# Prove $\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{2}\sqrt{\frac{\pi}{a}}\mathrm{e}^{-2\sqrt{ab}}$

This is Example 8.4.1 from Chapter 8 (The Normal Integral) of Irresistible Integrals by Boros and Moll. The authors outlined a solution method, which I will provide in full here. My question: is there another way to obtain this result?

Here is the solution of Boros and Moll with their notation.

Let $$\mathrm{L}(a,b) := \int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x \label{eq:iic8-1} \tag{1}$$ Making the substitution $t=x\sqrt{a}$ yields $$\int\limits_{0}^{\infty} \mathrm{exp}(-ax^{2}-\frac{b}{x^{2}}) \mathrm{d} x = \frac{1}{\sqrt{a}} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{ab}{t^{2}}) \mathrm{d} t \label{eq:iic8-2} \tag{2}$$ Letting $ab=c$ we call the integral in equation \eqref{eq:iic8-2} $f(c)$, $$f(c) = \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \mathrm{d} t \label{eq:iic8-3} \tag{3}$$ so that $$\mathrm{L}(a,b) = \frac{f(ab)}{\sqrt{a}} \label{eq:iic8-4} \tag{4}$$ In equation \eqref{eq:iic8-3} we let $y=\sqrt{c}/t$ $$f(c) = \sqrt{c} \int\limits_{0}^{\infty} \mathrm{exp}(-y^{2}-\frac{c}{y^{2}}) y^{-2} \mathrm{d} y \label{eq:iic8-5} \tag{5}$$

Combining equations \eqref{eq:iic8-3} and \eqref{eq:iic8-5}, we have $$f(c) = \frac{1}{2} \int\limits_{0}^{\infty} \mathrm{exp}(-t^{2}-\frac{c}{t^{2}}) \left(1+\frac{\sqrt{c}}{t^{2}} \right) \mathrm{d} t \label{eq:iic8-6} \tag{6}$$ Now we let $s = t - \sqrt{c}/t$ $$f(c) = \frac{\mathrm{e}^{- 2\sqrt{c}}}{2} \int\limits_{-\infty}^{\infty} \mathrm{exp}(-s^{2}) \mathrm{d} s = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \lim_{z \to \infty} \mathrm{erf}(z) = \frac{\sqrt{\pi}}{2} \mathrm{e}^{- 2\sqrt{c}} \label{eq:iic8-7} \tag{7}$$ Combining equations \eqref{eq:iic8-1}, \eqref{eq:iic8-4}, and \eqref{eq:iic8-7} yields our result.

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\int_{0}^{\infty}\exp\pars{-ax^{2} - {b \over x^{2}}}\,\dd x} = \int_{0}^{\infty}\exp\pars{-\root{ab}\bracks{\root{a \over b}x^{2} + \root{b \over a}{1 \over x^{2}}}}\,\dd x \end{align}

With $\ds{\pars{a \over b}^{1/4}\,\,x = \expo{\theta}}$: \begin{align} &\color{#f00}{\int_{0}^{\infty}\exp\pars{-ax^{2} - {b \over x^{2}}}\,\dd x} = \int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{2\theta}} \pars{b \over a}^{1/4}\expo{\theta}\,\dd\theta \\[4mm] = &\ \pars{b \over a}^{1/4} \int_{-\infty}^{\infty}\exp\pars{-2\root{ab}\cosh\pars{2\theta}} \bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta \\[5mm] = &\ 2\pars{b \over a}^{1/4} \int_{0}^{\infty}\exp\pars{-2\root{ab}\bracks{2\sinh^{2}\pars{\theta} + 1}} \cosh\pars{\theta}\,\dd\theta \\[5mm] \stackrel{t\ \equiv\ \sinh\pars{\theta}}{=}\,\,\,\,\,\,&\ 2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}} \int_{0}^{\infty}\exp\pars{-4\root{ab}t^{2}}\,\dd t \\[5mm] = &\ 2\pars{b \over a}^{1/4}\exp\pars{-2\root{ab}} \bracks{{1 \over \root{4\root{ab}}}\,{\root{\pi} \over 2}} = \color{#f00}{{\root{\pi} \over 2}\,{\expo{-2\root{ab}} \over \root{a}}} \end{align}

• Nice solution. I pursued differently and obtained an extra factor of $\frac12$. – Mark Viola Aug 31 '16 at 2:49
• @Dr.MV I'm reading carefully your answer to see the 'difference'. – Felix Marin Aug 31 '16 at 2:59
• Thank you for catching the typo. I've edited. Oh, and +1 for this nice solution. -Mark – Mark Viola Aug 31 '16 at 3:08
• @Dr.MV You're welcome. Thanks. – Felix Marin Aug 31 '16 at 3:29
• @FelixMarin, this was the type of solution I was seeking. – poweierstrass Aug 31 '16 at 14:30

A slick way is to exploit Glasser's master theorem. For any $c>0$,

$$\begin{eqnarray*} e^{2\sqrt{c}}\int_{-\infty}^{+\infty}\exp\left(-x^2-\frac{c}{x^2}\right)\,dx&=&\int_{-\infty}^{+\infty}\exp\left[-\left(x-\frac{\sqrt{c}}{x}\right)^2\right]\,dx\\&=&\int_{-\infty}^{+\infty}e^{-x^2}\,dx = \sqrt{\pi}. \end{eqnarray*}$$

• Cool tool, but I didn't understand the last identity: you called $\left(x - \frac{\sqrt{c}}{x}\right) = x$? – Von Neumann Aug 30 '16 at 23:27
• @FourierTransform: please have a look at Glasser's master theorem linked here and you will understand the magic beyond that. – Jack D'Aurizio Aug 30 '16 at 23:30
• @JackD'Aurizio, good solution and thanks for the introduction to Glasser's master theorem. I will have to add it to my list of integration tricks. – poweierstrass Aug 31 '16 at 10:50
• The theorem is fantastic! Thank you. – Babak Aug 31 '16 at 15:15

Some of the steps in @FelixMarin's answer were not obvious. Here I expand upon his answer so that it will be easier to follow.

1. $\mathrm{e}^{z} = \cosh(z) + \sinh(z)$
2. $$\int\limits_{-\infty}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z = \int\limits_{-\infty}^{0} \cosh(z) + \sinh(z) \mathrm{d} z + \int\limits_{0}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z$$ In the first integral, let $z=-y$, $$\int\limits_{-\infty}^{0} \cosh(z) + \sinh(z) \mathrm{d} z = \int\limits_{0}^{\infty} \cosh(y) - \sinh(y) \mathrm{d} y$$ Addition yields, $$\int\limits_{-\infty}^{\infty} \cosh(z) + \sinh(z) \mathrm{d} z = 2\int\limits_{0}^{\infty} \cosh(z) \mathrm{d} z$$