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 
 A: 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)
A: 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 $4!/8 = 3$.
A: \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}
A: $$
\begin{align}
%
 E(X^2)
&= \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} x^4 e^{-\frac{x^2} 2}\, dx
= \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} x^3 d\left(e^{-\frac{x^2}{2}} \right) \\
%
 &= \frac 1 {\sqrt{2\pi}} \int_{-\infty}^{+\infty} 3x^2\, dx 
= 3  E(X^2)  = 3 \operatorname{Var}(X) = 3
%
\end{align}
$$
There is another answer I found 
