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 

Let $W = X^2$. Then $E[X^4] = E[W^2]$. From the formula for variance,
$$E[W^2] = Var(W) + E[W]^2 $$
$$E[W]^2 = E[X^2]^2 = (Var(X) + E[X]^2)^2 = (1 + 0^2)^2 = 1$$
Note that $W$ is a $\chi^2$ r.v. with one degree of freedom. The variance of a chi-squared is twice its degrees of freedom, thus $Var(W)=2$.
Then: $$E[X^4] = E[W^2] = 2 + 1 = 3$$
(assuming you can use facts about the chi-squared distribution)
