How to evaluate the integral $\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}dx$? I am wondering how one would calculate the integral:
$$
\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}dx
$$
where $c$ is a constant. I have tried to reparametrize by letting $u = x^2$ to get:
$$
\int_0^{\infty} \frac{1}{2\sqrt{u}} e^{-\frac{1}{2}\left(u+\frac{c}{u}\right)}du
$$
and then trying to use integration by parts. However, I am getting nowhere with that approach. 

We can also use the Cauchy-Schlomilch transformation. For $a,b \gt 0$
\begin{equation}
\int\limits_{0}^{\infty} f\Big[\left(ax - \frac{b}{x} \right)^{2} \Big] \mathrm{d} x = \frac{1}{a} \int\limits_{0}^{\infty} f(y^{2}) \mathrm{d}y
\tag{1}
\label{eq:1}
\end{equation}
Expanding the term inside of the integrand, we have
\begin{equation}
\left(ax - \frac{b}{x} \right)^{2} = a^{2} x^{2} - 2ab + \frac{b^{2}}{x^{2}}
\end{equation}
\begin{equation}
-a^{2} x^{2} - \frac{b^{2}}{x^{2}} = -2ab - \left(ax - \frac{b}{x} \right)^{2}
\end{equation}
Matching variables with our problem, we have $a^{2} = 1/2$ and $b^{2} = c/2$ and the term in the
exponential of our problem becomes
\begin{equation}
-\frac{1}{2}x^{2} - \frac{c}{2}\frac{1}{x^{2}} = -\sqrt{c} - \left(\frac{x}{\sqrt{2}} - \frac{1}{x}\sqrt{\frac{c}{2}} \right)^{2}
\end{equation}
Substituting this into our integral, yields
\begin{align}
\int\limits_{0}^{\infty} \mathrm{exp}\left( -\frac{1}{2}x^{2} - \frac{c}{2}\frac{1}{x^{2}} \right) \mathrm{d} x 
&= \mathrm{e}^{-\sqrt{c}} \int\limits_{0}^{\infty} 
\mathrm{exp}\Big[-\left( \frac{x}{\sqrt{2}} - \frac{1}{x}\sqrt{\frac{c}{2}} \right)^{2} \Big] \mathrm{d} x \\
\tag{a}
&= \sqrt{2}\,\mathrm{e}^{-\sqrt{c}} \int\limits_{0}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d} y \\
\tag{b}
&= \sqrt{2}\,\mathrm{e}^{-\sqrt{c}} \,\frac{\sqrt{\pi}}{2} \mathrm{erf}(y) \Big|_{0}^{\infty} \\
&= \sqrt{\frac{\pi}{2}}\mathrm{e}^{-\sqrt{c}}
\end{align}
Notes:
a. Use the Cauchy-Schlomilch transformation, equation $\eqref{eq:1}$.
b. $\mathrm{erf}(z)$ is the error function.
