I have no idea ho to compute this integral:


I have put this integral into Wolfram Mathematica and the result is $$\int\limits_{0}^{\infty}e^{\large-\frac{1}{2}\left(y^2+\frac{t^2}{y^2}\right)}\,dy = \sqrt{\frac{\pi}{2}}e^{-\sqrt{t^2}}$$ But I would need to compute this integral manually. I have tried some usual "tricks" as to write the function under the integral as a integral of an other function or I have tried use derivation.

Any help will be appreciated.

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    $\begingroup$ This is a variation of the Gaussian integral. I am unsure about how to deal with THIS case, but the name might help you or another user to find the answer. $\endgroup$ – PackSciences Feb 17 at 10:15
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    $\begingroup$ Try completing the square in the exponent. $\endgroup$ – StubbornAtom Feb 17 at 10:19

First of all note that the integrand is an even function, so: $$I=\int_{0}^{\infty}e^{\large-\frac{1}{2}\left(y^2+\frac{t^2}{y^2}\right)}dy=\frac12\int_{-\infty}^{\infty}e^{\large-\frac{1}{2}\left(y^2+\frac{t^2}{y^2}\right)}\,dy$$ Now with some algebra: $$y^2 +\frac{t^2}{y^2}=\left(y-\frac{t}{y}\right)^2 +2t$$ One could have very well written $$y^2 +\frac{t^2}{y^2}=\left(y+\frac{t}{y}\right)^2 -2t$$ but the first one, also prepares to use this Glasser's Master Theorem. $$I=\frac12\int_{-\infty}^{\infty}e^{\large-\frac{1}{2}\left(\color{red}{\left(y-\frac{t}{y}\right)}^2 +2t\right)}\,dy=\frac12\int_{-\infty}^{\infty} e^{\large-\frac12 \left( \color{red}y^2 +2t\right)}dy$$ $$=\frac12 e^{-t} \int_{-\infty}^\infty e^{\large -\frac12 y^2}dy=\sqrt{\frac{\pi}{2}}e^{-t},\ t>0$$ Similarly for $t<0$.

Also note that the last integrand is just $\mathcal{N}(0,1)$ for the Normal Distribution, as seen from here that $$\Phi(x)=\frac{1}{\sqrt {2\pi}} \int_{-\infty }^x e^{\large-\frac12 y^2}dy$$ and well since it's a pdf we must must have that the total area equals to $1$ or $\lim\limits_{x\to \infty } \Phi(x)=1$.

For a proof just substitute $\frac12 y^2 =u^2$ and use this.


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