Help with Gaussian integration problem please

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SUMMARY

The forum discussion focuses on solving the improper integral \(\int^{\infty}_{-\infty}x^{2}e^{-x^2}dx\) using the known result \(\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}\). The user explores integration by parts but struggles to find a suitable substitution. A key insight is provided: the derivative \(\frac{d}{dx}e^{-x^2} = -2xe^{-x^2}\) can be utilized to reformulate the integral, leading to a potential application of integration by parts.

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rdioface
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Help with Gaussian integration problem please :)

Homework Statement


Compute the improper integral
\int^{\infty}_{-\infty}x^{2}e^{-x^2}dx
given
\int^{\infty}_{-\infty}e^{-x^2}dx=\sqrt{\pi}.

Homework Equations


Just the rule for doubly-improper integrals I guess:
\int^{\infty}_{-\infty}f(x)dx=\lim_{a\rightarrow-\infty}\lim_{b\rightarrow\infty}\int^{b}_{a}f(x)dx

erf(x) is beyond the scope of this course and thus cannot be utilized in any way.

The Attempt at a Solution


I can't see any substitutions that would make things easier, and integration by parts doesn't seem useful (choosing to derive u=e^{-x^2} and integrate dv=x^{2}dx will never simplify or isolate e^{-x^2}, and you can't choose to integrate dv=e^{-x^2}dx and derive u=x^2 because we are only given the special case of \int^{\infty}_{-\infty}e^{-x^2}dx and not a general antiderivative.
 
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I'll give you a little hint.

\frac{d}{dx}e^{-x^2} = -2xe^{-x^2}

Hence,

\begin{aligned} \int_{-\infty}^\infty x^2 e^{-x^2} \text{d}x & = -\frac{1}{2}\int_{-\infty}^\infty x (-2x e^{-x^2}) \text{d}x \\ & = - \frac{1}{2}\int_{-\infty}^\infty x \frac{d}{dx}e^{-x^2}dx\end{aligned}

Can you now use integration by parts?
 

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