$I = \int_{0}^\infty t^2 e^{-t^2/2} dt$ Q: Evaluate the integral $I = \int_\limits{0}^\infty t^2 e^{-t^2/2} dt$
Hint, write $I^2$ as the following iterated integral and convert to polar coordinates:
\begin{align*}
  I^2 &= \int_\limits{0}^\infty \int_\limits{0}^\infty x^2 e^{-x^2/2} \cdot y^2 e^{-y^2/2} \, dx \, dy \\
\end{align*}
I can see the final answer is $\frac{\pi}{2}$ but I don't see how to get this.
This problem is very similar to the Gaussian Integral: $I = \int_\limits{-\infty}^\infty e^{-x^2} dx = \sqrt{\pi}$. I can follow the derivation to this.
The Gaussian Integral technique of converting to polar coordinates doesn't seem to work as cleanly on this problem.
I can convert to polar coordinates:
\begin{align*}
  I^2 &= \int_\limits{0}^\infty \int_\limits{\pi/2}^\pi r^5 \sin^2 \theta \cos^2 \theta e^{-r^2/2} \, d\theta \, dr \\
\end{align*}
That doesn't look easy to evaluate.


I thought it might be instructive to present a way forward that relies on "Feynman's Trick" for differentiating under the integral sign.  

Proceeding, we let $I(a)$ be the integral given by
$$\begin{align}
I(a)&=\int_0^\infty e^{-at^2}\,dt\\\\
&=\frac12\sqrt{\frac{\pi}{a}} \tag 1
\end{align}$$
Differentiating $(1)$ reveals
$$-\int_0^\infty t^2 e^{-at^2}\,dt=-\frac14\frac{\sqrt\pi}{a^{3/2}} \tag 2$$
Setting $a=1/2$ in $(2)$ and multiplying by $-1$ yields the coveted result

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^2e^{-t^2/2}\,dt=\sqrt{\frac\pi{2}}}$$

And we are done!
