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Inegral evaluation

  1. Jan 24, 2010 #1
    This is not a homework question but I am trying to follow the proof on wolfram that [tex] \int_{-\infty }^{ \infty }{e}^{{x}^{-2}} dx = \sqrt{\pi} [/tex] and I am haveing trouble at one point where they state [tex] \int_{0 }^{ \infty }r{e}^{{r}^{-2}} dr = \left[- \frac{ 1}{ 2} {e }^{ {-r }^{2 } } \right ] \infty \rightarrow 0 [/tex].

    How the hell do they make this jump? The only way i would know to evaluate this integral is to use integration by parts and this would eventually leave you with another integral that is itself the gaussian. I am sure there is an easy were than parts. Does anyone have any insight here?

    btw the[tex] \infty \rightarrow 0 [/tex] is suppose to mean the integral is evaluated from 0 to infinity but i dont know how to do this proper in latex.
  2. jcsd
  3. Jan 24, 2010 #2


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    Hi dionysian! :smile:

    Use a substitution: u = r2. :wink:
  4. Jan 24, 2010 #3
    Thanks... I am a little rusty...
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