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How to calculate this improper integral $$ \int_{0}^{\infty}{\rm e}^{-\left(ax\ +\ b/x\right)^2}\,{\rm d}x\ {\large ?} $$

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Please refrain from using the displaystyle mode in the title. –  sos440 Mar 15 '13 at 1:47

2 Answers 2

up vote 8 down vote accepted

Note that

$$ \int_{0}^{\infty} e^{-\left( ax + \frac{b}{x} \right)^{2}} \, dx = e^{-4ab} \int_{0}^{\infty} e^{-\left( ax - \frac{b}{x} \right)^{2}} \, dx. $$

This shows that it suffices to consider the integral on the right-hand side. Associated to this we consider a more general situation. Let assume $a > 0, b > 0$ and $f$ is an integrable even function. With the substitution

$$ x = \frac{b}{at} \quad \Longrightarrow \quad dx = -\frac{b}{at^2} \, dt, $$

we obtain

$$ \int_{0}^{\infty} f\left( ax - \frac{b}{x} \right) \, dx = \int_{0}^{\infty} \frac{b}{at^2} f\left( at - \frac{b}{t} \right) \, dt. $$

Thus if we denote this common value by $I$, then

\begin{align*} 2aI = \int_{0}^{\infty} \left( a + \frac{b}{x^2} \right) f\left( ax - \frac{b}{x} \right) \, dx = \int_{-\infty}^{\infty} f (u) \, du, \end{align*}

where we used the substitution

$$ u = ax - \frac{b}{x}, \quad du = \left( a + \frac{b}{x^2} \right) \, dx. $$

Therefore we obtain the following identity.

$$ \int_{0}^{\infty} f\left( ax - \frac{b}{x} \right) \, dx = \frac{1}{2a} \int_{-\infty}^{\infty} f (x) \, dx $$

This gives us

$$ \int_{0}^{\infty} e^{-\left( ax + \frac{b}{x} \right)^{2}} \, dx = \frac{\sqrt{\pi}}{2a} e^{-4ab}. $$

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\infty}\expo{-\pars{ax\ +\ b/x}^{2}}\,\dd x:\ {\large ?}}$ \begin{align} &\color{#66f}{\Large\int_{0}^{\infty}\expo{-\pars{ax\ +\ b/x}^{2}}\,\dd x}\ =\ \overbrace{\int_{0}^{\infty} \exp\pars{-ab\bracks{\root{a \over b}x\ + \root{b \over a}\,{1 \over x}}^{2}} \,\dd x}^{\ds{\mbox{Set}\ \root{a \over b}x \equiv \expo{\theta}}} \\[3mm]&=\int_{-\infty}^{\infty}\exp\pars{-4ab\cosh^{2}\pars{\theta}} \root{b \over a}\expo{\theta}\,\dd\theta \\[3mm]&=\root{b \over a}\int_{-\infty}^{\infty} \exp\pars{-4ab\bracks{\sinh^{2}\pars{\theta} + 1}} \bracks{\cosh{\theta} + \sin\pars{\theta}}\,\dd\theta \\[3mm]&=2\root{b \over a}\expo{-4ab}\ \overbrace{\int_{0}^{\infty} \exp\pars{-4ab\sinh^{2}\pars{\theta}}\cosh{\theta}\,\dd\theta} ^{\ds{2\root{ab}\sinh\pars{\theta} \equiv t}} \\[3mm]&=2\root{b \over a}\expo{-4ab}\,{1 \over 2\root{ab}}\ \overbrace{\int_{0}^{\infty}\exp\pars{-t^{2}}\,\dd t}^{\ds{\root{\pi} \over 2}}\ =\ \color{#66f}{\Large{\root{\pi} \over 2a}\,\expo{-4ab}} \end{align}

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