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Can anyone help me on how to evaluate the following integral using differentiation with respect to a parameter $$ \int_0^\infty e^{-(x^2+\frac{1}{x^2})}dx $$ Any help would be greatly appreciated. Thanks.

From the comments:

"I think I need to make it in the form of $\int_0^\infty e^{-u} du$ because I know the value of that integral."

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  • $\begingroup$ Please share your thoughts so far :) $\endgroup$ – Shaun Feb 5 '15 at 14:10
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    $\begingroup$ I'm not entirely sure how to start the problem but I think I need to make it in the form of /int_0^/infty e^{-u} du because I know the value of that integral $\endgroup$ – Andrew Feb 5 '15 at 14:13
  • $\begingroup$ All you need to know is that $\int_{-\infty}^{+\infty}e^{-x^2}\,dx=\sqrt{\pi}$, then make some clever manipulations and substitutions. $\endgroup$ – Jack D'Aurizio Feb 5 '15 at 14:28
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We have: $$ I = \int_{0}^{1}\exp\left(-\left(x^2+\frac{1}{x^2}\right)\right)\,dx + \int_{1}^{+\infty}\exp\left(-\left(x^2+\frac{1}{x^2}\right)\right)\,dx $$ so: $$ I = \int_{0}^{1}\left(x+\frac{1}{x}\right)\exp\left(-\left(x^2+\frac{1}{x^2}\right)\right)\frac{dx}{x}\tag{1}$$ and by setting $y=x+\frac{1}{x}$, then $y=\sqrt{z}$, we get: $$\begin{eqnarray*} I &=& \int_{2}^{+\infty}\frac{y}{\sqrt{y^2-4}}e^{-(y^2-2)}\,dy = \frac{1}{2}\int_{4}^{+\infty}\frac{1}{\sqrt{z-4}}e^{-(z-2)}\,dz\\&=&\frac{1}{2e^2}\int_{0}^{+\infty}\frac{1}{\sqrt{z}}e^{-z}\,dz=\frac{1}{e^2}\int_{0}^{+\infty}e^{-w^2}\,dw\tag{2}=\color{red}{\frac{\sqrt{\pi}}{2e^2}}.\end{eqnarray*}$$

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  • $\begingroup$ Thanks so much. Could you explain how you get equation one. I understand everything from there but I can't seem to figure out how you got to equation 1 from the above $\endgroup$ – Andrew Feb 5 '15 at 14:44
  • $\begingroup$ @Andrew: in the second integral in the very first equation, replace $x$ with $\frac{1}{x}$. $\endgroup$ – Jack D'Aurizio Feb 5 '15 at 14:51

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