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Prove that $\displaystyle\int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha^3}}$.

You're allowed to use the formula $\displaystyle\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$.

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  • $\begingroup$ Something is wrong here: if $\alpha > 0$ the left-hand side is a divergent integral and if $\alpha < 0$ the right-hand side isn't real. $\endgroup$ – Umberto P. Jun 15 '15 at 13:37
  • $\begingroup$ @Umberto P.: Thanks. I forgot the minus sign. Edited. $\endgroup$ – Omega Force Jun 15 '15 at 13:45
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This requires Integration by parts. Note that

$$ \int u(x)v'(x) dx = u(x)v(x) - \int u'(x)v(x) dx $$

If we let $$ v'(x) = -2axe^{-ax^2}$$ And we let

$$ u(x) = -\frac{1}{2a}x $$

Then it is easy to see that

$$ u(x)v'(x) = x^2e^{-ax^2} $$

Note that by the chain rule

$$ v(x) = e^{-ax^2} $$

Thus

$$ \int_{-\infty}^{\infty} x^2e^{-ax^2} = -\frac{x}{2a}e^{-ax^2} | _{-\infty}^{\infty} - \frac{1}{2a} \int_{-\infty}^{\infty} e^{ax^2} dx $$

This is equal to

$$ 0+\frac{1}{2a} \sqrt{\frac{\pi}{a}} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}}$$

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Differentiate the allowed formula once with respect to $\alpha$.

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Taking the derivative of the given identiy with respect to $\alpha$: $$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$$ $$\implies {\partial\over\partial\alpha}\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = {\partial\over\partial\alpha}\sqrt{\frac{\pi}{\alpha}}$$ $$\implies \int_{-\infty}^{\infty} {\partial\over\partial\alpha}e^{-\alpha x^2} dx = -{1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$

$$\implies \int_{-\infty}^{\infty} -x^2e^{-\alpha x^2} dx = -{1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$

$$\implies \int_{-\infty}^{\infty} x^2e^{-\alpha x^2} dx = {1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$

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  • $\begingroup$ The step $$\frac{\partial}{\partial \alpha} \int_{-\infty}^\infty = \int_{-\infty}^\infty\frac{\partial}{\partial \alpha} $$ isn't trivial. $\endgroup$ – Umberto P. Jun 15 '15 at 13:55
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    $\begingroup$ well no, of course not, but it works! Feel free to edit the steps in if you like... $\endgroup$ – danimal Jun 15 '15 at 13:56

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