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I was wondering how to determine $\int x^2 e^{-x^2}\mathrm dx?$ I tried to move the exponential into the integrator, and apply integration by parts, but that will make the integrand more complicated. Thank you!

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    Actually, I seem to recall that there is a nice simple formula for $\int p(x) e^{-x^2}$ in terms of the polynomial $p$.2011-09-12

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Uhm.. of course you can't calculate it since your integral involves error function which is not a elementary function. Indeed you have $\int x^2e^{-x^2}\mathrm d x=\frac{1}{2}\int x\cdot 2x e^{-x^2}=-\frac{1}{2}xe^{-x^2}+\frac{1}{2}\int e^{-x^2}=-\frac{1}{2}xe^{-x^2}+\frac{1}{4}\sqrt \pi\text{ erf}(x).$

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    @Mark: I'$m$ just playi$n$g devil's advocate. I like the list of elementary functions; I think you're right - it's not really arbitrary. I can't explain why I feel this way though. Maybe it's like you say in that only finite processes are used. If you allow a finite-dimensional (2-D) space into the conversation, then maybe you can get trig functions defined too. Maybe we should post a soft question about it :)2011-09-14
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If you integrate by parts using $u=x, dv=x\exp(-x^2)\;dx$ you can get an integral in terms of the error function. You can see it at Wolfram Alpha

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An alternative is to set $y = x^2$ to convert the integral into $ \int_0^\infty x^2 e^{-x^2} dx = \frac{1}{2}\int_0^\infty \sqrt{y} e^{-y} dy $ which can be expressed as $\frac{1}{2}\Gamma(\frac{3}{2}) = \frac{1}{4}\Gamma(\frac{1}{2}) = \frac{\sqrt{\pi}}{4}$ if the definite integral from $0$ to $\infty$ is meant. Otherwise, as mentioned by others, there is no expression in elementary terms for the antiderivative of $x^2 e^{-x^2}$

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    That's actually how I was taught how to do it,Dilip. Works nicely as long as you change the limits when necessary.2011-09-12