TITLE: Compute integral of general form $ \int_0^\infty \left(\frac{x}{\sinh x}\right)^n d x $
QUESTION [11 upvotes]: I encountered the following integral and can't compute its value in general:
$$
I_n = \int_0^\infty \left(\frac{x}{\sinh x}\right)^n d x
$$
where $n\in \mathbb N$.
From integral tables/mathematica, I know that
$$I_1 = \frac{\pi^2}  4,\>\>\>\>\>I_2 = \frac{\pi^2}6$$
But, for general $n\in\mathbb N$, I don't know the result though. Any ideas?

REPLY [2 votes]: This is a remark continuing from J.G.'s hint and Q. Zhang's solution.
For even $n$ the coefficients $A_j^n$ of $\zeta(2k+2)$ relate to the "central factorial numbers" (OEIS A008955):
$$
2\begin{pmatrix}
I_2/2 \\
I_4/4 \\
I_6/6 \\
I_8/8 \\
I_{10}/10 \\
\vdots
\end{pmatrix}
=
\begin{pmatrix}
1&    \hfill0&    \hfill0&    \hfill0&    \hfill0 & \cdots \\
   1&   \hfill-1&   \hfill 0&  \hfill  0&  \hfill  0 \\
   1&   \hfill-5&   \hfill 4&  \hfill  0&  \hfill  0 \\
   1&  \hfill-14&   \hfill49&  \hfill-36&  \hfill  0 \\
   1&  -30&  273& -820&  576 \\
   \vdots &&&&&\ddots
\end{pmatrix}
\begin{pmatrix}
\zeta(2) \\
\zeta(4) \\
\zeta(6) \\
\zeta(8) \\
\zeta(10) \\
\vdots
\end{pmatrix}
$$
or generally $2I_{2m}/{(2m)} = \sum_{k=0}^{m-1} t(2m, 2m-2k)\zeta(2k+2)$ with $t(2n,2k)$ the even central factorial numbers (of the first kind), see OEIS A008955.
For odd $n$ such a relation holds as well, namely
$$
\frac{1}{2}
\begin{pmatrix}
I_1 /1 \\
I_3 /3 \\
I_5 /5 \\
I_7 /7 \\
I_9 /9 \\
\vdots
\end{pmatrix}
=
\begin{pmatrix}
1&    \hfill 0&    \hfill 0&   \hfill  0&   \hfill  0 & \cdots \\
     1&   \hfill -1&    \hfill 0&   \hfill  0&  \hfill   0 \\
     1&   \hfill-10&    \hfill 9&  \hfill   0&  \hfill   0 \\  
     1&   \hfill-35&  \hfill 259& \hfill -225&  \hfill   0 \\  
     1&   \hfill-84&  1974&-12916& 11025 \\
  \vdots &&&&&\ddots
\end{pmatrix}
\begin{pmatrix}
(1-2^{-2})\zeta(2) \\
(1-2^{-4})\zeta(4) \\
(1-2^{-6})\zeta(6) \\
(1-2^{-8})\zeta(8) \\
(1-2^{-10})\zeta(10) \\
\vdots
\end{pmatrix},
$$
i.e.,
$I_{2m+1}/(2m+1) = 2\sum_{k=0}^{m-1} 4^k t(2m+1,2m+1-2k)(1 - \frac{1}{2^{2k+2}})\zeta(2k+2)$ with $t(2n+1,2k+1)$ the odd central factorial numbers, see OEIS A008956.
This is of interest to me because $\sum_{n=1}^\infty I_n/n$ should exist and I would be interested if there's a closed form expression for it, as it relates to another integral I'm interested in, see this question.
To prove the relations above, one way is to note that Q. Zhang's coefficients can be expressed via Newton's identities as elementary symmetric polynomials. If $e_j^n$ is the $j$th elementary symmetric polynomial of degree $n$ evaluated at $1^2,\ldots,n^2$, the recursion $e_j^n = e_j^{n-1} + n^2 e_{j-1}^{n-1}$ holds. Comparing to the recursion formulae for the central factorial numbers (i.e., in OEIS or in Butzer et al.), the claim follows for even $n$. The case for odd $n$ can be shown along similar lines.