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

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


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

  • $\begingroup$ great answer thanks a lot $\endgroup$ – happymath Feb 28 '15 at 12:53
  • $\begingroup$ 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) $\endgroup$ – robjohn Feb 28 '15 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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