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

Given parameters $x\mathrm{\notin }{\mathbb{R}}^{-}\cup \{1\}$ and $\nu$, $\mathrm{R}\mathrm{e}(\nu )<0$, 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 $$\mathfrak{M}f(z)={\int }_0^1{e}^{-\frac{z}{x}{t}^{-\nu }(1-(1-x)t)}f(\frac{z}{x}{t}^{-\nu }(1-t))\frac{\mathrm{d}t}{t},z\in \mathbb{C},$$ for all $f \in \mathscr{H}0$\mathrm{.}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 $T = A(x,\nu) \cdot S \Leftrightarrow S = B(x,\nu) \cdot T$betweensequences$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 $$\{\begin{array}{l}{A}_{n,k}(x,\nu )={\displaystyle (-1{)}^k\left(\genfrac{}{}{0px}{}{n}{k}\right)F(k-n,-n\nu ;-n;x),B_{n,k}(x,\nu )={\displaystyle (-1{)}^k\left(\genfrac{}{}{0px}{}{n}{k}\right)F(k-n,k\nu ;k;x)}}\end{array}$$ 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 $${\mathfrak{L}}^{-1}f(z)=\frac{1-x}{2i\pi x}{e}^z{\int }_{(0+)}^1\frac{e^{-xtz}}{t(1-t)}f(xz(-t{)}^{\nu }(1-t{)}^{1-\nu })\mathrm{d}t,z\in \mathbb{C},$$ for the inverse ${\mathfrak{L}}^{-1}$ of operator $\mathfrak{L}$ on ${\mathcal{H}}_0$, where the integration contour encircles the point 0.