{2k}}\sum_{k=0}^{\infty}(-1)^k\int_{-\infty}^{\infty}{1\over sinh^2{u}}\cdot{u^{2k}\tanh^{2n}{u}}du=g(n)$$

I am shrugged here how to evaluate this integral at this point, I would some help please.

The first few values for $n=1,2,3,4,...$ are $1,4,22,{428\over 3},...$ From the look at its trend, $g(n)$ only seem to yield rational values?

As for the odd powers, we get zero as a result.[checked through wolfram integrator]

$$\int_{-1}^{1}{x^{2n-1}\over \pi^2+(2\tanh^{-1}{x})^2}dx=0$$

• the hyperbolic variants look like they can be tackeld via residue theorem Commented Jun 23, 2016 at 19:18
• i doubt that there is a nice closed form for arbritary $n$ because the residues become incredibly ugly, but maybe i'm wrong....^^ Commented Jun 23, 2016 at 19:34
• It is convenient to write $\left[\tanh\left(x\right)\right]^{\,-1}$ or/and $1/\tanh\left(x\right)$ or/and ${1 \over \tanh\left(x\right)}$ because it can be confused with $\mathrm{arctanh}$. Otherwise, it's a nice question. Commented Jun 24, 2016 at 0:49
• Wikipedia used artanh(x) and wolfram math world used $\tanh^{-1}$ and here I am suggested to use arctanh. I can't find it on the net.
– user348832
Commented Jun 24, 2016 at 1:45
• On my scientific calculator it also uses $tanh^{-1}$
– user348832
Commented Jun 24, 2016 at 1:46

The integral $$I(n)=\int_{-\infty}^{+\infty}\frac{1}{\sinh^2 u}\cdot\frac{\tanh^{2n}u}{\pi^2+4u^2}\,du$$ should not be terribly difficult to evaluate through the residue theorem. In the upper half-plane there is a pole with high multiplicity at $u=\frac{\pi i}{2}$ and simple poles at $\frac{3\pi i}{2},\frac{5\pi i}{2},\frac{7\pi i }{2},\ldots$. Moreover, the integral representation for Gregory coefficients
$$G_n=(-1)^{n+1}\int_{0}^{+\infty}\frac{dx}{(1+x)^n(\pi^2+\log^2 x)},\qquad \frac{z}{\log(1+z)}=1+\sum_{n\geq 1}G_n z^n$$ should be useful. In the simplest case $n=1$ I got:
$$I(1) = 2\pi i\cdot\left(-\frac{i(3+\pi^2)}{12\pi^3}+\sum_{k\geq 1}\frac{(2k+1)i}{4k^2(k+1)^2\pi^3}\right)=\color{red}{\frac{1}{6}}$$
Anyway you are probably not interested in this integral. The substitution $x=\tanh(u)$ leads to: $$dx=\text{sech}^2(u)\,du = \frac{du}{\color{red}{\cosh}^2 u}$$ but luckily the integrals $$J(n)=\int_{-\infty}^{+\infty}\frac{1}{\cosh^2 u}\cdot\frac{\tanh^{2n}u}{\pi^2+4u^2}\,du$$ can be evaluated with the same technique. For instance, $J(1)=\color{red}{\frac{2}{45}}$.