It says that you can use the symmetry of the normal standard distribution and that $ \int_{-\infty}^{\infty} \! \frac{1}{\sqrt{2\pi}}e^{-x^2/2} \, dx = 1$ to show that E(X)=$ \int_{-\infty}^{\infty} \! e^{-x^2} \, dx =\frac{\sqrt{\pi}}{2} $
I was trying to transform E(X)=$ \int_{0}^{\infty} \! 2xe^{-x^2} \, dx $ into
$\int_{0}^{\infty} \! \frac{1}{\sqrt{2\pi}}e^{\frac{e^{-x^2}}{2}} \, dx $ to use the symmetry but i couldn't make it.
Could you help me?