Let $I_\nu$ be the modified Bessel function of first kind. I wonder if there is some formulas for the integrals:
$\int_0^\infty e^{-cx} I_\nu(a \sqrt{x})I_\nu(b \sqrt{x})dx$,
$\int_0^\infty x e^{-cx} I_\nu(a \sqrt{x})I_\nu(b \sqrt{x})dx$,
where $a, b, c > 0$. Or is there any possible simplification?
After transforming the integration variable to $t=x^{1/2}$ and using http://dlmf.nist.gov/10.43.E28 one gets for the first integral ($\nu>-1$, $c>0$) $$ \int_0^\infty e^{-cx} I_\nu(a \sqrt{x}) \ I_\nu(b \sqrt{x}) \ dx = 2 \int_0^\infty t \ e^{-c t^2} I_\nu(a t) \ I_\nu(b t) \ dt = \frac{1}{c} \ \exp\left( \frac{a^2+b^2}{ 4 \ c} \right)\ I_{\nu}\left(\frac{a\ b}{2\ c}\right). $$ As @Michael Engelhardt commented, the second integral is just $-\frac{d}{dc}$ of the first one, hence $$ \int_0^\infty x \ e^{-cx} I_\nu(a \sqrt{x}) \ I_\nu(b \sqrt{x}) \ dx = -\frac{d}{d c} \int_0^\infty e^{-cx} I_\nu(a \sqrt{x}) \ I_\nu(b \sqrt{x}) \ dx = $$ $$\frac{1}{c^2} \ \exp\left( \frac{a^2+b^2}{ 4 \ c} \right) \ \left[ \left( 1+\frac{a^2+b^2}{4\ c}\right) \ \ I_{\nu}\left(\frac{a\ b}{2 \ c}\right) + \frac{a\ b}{2 \ c}\ I^{\prime}_{\nu}\left(\frac{a\ b}{2\ c}\right)\right]. $$
