Prove the equation Prove that
$$\int_0^{\infty}\exp\left(-\left(x^2+\dfrac{a^2}{x^2}\right)\right)\text{d}x=\frac{e^{-2a}\sqrt{\pi}}{2}$$
Assume that the equation is true for $a=0.$
 A: In general
$$
\begin{align}
\int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\
&=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\
&=\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx.
$$
Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, then
$$
I_t=\sqrt{\frac{b}{a}}\int_{t=0}^\infty \frac{\exp\left(-a\left(-\frac{1}{t}\sqrt{\frac{b}{a}}+t\right)^2\right)}{t^2}\,dt.
$$
Let $t=x\;\rightarrow\;dt=dx$, then
$$
I_t=\int_{t=0}^\infty \exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Adding the two $I_t$s yields
$$
2I=I_t+I_t=\int_{t=0}^\infty\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)\exp\left(-a\left(t-\frac{1}{t}\sqrt{\frac{b}{a}}\right)^2\right)\,dt.
$$
Let $s=t-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;ds=\left(1+\frac{1}{t^2}\sqrt{\frac{b}{a}}\right)dt$ and for $0<t<\infty$ is corresponding to $-\infty<s<\infty$, then
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}},
$$
where $I$ is a Gaussian integral. Thus
$$
\begin{align}
\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx
&=\large\color{blue}{\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}}.
\end{align}
$$
In our case, put $a=1$ and $b=a^2$.
A: $$I(a):=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\text{d}x$$
$$\frac{dI}{da}=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\left(-\frac{2a}{x^2}\right)\text{d}x$$
Now substitute $y=\frac{a}{x}$, so $dy=-\frac{a}{y^2}$:
$$\frac{dI}{da}=2\int_{\text{sgn}(a)\cdot\infty}^{0}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2\text{sgn}(a)\int_{0}^{\infty}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2\text{sgn}(a)I$$
To obtain $I$ you just have to solve the simple ODE:
$$\frac{dI}{da}=-2\text{sgn}(a)I$$
with initial condition given by
$$I(0)=\frac{\sqrt{\pi}}{2}\ .$$
This gives you 
$$I(a)=\frac{e^{-2|a|}\sqrt{\pi}}{2}\ .$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\exp\pars{-\bracks{x^{2} + {a^{2} \over x^2}}}\,\dd x
     ={\root{\pi} \over 2}\,\expo{-2\verts{a}}:\ {\large ?}}$

\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}\exp\pars{-\bracks{x^{2} + {a^{2} \over x^2}}}\,\dd x}\
=\
\overbrace{\int_{0}^{\infty}
\exp\pars{-\verts{a}\bracks{%
{x^{2} \over \verts{a}} + {\verts{a} \over x^2}}}\,\dd x}^{\ds{x \equiv \root{\verts{a}}\expo{\theta}}}
\\[3mm]&=\int_{-\infty}^{\infty}\expo{-2\verts{a}\cosh\pars{2\theta}}
\root{\verts{a}}\expo{\theta}\,\dd\theta
\\[3mm]&=\root{\verts{a}}\int_{-\infty}^{\infty}
\expo{-2\verts{a}\bracks{2\sinh^{2}\pars{\theta} + 1}}
\bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta
\\[3mm]&=\root{\verts{a}}\expo{-2\verts{a}}\
\overbrace{\int_{-\infty}^{\infty}
\expo{-4\verts{a}\sinh^{2}\pars{\theta}}\cosh\pars{\theta}\,\dd\theta}
^{\ds{t\ \equiv\ \sinh\pars{\theta}}}\
=\root{\verts{a}}\expo{-2\verts{a}}\int_{-\infty}^{\infty}
\expo{-4\verts{a}t^{2}}\,\dd t
\\[3mm]&=\root{\verts{a}}\expo{-2\verts{a}}\,{1 \over 2\root{\verts{a}}}\
\overbrace{\int_{-\infty}^{\infty}\expo{-t^{2}}\,\dd t}^{\ds{=\ \root{\pi}}}\
=\
\color{#66f}{\Large{\root{\pi} \over 2}\,\expo{-2\verts{a}}}
\end{align}

