Completely monotonic functions of the form \(s^{-b}(s^ 2+1)^{-a}\)
The author's aim is to determine the set of all \((a,b)\), in the first quadrant, for which the function \(f(s)\) of the title is completely monotonic (c.m.) on \((0,\infty)\), i.e. \((-1)^ nf^{(n)}(s)\geq 0\), \(s>0\), \(n=0,1,2,... \). This is equivalent to finding those a and b for which \(S_{a,b}(t)\), the inverse Laplace transform of \(f(s)\), is nonnegative on \((0,\infty)\). This problem has been discussed by a number of authors in connection with the positivity of hypergeometric functions and of integrals involving Bessel functions. Thus \textit{R. Askey} and \textit{H. Pollard} [SIAM J. Math. Anal. 5, 58-63 (1974; Zbl 0239.26010)] showed that f(s) is c.m. for \(b\geq a\geq 1\), while \textit{J. L. Fields} and \textit{M. E. Ismail} [ibid. 6, 551-559 (1975; Zbl 0301.33002)] showed it for \(b\geq 2a\). The present author uses a result of \textit{G. Gasper} [ibid. 6, 868- 881 (1975; Zbl 0313.33013)] to show that f(s) is completely monotonic for \(b\geq 1\) and \(0\leq a\leq 1\). He goes on to use an asymptotic formula for \(S_{a,b}(t)\) to show that f(s) is not c.m. for \(0<b<a\) or for \(0<a=b<1\). This leaves the question unanswered in the triangle with vertices (0,0), (1,1) and (1/2,1). Since if f(s) is c.m. for a particular b, it is also so for all larger b, there is a boundary curve within this triangle such that f(s) is c.m. on or above the curve and not c.m. below it. The author writes that numerical evidence indicates that this curve is increasing and concave down.
scientific article; zbMATH DE number 4057927