3
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.