Integrating a Bessel function I'm looking for help to show that
$$\int_{0}^{\infty} e^{-at}J_{\nu}(bt)t^{\mu -1} dt $$
can be expressed in terms of the hypergeometric function, where $J_{\nu}$ is the Bessel function of $\nu$ order. 
 A: $\int_0^\infty e^{-at}J_{\nu}(bt)t^{\mu-1}~dt$
$=\int_0^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^nb^{2n+\nu}t^{2n+\mu+\nu-1}e^{-at}}{n!\Gamma(n+\nu+1)2^{2n+\nu}}~dt$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\Gamma(2n+\mu+\nu)b^{2n+\nu}}{n!\Gamma(n+\nu+1)2^{2n+\nu}a^{2n+\mu+\nu}}$
A: Take a Mellin transform of the integrand with respect to $b$
$$
\int_0^\infty b^{s-1}e^{-at}J_{\nu}(bt)t^{\mu -1}\; db = \frac{2^{s-1}e^{-a t}t^{\mu-s-1}\Gamma(\frac{\nu}{2}+\frac{s}{2})}{\Gamma(1+\frac{\nu}{2}-\frac{s}{2})}
$$
Integrate this with respect to $t$ as originally desired
$$
\int_0^\infty \frac{2^{s-1}e^{-a t}t^{\mu-s-1}\Gamma(\frac{\nu}{2}+\frac{s}{2})}{\Gamma(1+\frac{\nu}{2}-\frac{s}{2})} \; dt = 2^{s-1}a^{s-\mu}\frac{\Gamma(m-s)\Gamma(\frac{\nu+s}{2})}{\Gamma(1+\frac{\nu}{2}-\frac{s}{2})}
$$
take the inverse Mellin transform (interpreted as a Barnes integral) of this result
$$
I = \frac{1}{2\pi i}\int_{c - i \infty}^{c + i \infty} b^{-s}2^{s-1}a^{s-\mu}\frac{\Gamma(\mu-s)\Gamma(\frac{\nu+s}{2})}{\Gamma(1+\frac{\nu}{2}-\frac{s}{2})}\;ds
$$
where
$$
I=\frac{ b^{\nu } \Gamma (\mu+\nu ) }{2^{\nu }a^{\mu}a^{\nu}\Gamma (\nu +1)}\; _2F_1\left(\frac{\mu+\nu }{2},\frac{1}{2} (\mu+\nu +1);\nu +1;-\frac{b^2}{a^2}\right)
$$
