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From Umbral Hyperbolic Integrals to a Cotangent Coefficient Formula for the Mittag Leffler Polynomials

Sep 2026 · 0 citations · 23 references
Mathematics

Abstract

Let \(g_n(x)\) be the Mittag--Leffler polynomials defined by $$ \sum_{n\ge0}g_n(x)t^n=\frac12\left(\frac{1+t}{1-t}\right)^x. $$ We derive the coefficient formula $$ g_n(x)=\frac1n\sum_{j=0}^{\lfloor (n-1)/2\rfloor} (-1)^j\frac{2^{n-2j-1}}{(n-2j-1)!} [u^{2j}](u\cot u)^n\,x^{n-2j}, $$ equivalently $$ [u^{n-r}](u\cot u)^n =(-1)^{(n-r)/2}\frac{n(r-1)!}{2^{r-1}}[x^r]g_n(x). $$ The identity is suggested by comparing an umbral representation of the hyperbolic tangent integral $$ K(a,b)=\int_0^\infty \frac{\tanh^a x}{x^b}\,dx $$ with a compact cotangent-coefficient formula, but is proved independently from the generating function by formal Lagrange--B\"urmann inversion. Substitution of the proved bridge back into the integral formula then yields a rigorous derivation of the umbral Mittag--Leffler representation. We also develop a Dirichlet--beta analogue for $$ I_{m,n}^{\beta}=\int_0^\infty\frac{\tanh^m x}{x^n\cosh x}\,dx. $$ Defining \(q_m(x)\) by $$ \sum_{m\ge0}q_m(x)t^m=(1-t^2)^{-1/2}\exp(2x\operatorname{artanh}t), $$ we prove $$ [u^{m-r}]\frac{u}{\sin u}(u\cot u)^m =(-1)^{(m-r)/2}\frac{r!}{2^r}[x^r]q_m(x). $$ This gives a parallel umbral representation for the beta integrals. The family \(q_m\) is a shifted \(c=-1,\beta=1\) specialization of the classical Meixner family, yielding a beta/Meixner counterpart to the zeta/Mittag--Leffler correspondence.

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