Integral of a function

Question

I have to solve a probability problem, and at a final step I arrive at the following integral:

$$\int_z^\infty xe^{-x^2/a^2}\displaystyle\frac{1}{\sqrt{x^2-z^2}}dx$$

I tried to integrate by parts, taking $f(x)=\displaystyle\frac{1}{\sqrt{x^2-z^2}}$ and $g'(x)= xe^{-x^2/a^2}$, but this choice leads to a more complicated integral $\int_z^\infty f'(x)g(x)dx$.

Is there an analytical method to solve this problem?

Answer 1

The integral is $$z\int_1^{\infty}\frac{x e^{-z^2x^2/a^2}}{\sqrt{x^2-1}}dx=\frac{z}{2}\int_1^{\infty}\frac{e^{-z^2x/a^2}}{\sqrt{x-1}}dx$$ $$=\frac{z}{8}e^{-z^2/2a^2}K_{1/4}(z^2/2a^2).$$

Answer 2

$\int_z^\infty xe^{-\frac{x^2}{a^2}}\dfrac{1}{\sqrt{x^2-z^2}}dx$

$=\int_z^\infty\dfrac{e^{-\frac{x^2}{a^2}}}{2\sqrt{x^2-z^2}}d(x^2)$

$=\int_{z^2}^\infty\dfrac{e^{-\frac{x}{a^2}}}{2\sqrt{x-z^2}}dx$

$=e^{-\frac{z^2}{a^2}}\int_{z^2}^\infty e^{-\frac{x-z^2}{a^2}}~d(\sqrt{x-z^2})$

$=e^{-\frac{z^2}{a^2}}\int_0^\infty e^{-\frac{x^2}{a^2}}~dx$

$=\dfrac{ae^{-\frac{z^2}{a^2}}\sqrt\pi}{2}$

Answer 3

Substitute $\mathrm{x^2=z^2+t^2}$, then, $\mathrm{xdx=tdt}$

and $$\mathrm{\int_z^{\infty}xe^{-x^2/a^2}\frac{1}{\sqrt{x^2-z^2}}dx=\int_0^{\infty}te^{-z^2/a^2}e^{-t^2/a^2}\frac{1}{t}dt}=e^{-z^2/a^2}\int_0^{\infty}e^{-t^2/a^2}dt=\frac{ae^{-z^2/a^2}\sqrt{\pi}}{2}$$