4
$\begingroup$

Question : $$\int_{0}^{\infty}\exp(-x^2-1/x^2)dx=?$$

I think the answer is $\sqrt{ \pi/2 \cdot \exp(-2)}$

If I change $-x^2-1/x^2$ to $-(x-1/x)^2-2$ then the above integral becomes $$\exp(-2)\int_{0}^{\infty} \exp(-(x-1/x)^2)dx$$

integral of $\exp(-x^2)dx$ from $0$ to infinity = square root of $\pi/2$, but then is the

integral of $\exp(-(x-1/x)^2)dx$ from $0$ to infinity=integral of $\exp(-x^2)dx$ from $0$ to infinity ?

  • 0
    @tony I think you are correct but the formatting is making it difficult to read! This might help you, http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#117 learning tex is easy!2012-02-08

1 Answers 1

11

The integral you want to evaluate is $\mathrm e^{-2}I$, where $I=\int\limits_0^{+\infty}\mathrm e^{-(x-1/x)^2}\mathrm dx$. Let us compute $I$. The change of variable $z=1/x$ yields $z\gt0$ and $\mathrm dz=z^2\mathrm dx$, hence $I=\int\limits_0^{+\infty}\mathrm e^{-(z-1/z)^2}\mathrm dz/z^2$. Summing these two expressions of $I$, one gets $2I=\int\limits_0^{+\infty}\left(1+1/x^2\right)\mathrm e^{-(x-1/x)^2}\mathrm dx$. The change of variable $u=x-1/x$ yields $u$ in the whole real line and $\mathrm du=\left(1+1/x^2\right)\mathrm dx$, hence $2I=\int\limits_{-\infty}^{+\infty}\mathrm e^{-u^2}\mathrm du$. Finally, this last integral is $\sqrt\pi$, hence $$ \color{red}{\int\limits_0^{+\infty}\mathrm e^{-x^2-1/x^2}\mathrm dx=\frac{\sqrt\pi}{2\mathrm e^2}}. $$

  • 0
    Again, thank you very much for your detailed explanation. I could never solved this without your help!!2012-02-08
  • 0
    Why $z = 1/x$ then $dz = z^2dx$ instead of $dz = -z^2dx$?2015-07-29
  • 0
    Because the sign is absorbed in the integral being from 0 to +oo, not from +oo to 0.2015-07-29