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What i tried was:

$\int_{-\infty}^{\infty} exp(-\sqrt{2\pi}x-\dfrac{(y-x)^2}{2})dy = -\dfrac{1}{y-x}exp(-\sqrt{2\pi}x-\dfrac{(y-x)^2}{2})|_{-\infty}^{\infty}$

The exp dominates the expression so it goes faster to $0$ for $y$ that approaches infinity or negative infinity, so the expression will be zero. However when i use wolfram to solve this integral i get: $\sqrt{2\pi}exp(-\sqrt{2\pi}x)$. Going to polar representation doesn't seem to be effective here. How did they come by this result?

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1 Answer 1

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$e^{-\sqrt 2\pi x} $ does not depend on $y$. So the answer is $e^{-\sqrt 2\pi x} \int e^{-(y-x)^{2}/2} dy=\sqrt {2 \pi} e^{-\sqrt {2\pi x}} $.

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  • $\begingroup$ Why does the second term equals to $\sqrt{2\pi}$? $\endgroup$
    – CodeHoarder
    Jun 18, 2020 at 6:30
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    $\begingroup$ @CodeHoarder Make the change of variable $z=y-x$. Do you know that $\int e^{-z^{2}/2}dz=\sqrt {2\pi}$? This is a standard integral. $\endgroup$
    – Kavi Rama Murthy
    Jun 18, 2020 at 6:32

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