Gaussian integral $\int\limits_\mathbb{R}e^{-\lambda x^{2}} x^2 dx = (\frac{\pi}{4\lambda^3})^\frac{1}{2}$ Any clues on how to show/prove that the following result holds when evaluating this Gaussian integral would be greatly appreciated ... 
$$\int\limits_\mathbb{R}e^{-\lambda x^{2}} x^2 dx =  (\frac{\pi}{4\lambda^3})^\frac{1}{2}$$
I'm doing this in the context of an exercise where I'm  asked to derive the normal gaussian distribution based on a maximum entropy procedure. 
If the bounds on the integral were the positive real line, i.e.  $\int\limits_{0}^\infty e^{-\lambda x^{2}} x^2 dx $ , I see how I could integrate this by parts, taking $u=x$ and $dv= e^{-\lambda x} x dx$. But when I integrate by parts and evaluate on the whole real line the result is not defined (giving $\infty$ on the $uv$ term). 
Thanks! 
 A: Let 
$I(\lambda)=
\int_\mathbb{R}e^{-\lambda x^{2}}dx $. Then,
$$I^2 (\lambda)=  \int_\mathbb{R^2}e^{-\lambda (x^{2}+y^{2})}\>dxdy 
= 2\pi \int_0^\infty e^{-\lambda r^2}rdr = \frac\pi{\lambda}$$
which leads to $I(\lambda) = \sqrt{ \frac\pi{\lambda} }$ and 
$$\int_\mathbb{R}e^{-\lambda x^{2}} x^2 dx 
=  -\frac{dI(\lambda)}{d\lambda}=\left(\frac{\pi}{4\lambda^3}\right)^\frac{1}{2}$$
A: Gamma function is defined as $\Gamma(t)= \int_{0}^{\infty} z^{t-1}e^{-z}dz$.
First using symmetry let's rewrite our integral as 
$$2\int_{0}^{\infty}e^{-\lambda x^{2}}x^{2}dx$$
Now in the desired integral lets change variable $u=x^2$ (so that $x=\sqrt{u}$), then we'll have $$2\int_{0}^{\infty}e^{-\lambda u}u \frac{1}{2\sqrt{u}}du=\int_{0}^{\infty}e^{-\lambda u}\sqrt{u}du$$
Let's multiply (and of course divide) with $\lambda ^{3/2}$ (one under d sign, and 1/2 to combine with $\sqrt{u}$).
we'll get $$\lambda^{-\frac{3}{2}}\int_{0}^{\infty}e^{-\lambda u}\sqrt{\lambda u} \ d\lambda u$$
Now taking $\lambda u = z$, we'll get
$$\lambda^{-\frac{3}{2}}\int_{0}^{\infty}e^{-z}\sqrt{z} \ dz = \lambda^{-\frac{3}{2}} \Gamma\left(\frac{3}{2}\right)$$
(3/2 in gamma's argument as we have that $t-1=1/2$).
Gamma has the following beautiful property as you know. $$\Gamma\left(\frac{3}{2}\right)=\frac{1}{2}\Gamma\left(\frac{1}{2}\right)$$.
And one more famous thing about gamma is that $\Gamma \left(\frac{1}{2}\right)=\sqrt{\pi}$.
So you get the answer.
This is noway the shortest method. However it seem to me to be straightforward enough, to deal with normal (Gaussian) integrals.
