Inversion of a Class of Singular Integral Operators on Entire Functions

Abstract

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 $\mathfrak{M}:{\mathcal{H}}_0\to {\mathcal{H}}_0$ defined by $$\mathfrak{M}f(z)={\int }_0^1{e}^{-z{t}^{-\nu }(1-(1-x)t)}f\left(z{t}^{-\nu }(1-t)\right)\frac{\mathrm{d}t}{t},z\in \mathbb{C},$$ for any $f\in {\mathcal{H}}_0$. 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 ${\mathcal{H}}_0$ has the integral representation $${\mathfrak{L}}^{-1}g(z)=\frac{1-x}{2i\pi x}{e}^z{\int }_1^{(0+)}\frac{e^{-xtz}}{t(t-1)}g\left(z(-t{)}^{\nu }(1-t{)}^{1-\nu }\right)\mathrm{d}t,z\in \mathbb{C},$$ for any $g\in {\mathcal{H}}_0$, 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.

16 pages. arXiv admin note: substantial text overlap with arXiv:1909.09694

References in corpus (1)

• Sojourn time in a $M^{[X]}/M/1$ Processor Sharing Queue with batch arrivals (II)