$I = \int_{0}^\infty t^2 e^{-t^2/2} dt$ 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.

*

*Why not integrate by parts from the initial integral? One has  $$   
   I=-\frac12\int_0^\infty x \cdot \left(-2xe^{-x^2}   
   \right)dx=-\frac12\int_0^\infty x \cdot \left(e^{-x^2} \right)'dx $$
then one may use the Gaussian integral to conclude.

*Another path is to differentiate the gaussian identity $$
   \int_0^\infty e^{-tx^2}dx=\frac{\sqrt{\pi }}{2 \sqrt{t}} \qquad t>0,
   $$  with respect to $t$ and put $t=1$.

