how to compute $E(X^4)$ when $X$ follow the normal law $N(0,1)$ Suppose $X$ follow the normal law $N(0,1)$ 
We have the density $f= \frac{1}{\sqrt{\sigma^2 2\pi}} e^{- \frac{(x-m)^2}{2\sigma^2}}$
We want to compute $E(X^4)$
We have by definition that $\displaystyle E(X^4) = \frac{1}{\sqrt{2\pi}} \int_\infty^\infty{e^{-x^2/2}} x^4\,dx $
But i dont know how to continue 
Thank you for helping me 

\begin{align}
& \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^4 e^{-x^2/2}\,dx \\[10pt]
= {} & 2\cdot\frac 1 {\sqrt{2\pi}} \int_0^\infty x^4 e^{-x^2/2} \, dx & & \text{since the integral is of an even} \\ & & & \text{function over an interval that} \\[8pt]
& & & \text{is symmetric about 0} \\[10pt]
= {} & \sqrt{\frac 2 \pi} \int_0^\infty x^3 e^{-x^2/2} (x\,dx) \\[10pt]
= {} & \sqrt{\frac 2 \pi} \int_0^\infty (2u)^{3/2} e^{-u} \, du \\[10pt]
= {} & \frac 4 {\sqrt\pi} \int_0^\infty u^{3/2} e^{-u} \, du \\[10pt]
= {} & \frac 4 {\sqrt\pi}\,\, \Gamma\left(\frac 5 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot\frac 3 2 \Gamma\left( \frac 3 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) \\[10pt]
= {} & \frac 4 {\sqrt\pi} \cdot \frac 3 2 \cdot \frac 1 2 \cdot \sqrt\pi \\[10pt]
= {} & 3.
\end{align}
