Integral of a Gaussian (continued)

Ok, so last post we derived (not so rigorously) that the integral of a standard Gaussian as;

$$\int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}$$

and using a substitution of 

$$x=\sqrt{a}y$$ and $$dx=\sqrt{a}dy$$

 we obtain; $$\int_{-\infty}^{\infty} e^{-ax^2} dx = \sqrt{\frac{\pi}{a}}$$ .

Now let's extend that to the more general form of;

$$I=\int_{-\infty}^{\infty} x^2 e^{-ax^2} dx$$ where a is some constant.

We can recognise (or at least be told to recognise) that the argument of the integral; $$x^2 e^{-ax^2}$$ is equal to

$$-\frac{\partial}{\partial a} e^{-ax^2}$$

Putting this together, we get;

$$I=\int_{-\infty}^{\infty}-\frac{\partial}{\partial a} e^{-ax^2} dx$$

We can swap the partial derivative and Integration sign (provided the variable we are integrating over is independent of a and the integrand is continuous)

We get; $$I= -\frac{\partial}{\partial a} \int_{-\infty}^{\infty} e^{-ax^2} dx = -\frac{\partial}{\partial a}\sqrt{\frac{\pi}{a}}$$

$$=\frac{1}{2a} \sqrt{\frac{\pi}{a}}$$

Therefore;  $$\int_{-\infty}^{\infty} x^2 e^{-ax^2} dx =\frac{1}{2a} \sqrt{\frac{\pi}{a}}$$

This technique can be extended for different powers of x - provided they are even.