Integral Evaluation: Jumping from 0 to ∞

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SUMMARY

The integral evaluation of \(\int_{0}^{\infty} r e^{-r^{-2}} dr\) simplifies to \(\left[-\frac{1}{2} e^{-r^2}\right]_{0}^{\infty} = 0\) through the substitution \(u = r^2\). This substitution transforms the integral into a more manageable form, allowing for straightforward evaluation without resorting to integration by parts. The discussion highlights the importance of recognizing substitution techniques in integral calculus.

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

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\infty \rightarrow 0 is suppose to mean the integral is evaluated from 0 to infinity but i don't know how to do this proper in latex.
 
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dionysian said:
… 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 …

Hi dionysian! :smile:

Use a substitution: u = r2. :wink:
 
Thanks... I am a little rusty...
 

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