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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

I think the answer is square root of $ \pi/2 \cdot \exp(-2)$. If I change $-x^2-1/x^2$ to $-(x-1/x)^2-2$ then the above integral becomes $\exp(-2) \cdot$ integral of $\exp(-(x-1/x)^2)dx$ from $0$ to infinity

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 ?

share|cite|improve this question
    
@tony I think you are correct but the formatting is making it difficult to read! This might help you, meta.math.stackexchange.com/questions/107/… learning tex is easy! – john w. Feb 8 '12 at 6:48
    
Thank you very much for your information!! – Tony Feb 8 '12 at 11:39
up vote 10 down vote accepted

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 $$ \color{red}{\int\limits_0^{+\infty}\mathrm e^{-x^2-1/x^2}\mathrm dx=\frac{\sqrt\pi}{2\mathrm e^2}}. $$

share|cite|improve this answer
    
Again, thank you very much for your detailed explanation. I could never solved this without your help!! – Tony Feb 8 '12 at 11:45
    
Why $z = 1/x$ then $dz = z^2dx$ instead of $dz = -z^2dx$? – GAVD Jul 29 '15 at 3:07
    
Because the sign is absorbed in the integral being from 0 to +oo, not from +oo to 0. – Did Jul 29 '15 at 6:14

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.