Integral over exponential involving reciprocial I want to show $$I := \int_{-\infty}^\infty \exp \left(-\left(x-\frac p x \right)^2\right) \, dx = \sqrt{\pi}$$ for any non-negative $p\geq 0$. I tried to prove $I^2=\pi$ using Fubini's theorem, but had no success. 
 A: Using the substitution
$$
x=\frac{u+\sqrt{u^2+4p}}2\implies\mathrm{d}x=\frac12\left(1+\frac{u}{\sqrt{u^2+4p}}\right)\mathrm{d}u
$$
we have $u=x-\frac px$ and, as $u$ varies from $-\infty$ to $+\infty$, $x$ varies from $0$ to $\infty$.
Since the integrand is even,
$$
\begin{align}
\int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,\mathrm{d}x
&=2\int_0^\infty e^{-\left(x-\frac px\right)^2}\,\mathrm{d}x\\
&=2\int_{-\infty}^\infty e^{-u^2}\frac12\left(1+\frac{u}{\sqrt{u^2+4p}}\right)\mathrm{d}u\\
&=\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u\\[6pt]
&=\sqrt\pi
\end{align}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{I \equiv \int_{-\infty}^{\infty}
\exp\pars{-\bracks{x - {p \over x}} ^{2}}\,\dd x = \root{\pi}:\ {\large }}$

\begin{align}
I &\equiv
\int_{-\infty}^{\infty}\exp\pars{-\bracks{x - {p \over x}} ^{2}}\,\dd x =
2\int_{0}^{\infty}
\exp\pars{-p\bracks{{x \over \root{p}} - {\root{p} \over x}} ^{2}}\,\dd x
\\[5mm] & =
2\root{p}\int_{0}^{\infty}
\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x
\\[5mm] & =
\root{p}\bracks{%
\int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x +
\int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x}
\\[5mm] & =
\root{p}\bracks{%
\int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}\,\dd x +
\int_{\infty}^{0}\exp\pars{-p\bracks{{1 \over x} - x} ^{2}}
\,\pars{-\,{1 \over x^{2}}}\dd x}
\\[5mm] & =
\root{p}
\int_{0}^{\infty}\exp\pars{-p\bracks{x - {1 \over x}} ^{2}}
\pars{1 + {1 \over x^{2}}}\,\dd x
\,\,\,\stackrel{x - 1/x\ \mapsto\ x}{=}\,\,\,
\int_{-\infty}^{\infty}\expo{-px^{2}}\root{p}\,\dd x
\\[5mm] & =
\int_{-\infty}^{\infty}\expo{-x^{2}}\,\dd x = \bbx{\root{\pi}}
\end{align}
A: I thought it might be instructive to present another approach from the ones already posted.  To that end, we now proceed.


There is nothing particularly special about the function $e^{-\left(x-\frac px\right)^2}$ in the development.  
In fact, using the substitutions $x=-\sqrt{p}e^{-t}$ for $x\in (-\infty, 0]$ and $x=\sqrt{p}e^{t}$ for $x\in [0,\infty)$ in the integral $\int_{-\infty}^\infty f\left(x-\frac px\right)\,dx$ yields
$$\begin{align}
\int_{-\infty}^\infty f\left(x-\frac px\right)\,dx&=\int_{-\infty}^0 f\left(x-\frac px\right)\,dx+\int_{0}^\infty f\left(x-\frac px\right)\,dx\\\\
&=\sqrt{p}\int_{-\infty}^\infty f\left(2\sqrt{p}\sinh(t)\right)\,e^{-t}\,dt+\sqrt{p}\int_{-\infty}^\infty f\left(2\sqrt{p}\sinh(t)\right)\,e^{t}\,dt\\\\
&=2\sqrt{p}\int_{-\infty}^\infty f(2\sqrt{p}\sinh(t))\,\cosh(t)\,dt\\\\
&=\int_{-\infty}^\infty f(u)\,du
\end{align}$$
which for $f(u)=e^{-u^2}$ is equal to $\sqrt{\pi}$ as expected!


