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.
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, $$ \begin{align} \int_0^\infty e^{-a^2y^2-\frac{b^2}{y^2}}\,\mathrm{d}y &=\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}(u+\sqrt{u^2+4ab})\\ &=\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}u +\frac1{2ae^{2ab}}\int_{-\infty}^\infty e^{-u^2}\,\mathrm{d}\sqrt{u^2+4ab}\\ &=\frac{\sqrt\pi}{2ae^{2ab}}+0 \end{align} $$
$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\expo{-ay^{2} - b/y^{2}}\,\dd y:\ {\large ?}.\qquad a, b > 0}$.
\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}
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 $$ \begin{align} \int_{v=0}^\infty \exp\left(-av^2-\frac{b}{v^2}\right)\,dv&=\exp(-2\sqrt{ab})\int_{v=0}^\infty \exp\left(-a\left(v-\frac{1}{v}\sqrt{\frac{b}{a}}\right)^2\right)\,dv\\ &=\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}. \end{align} $$
