Calculate the integral of x^2 e^(-x^2) from -infinity to + infinity

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SUMMARY

The integral of \( x^2 e^{-x^2} \) from \(-\infty\) to \(+\infty\) can be solved using differentiation under the integral sign. Specifically, the integral can be expressed as \(\frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} x^2 e^{\frac{-x^2}{2 \sigma^2}} dx\). By differentiating the Gaussian integral \(\int_{-\infty}^\infty e^{-\sigma x^2} dx = \sqrt{\frac{\pi}{\sigma}}\) with respect to \(\sigma\), one can derive the solution without resorting to integration by parts, which leads to the error function (Erf) that was not covered in the discussion.

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\frac{1}{\sqrt{2 \pi \sigma^2}} \int_{-\infty}^{\infty} x^2 e^{\frac{-x^2}{2 \sigma^2}} dx
I think just slving this would be fine too

\int_{-\infty}^{\infty} x^2 e^{-x^2} dx

what is the trick to solving this?/

Cnat integrate by parts because that would yeild Erf function which i have no been taught

I was told that there was some trick to differentiate both sides by sigma... but I am not relly sure...?
 
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You know:

\int_{-\infty}^\infty e^{-\sigma x^2} dx = \sqrt{\frac{\pi}{\sigma}}

Differentiate both sides with respect to sigma.
 

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