I am trying to learn about the properties of Hankel matrices, and they appear to have nice closed forms for quite a large class of sequences. The class I am interested is when the elements $a_n$ are given by

$a_n=\frac{\alpha n}{\sinh(\pi n \alpha)}$, for $\alpha$ some (very large) constant.

I understand that one place to start may be to try to solve the moment problem corresponding to this, that is, look for a function $f(x)$ satisfying $a_n=\int_a^b x^n f(x) dx$ on some interval $[a,b]$ (not necessarily finite). This to me looked like some form of inverse Mellin transform, but I cannot seem to solve it. (Since $\sinh(i \alpha n \tau)$ is just oscillating along a vertical line in the complex plane, the integral for the inverse transformation doesn't converge..?)

Another, maybe equivalent, way to proceed is to try to look for some basis in which the determinant can be diagonalised, which I guess amounts to finding the right kind of orthogonal polynomials.

Any help would be greatly appreciated! Thanks :)


The inverse Mellin transform is just the Laplace transform, and $$ \mathcal{L}\left(\frac{x}{\sinh x}\right) = \frac{1}{2}\,\psi'\left(\frac{s+1}{2}\right) $$ where $\psi'(z)=\frac{d^2}{dz^2}\log\Gamma(z)$ comes out by writing $\frac{x}{\sinh x}$ as a geometric series, performing termwise integration and exploiting $$ \sum_{n\geq 0}\frac{1}{(n+a)^2}=\psi'(a).$$

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    $\begingroup$ This to me is very confusing. Isn't the Mellin transform related to the (two-sided) Laplace transform, and hence the inverse should be related to the inverse Laplace transform, which is taken along a vertical contour in the complex plane, which would give me convergence issues..? $\endgroup$ – Lou May 27 '16 at 9:15

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