Given the identity
$$ \int^\infty\_0 K\_v\left(\alpha\sqrt{x^2+z^2}\right) \frac{x^{2\mu+1}}{\left(\sqrt{x^2+z^2}\right)^v}\:\mathrm{d}x = \frac{2^\mu \Gamma(\mu+1)}{\alpha^{\mu+1}z^{v-\mu-1}} K\_{v-\mu-1}(\alpha z), \quad \alpha>0,\quad \Re[\mu]>-1$$


how can I find a closed form for the integral:


$$ \int^\infty\_0 \exp\left(-\beta x^2\right) K\_v\left(\alpha\sqrt{x^2+z^2}\right) \frac{x^{2\mu+1}}{\left(\sqrt{x^2+z^2}\right)^v}\:\mathrm{d}x $$


I tried using series representation of the exponential function, but I got an infinite series.



Integral of Modified Bessel Function of the Second Type
Here is one situation when you can give a closed-form answer. Re-write the integral as


$$ I= e^{\beta z^2}\int\_0^\infty e^{-\beta(x^2+z^2)} K\_\nu(\alpha\sqrt{x^2+z^2}) x^{2\mu+1}(x^2+z^2)^{-\frac{v}{2}} dx $$


( $x:=zy$


$$ =z^{2\mu+2-v} e^{\beta z^2}\underbrace{\int\_0^\infty e^{-\beta z^2(y^2+1)} K\_\nu(\alpha z\sqrt{y^2+1}) y^{2\mu+1}(y^2+1)^{-\frac{v}{2}} dx}\_{=:A}. $$


To compute the integral $A$ use the change in variables $t=y^2+1$, $ y=(t-1)^{\frac{1}{2}} $ to reduce it to an integral of the form


$$ A = const \underbrace{\int\_1^\infty e^{-\beta z^2 t} K\_\nu(\alpha z t) (t-1)^\mu t^{-\frac{v}{2}} dt.}\_{=: B} $$
If $\beta z^2= \alpha z$, then you can find a description of $B$ in Gradshteyn and Ryzhik 6th Edition, formula 6.625 (9).


