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I'm looking for a function $f_p(x)$ with real parameter $p>0$ satisfying $$ \int_0^\infty f_p(x)x^{s-1}dx=e^{-p\psi(s)} $$ where $\psi(s)$ is the usual digamma function. The inverse Mellin formula is $$ f_p(x)=\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty}x^{-s}e^{-p\psi(s)}ds $$ Normally we would try the residue theorem; this is possible but there are essential singularities and it is not obvious (to me) how to find the residues. The preprint www.arxiv.org/abs/1312.1604 gives asymptotic expansions for the exponential of the digamma function that may be useful.

Now cross-posted to math.stackexchange

Edit: I think $f_p(x)$ is supported on the unit interval; note that the first integral is positive and does not grow exponentially with $s$. Expanding in the relevant Chebyshev polynomials, ie

$ f_p(x)=\sum_{n=0}^\infty F_{n,p}\frac{T_{2n}(\sqrt{x})}{\sqrt{x(1-x)}} $

and noting that the digamma at integer values is just the harmonic numbers, $\psi(n)=H_{n-1}-\gamma$, we have

$ e^{-p(H_n-\gamma)}=\sum_{m=0}^\infty F_{m,p}\int\frac{T_{2m}(\sqrt{x})}{\sqrt{x(1-x)}}x^ndx=\sum_{m=0}^n\frac{\pi F_{m,p}}{2^{2n-1}} \frac{n (2n-1)!}{(n+m)!(n-m)!} $

where $n(2n-1)!$ is taken to be unity when $n=0$. The integral was done on mathematica by choosing values of $n$ and observing a pattern (would be good to have a reference/proof). This gives an explicit set of equations to find the $F_{m,p}$. These can then be used to reconstitute $f_p(x)$ numerically, using 200 terms in the expansion. The plot shows $f_p(x)$ for $p\in\{1/3,2/3,\ldots,4\}$ from top to bottom near $x=1$, ie behaviour roughly as $c(1-x)^{p-1}$. Convergence starts to break down at both ends of this range of $p$ (the observed oscillations).

Numerical plot of $f_p(x)$

But the question remains as to whether $f_p(x)$ can be expressed explicitly in terms of known functions.

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  • $\begingroup$ I have a naive idea, but I got stuck with it. I am posting it as a comment in case someone can use it or show it won't work. First, find the inverse Mellin transform of $\mbox{exp}(m_k(-1)^ks^k)$ for each $k>0$ and any $m_k\in\mathbb{Q}$. Use expansion $\psi(s+1) = -\gamma- \sum_{k=1}^{\infty}\zeta(k+1)(-s)^k$ and multiplicative convolution for Mellin transform to arrive at a product $\mbox{exp}(-\gamma)\prod_{k=1}^{\infty}\mbox{exp}(-\zeta(k+1)(-z)^k) = \mbox{exp}(\psi(s+1))$. For example, the Mellin transform of $\mbox{exp}(-\pi^2/6)\delta(t-\mbox{exp}(-\pi^2/6))$ is $\exp(-(\pi^2/6)s)$. $\endgroup$
    – Andrew Stout
    Oct 23, 2015 at 18:55
  • $\begingroup$ I mean $m_k$ in $\mathbb{R}$. $\endgroup$
    – Andrew Stout
    Oct 23, 2015 at 19:02
  • $\begingroup$ and also the signs do not match up: $\mathcal{M}\{\exp(\pi^2/6)\delta(t-\exp(\pi^2/6))\}(s) = \exp((\pi^2/6)s)$ the first term in the product expansion. I do not know if it is possible to do this for $k> 1$ and then perhaps also this wouldn't give you the answer you want even if it does work. $\endgroup$
    – Andrew Stout
    Oct 23, 2015 at 19:11
  • $\begingroup$ @Andrew Thanks for your thoughts. If you can get something that is at least formally correct (I have doubts about using that $\psi(s+1)$ expansion outside its domain of convergence), it is probably detailed enough for an (incomplete) answer. $\endgroup$
    – user25199
    Oct 27, 2015 at 9:01
  • $\begingroup$ This likely requires the inverse of the digamma function $\endgroup$
    – Tyma Gaidash
    Dec 14, 2023 at 21:43

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