An inversion formula with hypergeometric polynomials and application to singular integral operators

TL;DR

This paper develops a new inversion formula involving hypergeometric polynomials to invert certain singular integral operators, providing explicit integral representations and functional relations for sequences and operators.

Contribution

It introduces a novel class of linear inversion formulas using hypergeometric polynomials and applies them to invert singular integral operators related to Volterra equations.

Findings

Derived explicit inverse formulas involving hypergeometric polynomials

Established integral representations for the inverse operators

Connected sequence relations with generating functions

Abstract

Given parameters $x \notin \mathbb{R}^- \cup \{1\}$ and $\nu$, $\mathrm{Re}(\nu) < 0$, and the space $\mathscr{H}_0$ of entire functions in $\mathbb{C}$ vanishing at $0$, we consider the family of operators $\mathfrak{L} = c_0 \cdot \delta \circ \mathfrak{M}$ with constant $c_0 = \nu(1-\nu)x/(1-x)$, $\delta = z \, \mathrm{d}/\mathrm{d}z$ and integral operator $\mathfrak{M}$ defined by $$ \mathfrak{M}f(z) = \int_0^1 e^{- \frac{z}{x}t^{-\nu}(1-(1-x)t)} \, f \left ( \frac{z}{x} \, t^{-\nu}(1-t) \right ) \, \frac{\mathrm{d}t}{t}, \qquad z \in \mathbb{C}, $$ for all $f \in \mathscr{H}_0$. Inverting $\mathfrak{L}$ or $\mathfrak{M}$ proves equivalent to solve a singular Volterra equation of the first kind. The inversion of operator $\mathfrak{L}$ on $\mathscr{H}_0$ leads us to derive a new class of linear inversion formulas $T = A(x,\nu) \cdot S \Leftrightarrow S = B(x,\nu) \cdot T$ between sequences $S = (S_n)_{n \in \mathbb{N}^*}$ and $T = (T_n)_{n \in \mathbb{N}^*}$, where the infinite lower-triangular matrix $A(x,\nu)$ and its inverse $B(x,\nu)$ involve Hypergeometric polynomials $F(\cdot)$, namely $$ \left\{ \begin{array}{ll} A_{n,k}(x,\nu) = \displaystyle (-1)^k\binom{n}{k}F(k-n,-n\nu;-n;x), B_{n,k}(x,\nu) = \displaystyle (-1)^k\binom{n}{k}F(k-n,k\nu;k;x) \end{array} \right. $$ for $1 \leqslant k \leqslant n$. Functional relations between the ordinary (resp. exponential) generating functions of the related sequences $S$ and $T$ are also given. These relations finally enable us to derive the integral representation $$ \mathfrak{L}^{-1}f(z) = \frac{1-x}{2i\pi x} \, e^{z} \int_{(0+)}^1 \frac{e^{-xtz}}{t(1-t)} \, f \left ( xz \, (-t)^{\nu}(1-t)^{1-\nu} \right ) \, \mathrm{d}t, \quad z \in \mathbb{C}, $$ for the inverse $\mathfrak{L}^{-1}$ of operator $\mathfrak{L}$ on $\mathscr{H}_0$, where the integration contour encircles the point 0.