Inversion of a Class of Singular Integral Operators on Entire Functions

Given constants $x,\nu \in \mathbb{C}$ and the space ${\mathcal{H}}_0$ of entire functions in $\mathbb{C}$ vanishing at $0$, we consider the integro-differential operator $$\mathfrak{L}=\left(\frac{x\nu (1-\nu )}{1-x}\right)\delta \circ \mathfrak{M},$$ with $\delta = z , $\mathrm{d}$/$\mathrm{d}$z$and$\mathfrak{M}:\mathscr{H}_0 \rightarrow \mathscr{H}_0$definedby$$ \mathfrak{M}f(z) = \int_0^1 e^{-z t^{-\nu}(1-(1-x)t)} , f \left (z , t^{-\nu}(1-t) \right ) , \frac{$\mathrm{d}$t}{t}, \qquad z \in $\mathbb{C}$, for any $f \in \mathscr{H}_0$. Operator $\mathfrak{L}$ originates from an inversion problem in Queuing Theory. Bringing the inversion of $\mathfrak{L}$ back to that of $\mathfrak{M}$ translates into a singular Volterra integral equation, but with no explicit kernel. In this paper, the inverse of operator $\mathfrak{L}$ is derived through a new inversion formula recently obtained for infinite matrices with entries involving Hypergeometric polynomials. For $x \notin \mathbb{R}^- \cup \{1\}$ and $\mathrm{Re}(\nu) < 0$, we then show that the inverse $\mathfrak{L}^{-1}$ of $\mathfrak{L}$ on $\mathscr{H}_0$ has the integral representation \mathfrak{L}^{-1}g(z) = \frac{1-x}{2i\pi x} , e^{z} \int_1^{(0+)} \frac{e^{-xtz}}{t(t-1)} , g \left (z , (-t)^{\nu}(1-t)^{1-\nu} \right ) , $\mathrm{d}$t, \qquad z \in $\mathbb{C}$, $$forany$g \in \mathscr{H}_0$, where the bounded integration contour in the complex plane starts at point 1 and encircles the point 0 in the positive sense. Other related integral representations of ${\mathfrak{L}}^{-1}$ are also provided.