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

Question

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

Answer 1

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

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

Answer 2

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

Answer 3

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

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

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