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

Abstract

Given parameters $x\notin {\mathbb{R}}^{-}\cup \{1\}$ and $\nu$, $\mathrm{Re}(\nu )<0$, and the space ${\mathcal{H}}_0$ 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 $$\mathfrak{M}f(z)={\int }_0^1{e}^{-\frac{z}{x}{t}^{-\nu }(1-(1-x)t)}f\left(\frac{z}{x}{t}^{-\nu }(1-t)\right)\frac{\mathrm{d}t}{t},z\in \mathbb{C},$$ for all $f\in {\mathcal{H}}_0$. 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 ${\mathcal{H}}_0$ leads us to derive a new class of linear inversion formulas $T=A(x,\nu )\cdot S\iff S=B(x,\nu )\cdot T$ between sequences $S=(S_n{)}_{n\in {\mathbb{N}}^{\ast }}$ and $T=(T_n{)}_{n\in {\mathbb{N}}^{\ast }}$, where the infinite lower-triangular matrix $A(x,\nu )$ and its inverse $B(x,\nu )$ involve Hypergeometric polynomials $F(\cdot )$, namely $$\{\begin{array}{l}{A}_{n,k}(x,\nu )=(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})F(k-n,-n\nu ;-n;x),B_{n,k}(x,\nu )=(-1{)}^k\left(\genfrac{}{}{0ex}{}{n}{k}\right)F(k-n,k\nu ;k;x)\end{array}$$ for $1⩽k⩽n$. 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 $${\mathfrak{L}}^{-1}f(z)=\frac{1-x}{2i\pi x}{e}^z{\int }_{(0+)}^1\frac{e^{-xtz}}{t(1-t)}f\left(xz(-t{)}^{\nu }(1-t{)}^{1-\nu }\right)\mathrm{d}t,z\in \mathbb{C},$$ for the inverse ${\mathfrak{L}}^{-1}$ of operator $\mathfrak{L}$ on ${\mathcal{H}}_0$, where the integration contour encircles the point 0.

29 pages, 2 figures. arXiv admin note: substantial text overlap with arXiv:1904.08283