# How to integrate $\int_0^{\infty}\frac{e^{-(t+\frac1t)}}{\sqrt t} dt$?

This is a problem given in my homework . I have to find the integral$$\int \limits_{0}^{\infty} \frac{e^{-(t+\frac{1}{t})}}{\sqrt t}dt$$

I am trying to use integral representation of the gamma function but I was not able to get it in the region of convergence i.e. $\int \limits_{0}^{\infty} \frac{e^{-t}}{\sqrt t}$ is clearly $\Gamma (\frac{1}{2})$ but the second factor is causing a problem. Any hints or suggestions are appreciated. Thanks.

Hint. Make the change of variable $t=x^2$ to obtain $$\int_0^{\infty}\frac{e^{-(t+\frac{1}{t})}}{\sqrt t}dt=2\int_0^{\infty}e^{ -x^2-1/x^2}dx=\int_{-\infty}^{\infty}e^{ -x^2-1/x^2}dx$$ You may then recall that, for any integrable function $f$, we have

$$\int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \tag1$$

Apply it to $f(x)=e^{-x^2}$, you get

$$\int_{-\infty}^{+\infty}e^{-(x-s/x)^2}\mathrm{d}x=\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x=\sqrt{\pi}, \quad s>0. \tag2$$

Thus

$$\int_{-\infty}^{+\infty}e^{-x^2-s^2/x^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2s}\tag3$$ then put $s=1$ to obtain your integral.

• great answer thanks a lot – happymath Feb 28 '15 at 12:53
• I hadn't seen the formula in the gray bar, and was computing the same thing in a messier way. This is cleaner. It is nice to know that formula. (+1) – robjohn Feb 28 '15 at 13:15