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. 
 A: Consider
$$F(c)=\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}dx.$$
Notice that for convergence we need that $c\geq 0$.
Then, for $c>0$,
$$F'(c)=-\frac{1}{2}\int_0^{\infty} e^{-\frac{1}{2}\left(x^2+ \frac{c}{x^2}\right)}\frac{dx}{x^2}.$$
On the other hand, by Daryl's comment, by letting $t=\sqrt{c}/x$ we get $dx=-\sqrt{c}dt/t^2$ and
$$F(c)=\sqrt{c}\int_0^{\infty} e^{-\frac{1}{2}\left(t^2+ \frac{c}{t^2}\right)}\frac{dt}{t^2}=-2\sqrt{c}F'(c).$$
Hence, by solving the differential equation in $(0,+\infty)$, we find
$$F(c)=F(0^+)\cdot e^{-\sqrt{c}}=\sqrt{\pi/2}\cdot e^{-\sqrt{c}}.$$
A: Assuming $c>0$, we may just apply Glasser's master theorem.
$$\int_{\mathbb{R}}\exp\left[-\frac{1}{2}\left(x^2+\frac{c}{x^2}\right)\right]\,dx = e^{-\sqrt{c}}\int_{\mathbb{R}}\exp\left[-\frac{1}{2}\left(x-\frac{\sqrt{c}}{x}\right)^2\right]\,dx $$
and the last integral equals
$$ e^{-\sqrt{c}}\int_{\mathbb{R}}\exp\left(-\frac{x^2}{2}\right)\,dx = \color{red}{\sqrt{2\pi}e^{-\sqrt{c}}}.$$
A: $t={1\over x}$, then 
$$I=\int\limits_0^\infty e^{-\left(x^2+ \frac1{x^2}\right)}dx=\int\limits_0^\infty e^{-\left({1\over t^2}+t^2\right)}{dt\over t^2}$$
Now rewrite the second integral by renaming its variable to $x$ (we can name our variables whatever we like, can't we?), and add them together:
$$ 2I=\int\limits_0^\infty e^{-\left(x^2+ \frac1{x^2}\right)}\left(1+{1\over x^2}\right)dx$$
Now comes another ingenious substitution: $t=x-{1\over x}$, so $dt=\left(1+{1\over x^2}\right)dx$, and
$$2I=\int\limits_{x=0}^{x=\infty} e^{-\left(x^2+ \frac1{x^2}\right)}dt=\int\limits_{t=-\infty}^{t=\infty} e^{-\left(x-\frac1x\right)^2-2}dt={1\over e^2}\int\limits_{-\infty}^\infty e^{-t^2}dt$$
which you supposedly know how to finish.
The minor details concerning introduction of some coefficients ($c$, etc.) I leave to you. Apparently, the first step should be modified according to Daryl's comment.
A: $$
\begin{align}
\int_0^\infty e^{-\frac12\left(x^2+\frac{c}{x^2}\right)}\,\mathrm{d}x
&=c^{1/4}e^{-\sqrt{c}}\int_0^\infty e^{-\frac{\sqrt{c}}2\left(x^2+\frac1{x^2}-2\right)}\,\mathrm{d}x\tag{1}\\
&=\frac12c^{1/4}e^{-\sqrt{c}}\int_{-\infty}^\infty e^{-\frac{\sqrt{c}}2y^2}\left(1+\frac{y}{\sqrt{y^2+4}}\right)\mathrm{d}y\tag{2}\\
&=\sqrt{\frac\pi2}\,e^{-\sqrt{c}}\tag{3}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto c^{1/4}x$
$(2)$: substitute $y=x-\frac1x\implies x=\frac{y+\sqrt{y^2+4}}2$
$(3)$: $\int_{-\infty}^\infty e^{-ay^2}\,\mathrm{d}y=\sqrt{\frac\pi a}$ and the integral of an odd function over $(-\infty,\infty)$ is $0$
A: 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.
