I've been struggling with this question for a while now.
Question is on a normal distribution $ f(x)= \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} $, where the $E(X)$ was asked to be found, and hence prove that $\operatorname{Var}(X) = 0$.
After digging around, I discovered that $\int e^{-x^2}\,\mathrm dx$ isn't an elementary function, however is it still possible to evaluate $\int_{-\infty}^{+\infty} x^2e^{-x^2}\,\mathrm dx$?
Yes. If $f(x)=e^{-x^2}$ then $f'(x)=-2xe^{-x^2}$ and $f''(x)=(-2+4x^2)e^{-x^2}$. Hence $$\int_{-a}^ax^2e^{-x^2}\,\mathrm dx = \int_{-a}^a(\tfrac14f''(x)+\tfrac12f(x))\,\mathrm dx=f'(a)-f'(-a)+ \frac12\int_{-a}^af(x)\,\mathrm dx.$$ As we let $a\to\infty$, all that remains is $$ \int_{-\infty}^\infty x^2e^{-x^2}\,\mathrm dx = \frac12 \int_{-\infty}^\infty e^{-x^2}\,\mathrm dx=\sqrt{\frac\pi2}$$