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I'm struggling with this integral:

$\int_{-\infty}^{\infty}x^{2}e^{-ax^{2}}dx = \frac{\sqrt{\pi}}{2a^{3/2}}$

Is there a way to do this integration without using integration by parts and then explicitly relying on the Gaussian integral, instead using the Gamma function $\Gamma(a)=\int_{0}^{\infty}x^{a-1}e^{-x} dx$ ? I had tried the substitution $u=ax^2$ to give $\frac{2}{a}\int_{0}^{\infty}ue^{-u}du$ but that doesn't seem to get me far.

Thanks in advance for any help or hints !

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    Hi KCd. Thanks. I am curious. So we would evaluate this: $\int_{-\infty}^{\infty}\frac{\partial }{\partial a}x^{2}e^{-ax^{2}}dx$ ?2012-03-27

2 Answers 2

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Summarising the comments, after substitution of $u=ax^2$ and $du=2axdx$, you'll get $ \int_0^\infty x^2 e^{-ax^2}dx=\frac{1}{2a^{3/2}}\int_0^\infty u^{3/2-1}e^{-u} du=\frac{\Gamma(3/2)}{2a^{3/2}}=\frac{\sqrt{\pi}}{4a^{3/2}}. $

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You can write the integral in this form: $d/da \int{\exp{(-ax^2)}dx}$ And we know that $\int{\exp{(-ax^2)}dx}=\sqrt{\pi/a}$ So $\int{-x^2\exp{(-ax^2)}dx}=d/da (\sqrt{\pi/a})=\sqrt{\pi}/2a^{3/2}$