7
$\begingroup$

We know that $$\Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}dx=\sqrt{\pi} $$ but it seems that, for every $a>0 $ we have $$\int_{0}^{\infty}\frac{e^{-\left(\sqrt{x}-a/\sqrt{x}\right)^{2}}}{\sqrt{x}}dx=\sqrt{\pi}$$ so the additional term $a/\sqrt{x} $ doesn't change the value of the integral. How we can prove (or disprove) it? Thank you.

$\endgroup$
  • $\begingroup$ Are you sure you wrote your second integral properly? Plugging in a=0 gives 2. (Is the square root on x intentional?) $\endgroup$ – Omry Jul 11 '15 at 19:29
  • $\begingroup$ @Omry There was a typo, thank you. $\endgroup$ – User Jul 11 '15 at 19:32
  • 1
    $\begingroup$ This identity has been used today in proof of an amazing identity by Ramanujan. See math.stackexchange.com/a/1357872/72031 $\endgroup$ – Paramanand Singh Jul 12 '15 at 6:43
7
$\begingroup$

Hint. Make the change of variable $u=\sqrt{x}$, $du=\dfrac{dx}{2\sqrt{x}} $, to obtain $$ \int_0^{+\infty}\frac{e^{-\left(\sqrt{x}-a/\sqrt{x}\right)^{2}}}{\sqrt{x}}dx=2\int_0^{+\infty}e^{-\left(u-a/u\right)^{2}}du=\int_{-\infty}^{+\infty}e^{-\left(u-a/u\right)^{2}}du \tag1 $$ One may then recall that, for any integrable function $f$, we have (see here for a proof):

$$ \int_{-\infty}^{+\infty}f\left(u-\frac{a}{u}\right)\mathrm{d}u=\int_{-\infty}^{+\infty} f(u)\: \mathrm{d}u, \quad a>0. \tag2 $$

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

$$ \int_{-\infty}^{+\infty}e^{-(u-a/u)^2}\mathrm{d}u=\int_{-\infty}^{+\infty} e^{-u^2} \mathrm{d}u=\color{blue}{\sqrt{\pi}}, \quad a>0, \tag3 $$ giving the desired result.

$\endgroup$
  • $\begingroup$ Thank you! Where can I find a proof of $(1)$? $\endgroup$ – User Jul 11 '15 at 19:42
  • $\begingroup$ I found it! Thank you again! $\endgroup$ – User Jul 11 '15 at 19:46
  • $\begingroup$ @Elajan Can you link to the proof you found? $\endgroup$ – Omry Jul 11 '15 at 19:48
  • 1
    $\begingroup$ @Omry math.stackexchange.com/questions/457231/… the proof of Anastasiya-Romanova. $\endgroup$ – User Jul 11 '15 at 19:49

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.