I have to find $$I=\int_{0}^{\infty} \mathrm{e}^{-x^2-x^{-2}}\, dx $$ I think we could use $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2} $$ But I don't know how. Thanks.

  • $\begingroup$ This integral has definitely come up before on MSE but I cannot seem to find it. Maybe someone else can. The answer is $$\frac{\sqrt{\pi}}{2e^2}$$ according to Mathematica. $\endgroup$ – Cameron Williams Oct 7 '14 at 15:57
  • $\begingroup$ My hunch is that the substitution $y=x+1/x$ ought to be relevant.... $\endgroup$ – Greg Martin Oct 7 '14 at 16:21

Consider \begin{align} x^{2} + \frac{1}{x^{2}} = \left( x - \frac{1}{x} \right)^{2} +2 \end{align} for which \begin{align} I = \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = e^{-2} \, \int_{0}^{\infty} e^{-\left(x - \frac{1}{x}\right)^{2}} \, dx. \end{align} Now make the substitution $t = x^{-1}$ to obtain \begin{align} e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \, \frac{dt}{t^{2}}. \end{align} Adding the two integral form leads to \begin{align} 2 e^{2} I = \int_{0}^{\infty} e^{- \left( t - \frac{1}{t} \right)^{2}} \left(1 + \frac{1}{t^{2}} \right) \, dt = \int_{-\infty}^{\infty} e^{- u^{2}} \, du = 2 \int_{0}^{\infty} e^{- u^{2}} \, du = \sqrt{\pi}, \end{align} where the substitution $u = t - \frac{1}{t}$ was made. It is now seen that \begin{align} \int_{0}^{\infty} e^{-\left(x^{2} + \frac{1}{x^{2}}\right)} \, dx = \frac{\sqrt{\pi}}{2 e^{2}}. \end{align}


Let \begin{equation} J=\int_{0}^{\infty}e^{-x^2-x^{-2}}dx\\ =\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}dx\\ \end{equation} Let $x=\frac{1}{t}$, then we have \begin{equation} J=\frac{1}{e^2}\int_{0}^{\infty}e^{-(t-\frac{1}{t})^2}\frac{1}{t^2}dt\\ =\frac{1}{e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}\frac{1}{x^2}dx \end{equation} So taking the average we have \begin{equation} J=\frac{1}{2e^2}\int_{0}^{\infty}e^{-(x-\frac{1}{x})^2}(1+\frac{1}{x^2})dx \end{equation} Letting $z=x-\frac{1}{x}$, we have \begin{equation} J=\frac{1}{2e^2}\int_{0}^{\infty}e^{-z}dz=\frac{\sqrt{\pi}}{2e^2} \end{equation}

  • $\begingroup$ I think your proof is quite nearly the same as the first answer given. $\endgroup$ – Cameron Williams Oct 7 '14 at 16:33
  • $\begingroup$ @Cameron Williams, yes but the first answer was posted while I was typing the answer..........So I didn't see it........ $\endgroup$ – Raven Oct 7 '14 at 16:35
  • $\begingroup$ Fair enough. $\,$ $\endgroup$ – Cameron Williams Oct 7 '14 at 16:37

First recall that, for any 'nice' function, we have

$$ \int_{-\infty}^{+\infty}f\left(x-\frac{s}{x}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x, \quad s>0. \tag1 $$

Apply it to $f(x)=e^{-x^2}$, you get

$$ \int_{-\infty}^{+\infty}e^{-(x-s/x)^2}\mathrm{d}x=\int_{-\infty}^{+\infty} e^{-x^2} \mathrm{d}x=\sqrt{\pi}, \quad s>0. \tag2 $$


$$ \int_{-\infty}^{+\infty}e^{-x^2-s^2/x^{2}}\mathrm{d}x=\sqrt{\pi}\:e^{-2s}\tag3 $$ then put $s=1$ and use parity to obtain your integral.


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