calculating a definite integral of gaussian-like form As part of a homework question in the course "Introduction to Probability" I take, I was given the following formula:  
$$\int_0^\infty \exp\left(-x^2-\frac{a^2}{x^2}\right)dx = \frac{\sqrt{\pi}}{2}\cdot\exp(-2|a|).$$
The question is how do I get this result?  
I tried to integrate by using the substitution $u = x+a/x$, but I can't handle the Jacobian of this substitution. Any suggestions?
 A: Let
$$
f(a) = \int_0^\infty \exp(-x^2-a^2/x^2) dx.
$$
With $x=a/t$ we get that
$$
f(a) = a \int_0^\infty \exp(-t^2-a^2/t^2)/t^2 dt,
$$
and differentiating under the integral sign yields
$$
f'(a) = -2a \int_0^\infty \exp(-x^2 - a^2/x^2) / x^2 dx.
$$
Hence $f'(a) = -2f(a)$ and $f(0)$ is well-known. Solve this and you get the result.
Edit: Above is for $a > 0$.
A: Let
$$
f(a)=\int_0^{+\infty}\exp(-x^2-a^2/x^2)\,dx.
$$
We would like to show that $f(a)=\frac{\sqrt{\pi}}{2}\exp(-2|a|)$. Let us apply the Fourier transform to both sides. A standard transform is
$$
\mathcal{F}\,\Bigl(\frac{\sqrt{\pi}}{2}\exp(-2|a|)\Bigr)(b)=\frac{2\sqrt{\pi}}{4+b^2}.
$$
Now, the Fourier transform of $\exp(-a^2/x^2)$ (with respect to $a$, and assuming $x>0$) is
$$
\sqrt{\pi}x\exp\bigl(-\tfrac{1}{4}b^2x^2\bigr)
$$
Thus, the right-hand side becomes (if we believe that we can switch the order of integration and so on, I leave that to you)
$$
\begin{align}
\mathcal{F}(f(a))(b)&=\int_0^{+\infty}\sqrt{\pi}x\exp\bigl(-(1+\tfrac{1}{4}b^2)x^2\bigr)\,dx\\
&=\Bigl[ -\frac{2\sqrt{\pi}}{4+b^2}\exp\bigl(-(1+\tfrac{1}{4}b^2)x^2\bigr)\Bigr]_0^{+\infty}\\
&= \frac{2\sqrt{\pi}}{4+b^2}
\end{align}
$$
The uniqueness theorem for Fourier transforms give the desired conclusion.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}\exp\pars{-x^{2} - {a^{2} \over x^{2}}}\,\dd x
     ={\root{\pi} \over 2}\,\exp\pars{-2\verts{a}}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\exp\pars{-x^{2} - {a^{2} \over x^{2}}}\,\dd x}
=\int_{0}^{\infty}\exp\pars{%
-\verts{a}\bracks{{x^{2} \over \verts{a}} + {\verts{a} \over x^{2}}}}\,\dd x
\\[5mm]&=\root{\verts{a}}\int_{0}^{\infty}\exp\pars{%
-\verts{a}\bracks{x^{2} + {1 \over x^{2}}}}\,\dd x
\\[5mm]&=\root{\verts{a}}\int_{0}^{\infty}\exp\pars{%
-\verts{a}\braces{\bracks{x - {1 \over x}}^{2} + 2}}\,\dd x
\\[5mm]&=\ \underbrace{\root{\verts{a}}\expo{-2\verts{a}}\ \overbrace{%
\int_{0}^{\infty}\exp\pars{-\verts{a}\bracks{x - {1 \over x}}^{2}}\,\dd x}
^{\ds{\dsc{x}\ \mapsto\ \dsc{1 \over x}}}}
_{\ds{\equiv\dsc{I_{1}}}}
\\[5mm]&=\ \underbrace{\root{\verts{a}}\expo{-2\verts{a}}
\int_{0}^{\infty}\exp\pars{-\verts{a}\bracks{x - {1 \over x}}^{2}}\,{\dd x \over x^{2}}}_{\ds{\equiv\ \dsc{I_{2}}}}
\end{align}

Then,
\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}\exp\pars{-x^{2} - {a^{2} \over x^{2}}}\,\dd x}
={\dsc{I_{1}} + \dsc{I_{2}} \over 2}
\\[5mm]&=\ \overbrace{\half\,\root{\verts{a}}\expo{-2\verts{a}}\int_{0}^{\infty}
\exp\pars{-\verts{a}\bracks{x - {1 \over x}}^{2}}\pars{1 + {1 \over x^{2}}}
\,\dd x}
^{\ds{\dsc{x - {1 \over x}}\ \mapsto\ \dsc{x}}}
\\[5mm]&=\half\,\root{\verts{a}}\expo{-2\verts{a}}\int_{-\infty}^{\infty}
\exp\pars{-\verts{a}x^{2}}\,\dd x
=\half\,\root{\verts{a}}\expo{-2\verts{a}}{1 \over \root{\verts{a}}}\ 
\overbrace{\int_{-\infty}^{\infty}\exp\pars{-x^{2}}\,\dd x}^{\dsc{\root{\pi}}}
\\[5mm]&=\color{#66f}{\large{\root{\pi} \over 2}\,\exp\pars{-2\verts{a}}}
\end{align}
A: My first attempt would be to arrange the equation like this and then attempt to use "Integration by Parts".
$
\int_{0}^{\infty} e^{-x^2}e^{-a^2/x^2}\,\mathrm{d}x
$
You will encounter some unusual integrals along the way, but those integrals are provided on Wikipedia:
http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions
The two I would focus on at first glance are:
$
\int_{0}^{\infty} e^{-ax^2}\,\mathrm{d}x=\frac{1}{2} \sqrt{\pi \over a} \quad (a>0)
$
$
\int_0^\infty e^{-ax^b} dx = \frac{1}{b}\ a^{-\frac{1}{b}} \, \Gamma\left(\frac{1}{b}\right)
$
Where $\Gamma(-\tfrac{1}{2})  = -2\sqrt{\pi}$
This is not a typical integration problem that you normally see in the first two years of technical schooling. It's more difficult.
