how to prove this formula $\int_{0}^{\infty}\frac{e^{-\frac{x^2}{2}-\frac{z^2}{2x^2}}}{x^2}dx=\sqrt{\frac{\pi }{2}}\frac{ze^{-|z|}}{|z|}$? i have tried residue theorem, but it seems not to work
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Does this answer your question (after substituting $x=1/y$)? Given $\int_0^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}$, evaluate $\int_0^{\infty}e^{-a^2x^2-\frac{b^2}{x^2}}dx $ – metamorphy Sep 16 '20 at 07:54
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The LHS is invariant under the transformation $z\to -z$ but the RHS is not. – Gary Sep 16 '20 at 07:55
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@Gary Indeed the OP has forgotten to place the first $z$ between absolute values; but this would mean that a concellation occurs ?? – Jean Marie Sep 16 '20 at 07:58
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1In fact there seems to be a lot of duplicates. E.g. (1), (2), (3)... – metamorphy Sep 16 '20 at 07:59