Evaluate Gauss-like Integral Evaluate Integral
$$\int_0^\infty e^{-ay^{2}-\frac{b}{y^2}}dy $$
Where a and b are real and positive.
This integral is eerily similar to the Gaussian integral 
$$\int_0^\infty e^{-\alpha x^2}dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha}}$$
This is an integral I have come across as a step in a problem doing some homework for Advanced Statistics... Not sure where to begin.
 A: Substitute $u=ay-\frac by$, then $a^2y^2+\frac{b^2}{y^2}=2ab+u^2$. Furthermore, $y=\frac{u+\sqrt{u^2+4ab}}{2a}$. Therefore,
$$
\begin{align}
\int_0^\infty e^{-a^2y^2-\frac{b^2}{y^2}}\,\mathrm{d}y
&=\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}(u+\sqrt{u^2+4ab})\\
&=\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u
+\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}\sqrt{u^2+4ab}\\
&=\frac{\sqrt\pi}{2ae^{2ab}}+0
\end{align}
$$
A: This answer is taken from my answer here.
$$
\begin{align}
\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\int_{v=0}^\infty \exp\left(-a\left(v^2+\frac{b}{av^2}\right)\right)\,dv\\
&=\int_{v=0}^\infty \exp\left(-a\left(v^2-2\sqrt{\frac{b}{a}}+\frac{b}{av^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dv\\
&=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dv\\
&=\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
\end{align}
$$
The trick to solve the last integral is by setting
$$
I=\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv.
$$
Let $t=-\frac{1}{v}\sqrt{\frac{b}{a}}\;\rightarrow\;v=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dv=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, 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 $t=v\;\rightarrow\;dt=dv$, 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
$$
I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}.
$$
Thus
$$
\begin{align}
\int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\
&=\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}.
\end{align}
$$
A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{0}^{\infty}\expo{-ay^{2} - b/y^{2}}\,\dd y:\ {\large ?}.\qquad
     a, b > 0}$.

\begin{align}&\color{#66f}{\large\int_{0}^{\infty}\expo{-ay^{2} - b/y^{2}}\,\dd y}
\ =\ \overbrace{\int_{0}^{\infty}
\exp\pars{-\root{ab}\bracks{\root{a \over b}y^{2} + \root{b \over a}y^{-2}}}
\,\dd y}^{\ds{\mbox{Set}\quad\pars{a \over b}^{1/4}y \equiv \expo{\theta}}}
\\[3mm]&=\int_{-\infty}^{\infty}
\exp\pars{-\root{ab}\bracks{\expo{2\theta} + \expo{-2\theta}}}
\pars{b \over a}^{1/4}\expo{\theta}\,\dd\theta
\\[3mm]&=\pars{b \over a}^{1/4}\int_{-\infty}^{\infty}
\exp\pars{-\root{ab}\bracks{2\cosh\pars{2\theta}}}
\bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta
\\[1cm]&=2\pars{b \over a}^{1/4}\ \overbrace{\int_{0}^{\infty}
\exp\pars{-2\root{ab}\bracks{2\sinh^{2}\pars{\theta} + 1}}
\cosh\pars{\theta}\,\dd\theta}^{\ds{\mbox{Set}\quad t \equiv \sinh\pars{\theta}}}
\\[1cm]&=2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}}
\int_{0}^{\infty}\exp\pars{-4\root{ab}t^{2}}\,\dd t
\\[1cm]&=2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}}
\bracks{{1 \over 2\pars{ab}^{1/4}}
\ \overbrace{\int_{0}^{\infty}\exp\pars{-t^{2}}\,\dd t}^{\ds{\root{\pi} \over 2}}} 
=\ \color{#66f}{\large\half\,\root{\pi \over a}\expo{-2\root{ab}}}
\end{align}

