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I have been able to calculate the integral of

$$\int^\infty_\infty x^2e^{-x^2/2}$$

and there is a lot of information online about integration with even powers of $x$.
However I have been unable to calculate:

$$\int^\infty_\infty x^3e^{-x^2/2}.$$

The closest I have come to finding a solution is
$$\int^\infty_0 x^{2k+1}e^{-x^2/2} = \frac{k!}{2}$$

Which I found here.

Any help with solving this integral would be great.

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    $\begingroup$ Substitute $u=x^2/2$, and integrate by parts $\endgroup$ – Jakobian Oct 31 '18 at 15:42
  • $\begingroup$ @Jakobian got it, thank you I was thinking too much in terms of Gaussian integrals and missed the obvious. $\endgroup$ – Matthew Oct 31 '18 at 15:45
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Do you mean $\int_{-\infty}^\infty x^3e^{-\frac{x^2}2}\,\mathrm dx$? It is $0$, since the function is an odd function and integrable (it is the product of a polynomial function with $e^{-\frac{x^2}2}$).

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    $\begingroup$ Just because it's odd, isn't enough to say that the integral is $0$. The integral doesn't have to exists $\endgroup$ – Jakobian Oct 31 '18 at 15:45
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    $\begingroup$ @Jakobian I've edited my answer. I hope the you find it acceptable now. $\endgroup$ – José Carlos Santos Oct 31 '18 at 15:47
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Substitute $$u=x^2$$ then we get $$\frac{1}{2}\int e^{-u/2}udu$$ and then use Integration by parts.

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