Laplace Transform of Confluent Hypergeometric

Question

What would be the closed form of a Laplace transform for a hypergeometric of the form $$\int_0^\infty e^{-sx}{_1}F_1(a;b;x)dx$$ b>a>0 and s>1. Just curious what this ends up as. If I use an integral representation for ${_1}F_1$ it ends up a bit messy but is that the right way to proceed?

Answer 1

Using the series $$ {}_1F_1(a;b;x) = \sum_{k=0}^\infty \frac{\Gamma(a+k) \Gamma(b)}{\Gamma(a) \Gamma(b+k) k!} x^k$$

and absolute convergence to exchange sum and integral, your Laplace transform becomes

$$\eqalign{L(s) &= \int_0^\infty {}_1F_1(a;b;x) \exp(-sx)\; dx \cr &= \sum_{k=0}^\infty \frac{\Gamma(a+k)\Gamma(b) }{\Gamma(a)\Gamma(b+k) s^{k+1}}\cr &= \frac{ {}_2F_1(1,a; b; 1/s)}{s}} $$