$\displaystyle \int_{0}^{\infty}x^2e^{-x^2/(2\theta^2)}dx$
Hello,
Note that this integral is half $\displaystyle \int_{\mathbb{R}} x^2 e^{-x^2/2\sigma^2} ~dx$
but then you can recognize $\displaystyle e^{-x^2/2\sigma^2}$ as nearly the pdf of a random variable Z following $\displaystyle N(0,\sigma^2)$
Try to find what $\displaystyle \sigma^2=E[Z^2]$ is, in terms of an integral and compute that integral.