<s>[INST] Integral of a function 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?
 [/INST] $\int_z^\infty xe^{-\frac{x^2}{a^2}}\dfrac{1}{\sqrt{x^2-z^2}}dx$
$=\int_z^\infty\dfrac{e^{-\frac{x^2}{a^2}}}{2\sqrt{x^2-z^2}}d(x^2)$
$=\int_{z^2}^\infty\dfrac{e^{-\frac{x}{a^2}}}{2\sqrt{x-z^2}}dx$
$=e^{-\frac{z^2}{a^2}}\int_{z^2}^\infty e^{-\frac{x-z^2}{a^2}}~d(\sqrt{x-z^2})$
$=e^{-\frac{z^2}{a^2}}\int_0^\infty e^{-\frac{x^2}{a^2}}~dx$
$=\dfrac{ae^{-\frac{z^2}{a^2}}\sqrt\pi}{2}$
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