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

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 $\mathfrak{L} = c0 \cdot \delta \circ \mathfrak{M}$ with constant $c0 = \nu(1-\nu)x/(1-x)$, $\delta = z \, \mathrm{d}/\mathrm{d}z$ and integral operator $\mathfrak{M}$ defined by $$ \mathfrak{M}f(z) = \int0¹ e^{- \frac{z}{x}t^{{-\nu}(1-(1-x)t)}} \, f \left ( \frac{z}{x} \, t^{-\nu}(1-t) \right ) \, \frac{\mathrm{d}t}{t}, \qquad z \in \mathbb{C}, $$ 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 $S = (Sn){n \in \mathbb{N}^*}$ and $T = (Tn){n \in \mathbb{N}^*}$, 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.