What is a good method for evaluating $\int_{-\infty}^{\infty} x^2 e^{-bx^2}dx$ only with elementary functions? I want to evaluate
$$\mathbf{I} = \int_{-\infty}^{\infty} x^2e^{-bx^2}dx \quad \quad (1)$$
If possible with real methods. I know that it can be evaluated with integration by parts but this uses non-elementary functions. Since this is in an example of a physics textbook that calls only for differential equations and linear algebra (Griffiths Quantum Mechanics) i suspect that there is a easier way to solve this. 
I tried defining a function
$$g(y) := \int_{-\infty}^{\infty}x^2e^{-bx^2}e^{\frac{-y}{x^2}}dx$$
And then i would want to evaluate $g(0)$. I tried to do this by differenting:
$$g'(y) = - \int_{-\infty}^{\infty} e^{-bx^2}e^{\frac{-y}{x^2}}dx \quad \quad (2)$$
And as we can see, $g(y) \rightarrow 0$ as $y \rightarrow \infty$, and so
$$\int_{0}^{\infty} g'(y)dy = g(\infty) - g(0) = - \mathbf{I}$$
The problem with this method is that the integral $(2)$ is not easy (in fact, i think it is harder than $(1)$) to evaluate. So, is there any easier way, using elementary functions, to solve this?
Thanks.
 A: 
I thought it would be instructive to present a way forward that does not rely on differentiating under the integral, which requires justification. 
For example, we can show that the integral $\int_{-\infty}^\infty x^2e^{-bx^2}\,dx$ converges uniformly for $b\ge \delta>0$ and appeal to a theorem for differentiating under the integral for improper integrals.


We can avoid the justification of differentiating under the integral and instead, integrate by parts with $u=x$ and $v=-\frac{e^{-bx^2}}{2b}$.   Proceeding, we find that for $b>0$
$$\begin{align}
\int_{-\infty}^\infty x^2e^{-bx^2}\,dx&=\underbrace{\left.\left(-\frac{xe^{-bx^2}}{2b}\right)\right|_{-\infty}^{\infty}}_{=0}+\frac1{2b}\underbrace{\int_{-\infty}^\infty e^{-bx^2}\,dx}_{=\sqrt{\frac\pi b}}\\\\
&=\frac{\sqrt\pi}{2b^{3/2}}
\end{align}$$
And we are done!  

Note that there was no appeal to special functions.

A: Define $$G(b) := \int_{-\infty}^{\infty} e^{-bx^2}~\mathrm dx = \sqrt{\frac{\pi}{b}}.$$
Now find $-G'(b)$.
A: Through Feynman trick we can write
$$\int_{-\infty}^{+\infty} -\frac{d}{db} e^{-bx^2}\ dx = -\frac{d}{db}\int_{-\infty}^{+\infty} e^{-bx^2}\ dx$$
The latter integral has to be known, which is $-\sqrt{\frac{\pi}{b}}$.
From this, 
$$\frac{d}{db}\ \left(-\sqrt{\frac{\pi}{b}}\right) = \frac{\sqrt{\pi }}{2 b^{3/2}}$$
Provided that $\Re(b) >0$
