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 

If you know about the moment-generating function for the standard normal, which is $M_X(t) = e^{t^2/2}$, then you have $E(X^4) = M_X^{(4)}(0)$ (i. e. the fourth derivative of $M_X$ evaluated at 0.)  But by the usual series expansion you have
$$M_X(t) = 1 + {t^2 \over 2} + {1 \over 2!} \left( {t^2 \over 2} \right)^2  + \cdots $$
and so the $t^4$ term of that series is $1/8$; thus the fourth derivative is $4!/8 = 3$.
