2021/01/04 by Ridha Nasri, Nasri, Ridha, Alain Simonian +3
Mathematics · #Algebraic and Geometric Analysis #Classical Analysis and ODEs (math.CA) #Complex Variables (math.CV) #Differential Equations and Boundary Problems #FOS: Mathematics #Functional Analysis (math.FA) #Mathematical functions and polynomials #math.CA #math.CV #math.FA
paper · pdf · doi:10.48550/arxiv.2101.00831
16 pages. arXiv admin note: substantial text overlap with arXiv:1909.09694
arxiv created 2021/01/04 · openalex publication_date 2021/01/04 · arxiv updated 2021/01/05 · openalex created_date 2025/10/10 · openalex updated_date 2026/07/28
Given constants x, ν∈ ℂ and the space \mathscrH0 of entire functions in ℂ vanishing at 0, we consider the integro-differential operator \mathfrakL = ( (x ν(1-ν))/(1-x) ) δ∘ \mathfrakM , with δ= z d/dz and \mathfrakM:\mathscrH0 → \mathscrH0 defined by \mathfrakMf(z) = ∫01 e^-z t-ν(1-(1-x)t) f (z t-ν(1-t) ) (dt)/(t), z ∈ ℂ, for any f ∈ \mathscrH0. Operator \mathfrakL originates from an inversion problem in Queuing Theory. Bringing the inversion of \mathfrakL back to that of \mathfrakM translates into a singular Volterra integral equation, but with no explicit kernel. In this paper, the inverse of operator \mathfrakL is derived through a new inversion formula recently obtained for infinite matrices with entries involving Hypergeometric polynomials. For x ∉ ℝ- ∪ \1\ and Re(ν) < 0, we then show that the inverse \mathfrakL-1 of \mathfrakL on \mathscrH0 has the integral representation \mathfrakL-1g(z) = (1-x)/(2iπx) ez ∫1(0+) \frace-xtzt(t-1) g (z (-t)ν(1-t)1-ν ) dt, z ∈ ℂ, for any g ∈ \mathscrH0, 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 \mathfrakL-1 are also provided.