Prove that $\int_{-\infty}^{\infty} x^2 e^{-\alpha x^2} dx = \frac{1}{2} \sqrt{\frac{\pi}{\alpha^3}}$ 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 $\displaystyle\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$.
 A: Differentiate the allowed formula once with respect to $\alpha$.
A: 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
$$ 0+\frac{1}{2a} \sqrt{\frac{\pi}{a}} = \frac{1}{2} \sqrt{\frac{\pi}{a^3}}$$
A: Taking the derivative of the given identiy with respect to $\alpha$:
$$\int_{-\infty}^{\infty} e^{-\alpha x^2} dx = \sqrt{\frac{\pi}{\alpha}}$$
$$\implies {\partial\over\partial\alpha}\int_{-\infty}^{\infty} e^{-\alpha x^2} dx =  {\partial\over\partial\alpha}\sqrt{\frac{\pi}{\alpha}}$$
$$\implies \int_{-\infty}^{\infty} {\partial\over\partial\alpha}e^{-\alpha x^2} dx =  -{1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$
$$\implies \int_{-\infty}^{\infty} -x^2e^{-\alpha x^2} dx =  -{1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$
$$\implies \int_{-\infty}^{\infty} x^2e^{-\alpha x^2} dx = {1\over2}\sqrt{\frac{\pi}{\alpha^3}}$$
