# Prove the equation

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 $a=0.$

$$I(a):=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\text{d}x$$ $$\frac{dI}{da}=\int_0^{\infty}e^{-\left(x^2+\dfrac{a^2}{x^2}\right)}\left(-\frac{2a}{x^2}\right)\text{d}x$$ 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 $$I(a)=\frac{e^{-2|a|}\sqrt{\pi}}{2}\ .$$

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

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 $b=a^2$.


\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}