# Integration with respect to a parameter

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.

"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."
• All you need to know is that $\int_{-\infty}^{+\infty}e^{-x^2}\,dx=\sqrt{\pi}$, then make some clever manipulations and substitutions. – Jack D'Aurizio Feb 5 '15 at 14:28
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: $$\begin{eqnarray*} I &=& \int_{2}^{+\infty}\frac{y}{\sqrt{y^2-4}}e^{-(y^2-2)}\,dy = \frac{1}{2}\int_{4}^{+\infty}\frac{1}{\sqrt{z-4}}e^{-(z-2)}\,dz\\&=&\frac{1}{2e^2}\int_{0}^{+\infty}\frac{1}{\sqrt{z}}e^{-z}\,dz=\frac{1}{e^2}\int_{0}^{+\infty}e^{-w^2}\,dw\tag{2}=\color{red}{\frac{\sqrt{\pi}}{2e^2}}.\end{eqnarray*}$$
• @Andrew: in the second integral in the very first equation, replace $x$ with $\frac{1}{x}$. – Jack D'Aurizio Feb 5 '15 at 14:51
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