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Question : integral of $\exp(-x^2-1/x^2)dx$ from $0$ to infinity

I think the answer is square root of $ \pi/2 \cdot \exp(-2)$. If I change $-x^2-1/x^2$ to $-(x-1/x)^2-2$ then the above integral becomes $\exp(-2) \cdot$ integral of $\exp(-(x-1/x)^2)dx$ from $0$ to infinity

integral of $\exp(-x^2)dx$ from $0$ to infinity = square root of $\pi/2$, but then is the

integral of $\exp(-(x-1/x)^2)dx$ from $0$ to infinity=integral of $\exp(-x^2)dx$ from $0$ to infinity ?

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@tony I think you are correct but the formatting is making it difficult to read! This might help you,… learning tex is easy! – john w. Feb 8 '12 at 6:48
Thank you very much for your information!! – Tony Feb 8 '12 at 11:39
up vote 10 down vote accepted

The integral you want to evaluate is $\mathrm e^{-2}I$, where $I=\int\limits_0^{+\infty}\mathrm e^{-(x-1/x)^2}\mathrm dx$. Let us compute $I$. The change of variable $z=1/x$ yields $z\gt0$ and $\mathrm dz=z^2\mathrm dx$, hence $I=\int\limits_0^{+\infty}\mathrm e^{-(z-1/z)^2}\mathrm dz/z^2$. Summing these two expressions of $I$, one gets $2I=\int\limits_0^{+\infty}\left(1+1/x^2\right)\mathrm e^{-(x-1/x)^2}\mathrm dx$. The change of variable $u=x-1/x$ yields $u$ in the whole real line and $\mathrm du=\left(1+1/x^2\right)\mathrm dx$, hence $2I=\int\limits_{-\infty}^{+\infty}\mathrm e^{-u^2}\mathrm du$. Finally, this last integral is $\sqrt\pi$, hence $$ \color{red}{\int\limits_0^{+\infty}\mathrm e^{-x^2-1/x^2}\mathrm dx=\frac{\sqrt\pi}{2\mathrm e^2}}. $$

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Again, thank you very much for your detailed explanation. I could never solved this without your help!! – Tony Feb 8 '12 at 11:45
Why $z = 1/x$ then $dz = z^2dx$ instead of $dz = -z^2dx$? – GAVD Jul 29 '15 at 3:07
Because the sign is absorbed in the integral being from 0 to +oo, not from +oo to 0. – Did Jul 29 '15 at 6:14

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