Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Prove that $$\int_0^{\infty}\exp\left(-\left(x^2+\dfrac{a^2}{x^2}\right)\right)\text{d}x=\frac{e^{-2a}\sqrt{\pi}}{2}$$ Assume that the equation is true for $a=0.$

share|improve this question
add comment

3 Answers 3

up vote 12 down vote accepted

$$I(a):=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\text{d}x$$ $$\frac{dI}{da}=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\left(-\frac{2a}{x^2}\right)\text{d}x$$ Now substitute $y=\frac{a}{x}$, so $dy=-\frac{a}{y^2}$: $$\frac{dI}{da}=2\int_{\infty}^{0}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2\int_{0}^{\infty}e^{-\left(\dfrac{a^2}{y^2}+y^2\right)}\text{d}y=-2I$$ To obtain $I$ you just have to solve the simple ODE: $$\frac{dI}{da}=-2I$$ with initial condition given by $$I(0)=\frac{\sqrt{\pi}}{2}\ .$$ This gives you $$I(a)=\frac{e^{-2a}\sqrt{\pi}}{2}\ .$$

share|improve this answer
    
+1. This is a nice answer. –  Tunk-Fey Jun 24 at 13:58
    
Thanks. It is very clever. But can you elaborate a bit more on how you came up with the solution translating the original problem into solving an ODE? –  lovelesswang Jun 24 at 14:06
    
Do you mean the explicit resolution of the ODE? –  Dario Jun 24 at 14:08
    
No. I can read your solution. What I asked was the idea behind your solution. –  lovelesswang Jun 24 at 14:09
    
The idea behind the solution is the technique of differentiation under the integral sign(en.wikipedia.org/wiki/…): when you have an integral that depend on a parameter, you can derive the integral w.r.t. that parameter and sometimes you obtain a much simpler integral. In this particular case the trick is to notice that after the derivation, getting rid of the extra factor you obtain by a change of variable you get again the first integral. This allow you to find a relation between $I$ and its derivative, i.e. the ODE. –  Dario Jun 24 at 14:13
show 1 more comment

In general $$ \begin{align} \int_{x=0}^\infty \exp\left(-ax^2-\frac{b}{x^2}\right)\,dx&=\int_{x=0}^\infty \exp\left(-a\left(x^2+\frac{b}{ax^2}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x^2-2\sqrt{\frac{b}{a}}+\frac{b}{ax^2}+2\sqrt{\frac{b}{a}}\right)\right)\,dx\\ &=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2-2\sqrt{ab}\right)\,dx\\ &=\exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx\\ \end{align} $$ The trick to solve the last integral is by setting $$ I=\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx. $$ Let $t=-\frac{1}{x}\sqrt{\frac{b}{a}}\;\rightarrow\;x=-\frac{1}{t}\sqrt{\frac{b}{a}}\;\rightarrow\;dx=\frac{1}{t^2}\sqrt{\frac{b}{a}}\,dt$, 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 $t=x\;\rightarrow\;dt=dx$, 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 $$ I=\frac{1}{2}\int_{s=-\infty}^\infty e^{-as^2}\,ds=\frac{1}{2}\sqrt{\frac{\pi}{a}}, $$ where $I$ is a Gaussian integral. Thus $$ \begin{align} \exp(-2\sqrt{ab})\int_{x=0}^\infty \exp\left(-a\left(x-\frac{1}{x}\sqrt{\frac{b}{a}}\right)^2\right)\,dx &=\large\color{blue}{\frac12\sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}}. \end{align} $$ In our case, put $a=1$ and $b=a^2$.

share|improve this answer
add comment

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

\begin{align} &\color{#66f}{\large\int_{0}^{\infty}\exp\pars{-\bracks{x^{2} + {a^{2} \over x^2}}}\,\dd x}\ =\ \overbrace{\int_{0}^{\infty} \exp\pars{-\verts{a}\bracks{% {x^{2} \over \verts{a}} + {\verts{a} \over x^2}}}\,\dd x}^{\ds{x \equiv \root{\verts{a}}\expo{\theta}}} \\[3mm]&=\int_{-\infty}^{\infty}\expo{-2\verts{a}\cosh\pars{2\theta}} \root{\verts{a}}\expo{\theta}\,\dd\theta \\[3mm]&=\root{\verts{a}}\int_{-\infty}^{\infty} \expo{-2\verts{a}\bracks{2\sinh^{2}\pars{\theta} + 1}} \bracks{\cosh\pars{\theta} + \sinh\pars{\theta}}\,\dd\theta \\[3mm]&=\root{\verts{a}}\expo{-2\verts{a}}\ \overbrace{\int_{-\infty}^{\infty} \expo{-4\verts{a}\sinh^{2}\pars{\theta}}\cosh\pars{\theta}\,\dd\theta} ^{\ds{t\ \equiv\ \sinh\pars{\theta}}}\ =\root{\verts{a}}\expo{-2\verts{a}}\int_{-\infty}^{\infty} \expo{-4\verts{a}t^{2}}\,\dd t \\[3mm]&=\root{\verts{a}}\expo{-2\verts{a}}\,{1 \over 2\root{\verts{a}}}\ \overbrace{\int_{-\infty}^{\infty}\expo{-t^{2}}\,\dd t}^{\ds{=\ \root{\pi}}}\ =\ \color{#66f}{\Large{\root{\pi} \over 2}\,\expo{-2\verts{a}}} \end{align}

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.