4
$\begingroup$

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 $a=0.$

$\endgroup$
12
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – lovelesswang Jun 24 '14 at 14:06
  • $\begingroup$ Do you mean the explicit resolution of the ODE? $\endgroup$ – Dario Jun 24 '14 at 14:08
  • $\begingroup$ No. I can read your solution. What I asked was the idea behind your solution. $\endgroup$ – lovelesswang Jun 24 '14 at 14:09
  • $\begingroup$ 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. $\endgroup$ – Dario Jun 24 '14 at 14:13
  • $\begingroup$ I get it. Cheers. $\endgroup$ – lovelesswang Jun 24 '14 at 14:15
2
$\begingroup$

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 $b=a^2$.

$\endgroup$
1
$\begingroup$

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

\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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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