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.
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Are you sure you wrote your second integral properly? Plugging in a=0 gives 2. (Is the square root on x intentional?) – Omry Jul 11 '15 at 19:29
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@Omry There was a typo, thank you. – User Jul 11 '15 at 19:32
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1This identity has been used today in proof of an amazing identity by Ramanujan. See http://math.stackexchange.com/a/1357872/72031 – Paramanand Singh Jul 12 '15 at 06:43
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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.
Olivier Oloa
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1@Omry http://math.stackexchange.com/questions/457231/how-to-prove-int-infty-infty-fxdx-int-infty-infty-f-left?rq=1 the proof of Anastasiya-Romanova. – User Jul 11 '15 at 19:49