Dealing w/slight modification of Gaussian integral?

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SUMMARY

The discussion focuses on evaluating the integral \(\int_{-\infty}^{\infty}x^2e^{-\alpha x^2} dx\) where \(\alpha > 0\). The Gaussian integral is known to equal \(\sqrt{\pi}\), but the challenge arises in handling the parameter \(\alpha\). A recommended approach is to use the substitution \(y=\sqrt{\alpha} x\) to simplify the integral, followed by applying integration by parts to arrive at the correct solution.

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Zacarias Nason
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Hey, folks. I'm doing a problem wherein I have to evaluate a slight variation of the Gaussian integral for the first time, but I'm not totally sure how to go about it; this is part of an integration by parts problem where the dv is similar to a gaussian integral:

\int_{-\infty}^{\infty}x^2e^{-\alpha x^2} dx, \ \text{where} \ \alpha > 0. My question is, how should I deal with the alpha in the exponent? I know the Gaussian integral equals
\sqrt{\pi}

But how do I deal with that alpha?

I thought I found my answer in a handout, but when I applied it, I ended up still being wrong; I attempted to integrate by parts and I got zero, the two terms cancelling each other out; what gives?
 
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Zacarias Nason said:
Hey, folks. I'm doing a problem wherein I have to evaluate a slight variation of the Gaussian integral for the first time, but I'm not totally sure how to go about it; this is part of an integration by parts problem where the dv is similar to a gaussian integral:

\int_{-\infty}^{\infty}x^2e^{-\alpha x^2} dx, \ \text{where} \ \alpha > 0. My question is, how should I deal with the alpha in the exponent? I know the Gaussian integral equals
\sqrt{\pi}

But how do I deal with that alpha?

I thought I found my answer in a handout, but when I applied it, I ended up still being wrong; I attempted to integrate by parts and I got zero, the two terms cancelling each other out; what gives?
Try a substitution to get back to the basic form: ##y=\sqrt{\alpha} x##
Then integration by parts.
 
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