Using that the Gaussian integral $\int_0^{\infty}e^{-x^2}\;dx$ is equal to $\sqrt{\pi}/2$, compute the following integral $$ \int_0^{\infty}e^{-\left(x^2+\frac{a^2}{x^2}\right)}\;dx$$ where $a>0$ is a parameter.
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Let $u:=x-a/x$ so, by Glasser's master theorem, your integral is$$\frac12e^{-2a}\int_{\Bbb R}e^{-u^2}dx=\frac12e^{-2a}\int_{\Bbb R}e^{-u^2}du=\frac{\sqrt{\pi}}{2}e^{-2a}.$$
J.G.
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I never heard of that Theorem in my life ... but that is sick! – kvantour Sep 29 '20 at 06:43
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A french reference for that theorem: Exercice 1.1 p.12 of "Oraux X-Ens 5" (nouvelle édition 2023), Francinou, Gianella, Nicolas – Noix07 Aug 04 '23 at 10:30