I want to evaluate the following integral: $$\int_{a}^{\infty}\frac{x\exp(-\frac{x^2}{2})}{\sqrt{x^2-a^2}}dx.$$ I know the solution is $\sqrt{\frac{\pi}{2}}\exp(-a^2/2)$, but I am unsure how to arrive there. I tried integrating by parts, but this just made the integration messier.
Thanks.
This is a rather straightforward integral, and J.G.'s comment already solved it. I would use the same approach, with the last substitution:
$$y = \sqrt{x^2-a^2} \qquad \text{d}y = \frac{x}{\sqrt{x^2 - a^2}}\ \text{d}x \qquad x^2 = y^2 + a^2$$
Don't forget about the extrema of the integral which now runs from $0$ to $+\infty$ that is:
$$\int_0^{+\infty} e^{-(y^2+a^2)/2}\ \text{d}y$$
Now there is no remorse in evaluating this as a Gaussian (what it is) integral. We use exponential property, rewriting:
$$e^{-a^2/2} \int_0^{+\infty} e^{-y^2/2}\ \text{d}y$$
Now I really want to assume that you do know how to calculate such integral, whose result is $\sqrt{\frac{\pi}{2}}$, whence the final resul
$$\boxed{\sqrt{\frac{\pi}{2}}e^{-a^2/2}}$$