We could have approached the problem of specific interest with a slight modification of the general development in the highlighted section. 
First we let $x=\sqrt{p}\,e^{t}$.  Then, we have $x-\frac px =2\sqrt{p}\sinh(t)$ and 
$$\begin{align}
\int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\int_{0}^\infty e^{-\left(x-\frac px\right)^2}\,dx\\\\
&=2\sqrt p\int_{-\infty}^\infty e^{-4p\sinh^2(t)}\,e^t\,dt\tag 1
\end{align}$$

Next, we let $x=\sqrt {p}\,e^{-t}$.  Then, we have $x-\frac px=-2\sqrt{p}\sinh(t)$ and
$$\begin{align}
\int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\int_{0}^\infty e^{-\left(x-\frac px\right)^2}\,dx\\\\
&=2\sqrt p\int_{-\infty}^\infty e^{-4p\sinh^2(t)}\,e^{-t}\,dt\tag 2
\end{align}$$

Adding $(1)$ and $(2)$ and dividing by $2$, we obtain
$$\begin{align}
\int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\sqrt{p}\int_0^\infty e^{-4p\sinh^2(x)}\,\cosh(x)\,dx\tag3
\end{align}$$

Finally, enforcing the substitution $u=\sinh(t)$ in $(3)$ yields
$$\begin{align}
\int_{-\infty}^\infty e^{-\left(x-\frac px\right)^2}\,dx&=2\sqrt{p}\int_0^\infty e^{-4pu^2}\,\,du\\\\
&=\sqrt{\pi}
\end{align}$$
as expected!
A: The key is to essentially make the substitution $u = x - p/x$. The problem is that this isn't an invertible function--it has a positive and a negative root for $x$. To fix, this we divide the integral into a positive and negative side and make the substitutions
\begin{eqnarray}
x(u) &=& \frac{1}{2}\left(u \pm \sqrt{4p+u^2}\right) \\
dx &=& \frac{1}{2}\left(1 \pm \frac{u}{\sqrt{4p+u^2}}\right)
\end{eqnarray}
This gives
\begin{multline}
\int_{-\infty}^\infty \exp\left[-\left(x-\frac{p}{x}\right)^2\right]\,dx =\int_{-\infty}^0 \exp\left[-\left(x-\frac{p}{x}\right)^2\right] \, dx+\int_0^\infty \exp\left[-\left(x-\frac{p}{x}\right)^2\right]dx
\\= \int_{-\infty}^\infty \frac{e^{-u^2}}{2}\left(1 - \frac{u}{\sqrt{4p+u^2}}\right) \, du + \int_{-\infty}^\infty \frac{e^{-u^2}}{2}\left(1 + \frac{u}{\sqrt{4p+u^2}}\right) \, du
\\= \int_{-\infty}^\infty e^{-u^2}du = \sqrt{\pi}
\end{multline}
A: Hint: Substitute $x-\frac{p}{x}$ with $t$ and this looks suspiciously similar to the error function.
A: For $a=\sqrt p$, tet $x=at$ and then
$$ x-\frac{p}{x}=at-\frac{p}{at}=\sqrt p(t-\frac1t).$$
So
\begin{equation}
\begin{aligned}
I & =\int_{-\infty}^\infty \exp \left(-\left(x-\frac p x \right)^2\right) \, dx=\sqrt p\int_{-\infty}^\infty \exp \left(-p\left(t-\frac1t \right)^2\right) \, dt \\[10pt]
& =2\sqrt p\int_0^\infty \exp \left(-p\left(t-\frac1t \right)^2\right) \, dt.
\end{aligned}
\tag 1
\end{equation}
Using $t\to\frac1t$, one has
$$ I=2\sqrt p\int_{0}^\infty \frac1{t^2}\exp \left(-p\left(t-\frac1t \right)^2\right) \, dt.\tag{2}$$
Then adding (1) and (2) and under $u=t-\frac1t$, one has
$$ I=\sqrt p\int_{0}^\infty (1+\frac1{t^2})\exp \left(-\left(t-\frac1t \right)^2\right) \, dt=\sqrt p\int_{-\infty}^\infty\exp(pu^2) \, du=\sqrt\pi.$$
