How to show $e^{-\sqrt{2\lambda }x}=\int_0^{\infty} e^{-\lambda u} (2\pi u^3)^{-1/2} x \exp(-\frac{x^2}{2u}) du$?

Question

I want to show the following integral:

$$e^{-\sqrt{2\lambda }x}=\int_0^{\infty} e^{-\lambda u} (2\pi u^3)^{-1/2} x \exp(-\frac{x^2}{2u}) du$$

Could anyone give me some hints on evaluating the right hand side?

Answer 1

By the change of variable $v=\dfrac{ x}{\sqrt{2}\cdot u^{1/2}}$ we obtain $dv=-\dfrac{x}{2\sqrt{2}}\cdot (u^{3})^{-1/2}du$ and $$ \begin{align} \int_0^{\infty}\! e^{-\lambda u} (2\pi u^3)^{-1/2} x \exp(-\frac{x^2}{2u}) du&=\frac{2}{\sqrt{\pi}}\!\!\int_0^{\infty}e^{ -v^2-\lambda x^2/(2v^2)}dv \\&=\frac{1}{\sqrt{\pi}}\!\int_{-\infty}^{\infty}e^{ -v^2-\lambda x^2/(2v^2)}dv \tag1 \end{align} $$ one may then recall that, for any integrable function $f$, we have (see here)

$$ \int_{-\infty}^{+\infty}f\left(v-\frac{a}{v}\right)\mathrm{d}v=\int_{-\infty}^{+\infty} f(v)\: \mathrm{d}v, \quad a>0. \tag2 $$

Applying it to $f(v)=e^{-v^2}$, gives

$$ \int_{-\infty}^{+\infty}e^{-(v-a/v)^2}\mathrm{d}v=\int_{-\infty}^{+\infty} e^{-v^2} \mathrm{d}v=\sqrt{\pi}, \quad a>0. \tag3 $$

Thus from $(3)$, expanding the square, we get

$$ \int_{-\infty}^{+\infty}e^{-v^2-a^2/v^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2a}\tag4 $$ then, using $(1)$, we deduce the announced result.

Answer 2

$\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{\quad u \equiv A\expo{2\theta}\,,\quad A > 0.\quad}$ The $\ds{A}$-constant will be chosen in a 'convenient way' as shown below.

\begin{align} &\color{#f00}{\int_{0}^{\infty}\expo{-\lambda u}\pars{2\pi u^{3}}^{-1/2}\,x\, \exp\pars{-\,{\phantom{^{2}}x^{2} \over 2u}}\,\dd u} \\[5mm] = &\ {x \over \root{2\pi}}\int_{0}^{\infty}A^{-3/2}\expo{-3\theta} \exp\pars{-\lambda A\expo{2\theta} -\,{\phantom{^{2}}x^{2} \over 2}\,A^{-1} \expo{-2\theta}}\,\pars{2A\expo{2\theta}}\,\dd\theta\label{1}\tag{1} \end{align} We'll choose $\ds{A > 0}$ to satisfies $\ds{\lambda A = x^{2}A^{-1}/2 \implies A = \verts{x}/\root{2\lambda}}$. \eqref{1} is reduced to: \begin{align} &\color{#f00}{\int_{0}^{\infty}\expo{-\lambda u}\pars{2\pi u^{3}}^{-1/2}\,x\, \exp\pars{-\,{\phantom{^{2}}x^{2} \over 2u}}\,\dd u} \\[5mm] = &\ {2x \over \root{2\pi}}\pars{\verts{x} \over \root{2\lambda}}^{-1/2} \int_{-\infty}^{\infty} \exp\pars{-\lambda\,{\verts{x} \over \root{2\lambda}}\,\bracks{2\cosh\pars{2\theta}}}\expo{-\theta}\,\dd\theta \\[5mm] = &\ 2^{3/4}{\lambda^{1/4} \over \root{\pi}}{x \over \verts{x}^{1/2}} \int_{-\infty}^{\infty} \exp\pars{-\root{2\lambda}\verts{x}\bracks{2\sinh^{2}\pars{\theta} + 1}} \bracks{\cosh\pars{\theta} - \sinh\pars{\theta}}\,\dd\theta \end{align}

Note that $\ds{\quad\cosh\pars{2\theta} = 2\sinh^{2}\pars{\theta} + 1\quad}$ and $\ds{\quad\expo{-\theta} = \cosh\pars{\theta} - \sinh\pars{\theta}}$.

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Then, \begin{align} &\color{#f00}{\int_{0}^{\infty}\expo{-\lambda u}\pars{2\pi u^{3}}^{-1/2}\,x\, \exp\pars{-\,{\phantom{^{2}}x^{2} \over 2u}}\,\dd u} \\[5mm] = &\ 2^{3/4}{\lambda^{1/4} \over \root{\pi}}\,\mrm{sgn}\pars{x}\verts{x}^{1/2} \exp\pars{-\root{2\lambda}\verts{x}}\times \\[3mm] &\ \bracks{2\int_{0}^{\infty} \exp\pars{-2\root{2\lambda}\verts{x}\sinh^{2}\pars{\theta}}\cosh\pars{\theta} \,\dd\theta} \\[1cm] \stackrel{t\ =\ \sinh\pars{\theta}}{=}\,\,\,&\ 2^{7/4}{\lambda^{1/4} \over \root{\pi}}\,\mrm{sgn}\pars{x}\verts{x}^{1/2} \exp\pars{-\root{2\lambda}\verts{x}} \int_{0}^{\infty}\exp\pars{-2\root{2\lambda}\verts{x}t^{2}}\,\dd t \\[5mm] = &\ 2^{7/4}{\lambda^{1/4} \over \root{\pi}}\,\mrm{sgn}\pars{x}\verts{x}^{1/2} \exp\pars{-\root{2\lambda}\verts{x}} \bracks{\root{\pi} \over 2^{7/4}\lambda^{1/4}\verts{x}^{1/2}} \\[5mm] = &\ \color{#f00}{\mrm{sgn}\pars{x}\exp\pars{-\root{2\lambda}\verts{x}}} \end{align}