$E(x^4)$ for a normal distribution I have to show that $E(x^4)= 3t^2$ if $x$ is a normally distributed random variable with distribution $N(0,t)$.
What I know is that
$$E(x^4) = \int_{-\infty}^{\infty}x^4\frac{1}{\sqrt{2\pi t}}e^{-x^2/2t}\;dx$$
I've tried solving numerous times it by parts and then taking limits but I keep getting $0$ and not $3t^2$! 
Can somebody give me a better direction?
My integration by parts yields to this: 

where F(x) is  (the density function) so the last term of the integration by part is E(x) for a Normal(0,t) which is 0, and taking the limit of each term i get to 0!
 A: Integration by parts should work and you should probably show your attempt so someone can have a look over it and see where it is going wrong, however another fun way of getting the result that is slightly too long for a comment is to write the density as
$$
\varphi(x)=\frac{1}{\sqrt{2\pi t}}e^{-\frac{x^2}{2t} }
$$
then it satisfies the ODE
$$
t\varphi'(x) + x\varphi(x) = 0,
$$
which suggests, with a bit of hand waving, that for suitable functions $f(x)$ we should have
$$
t\int_{\mathbb{R}} f'(x) \varphi(x) dx = \int_{\mathbb{R}} x f(x) \varphi(x) dx,
$$
or
$$
t\mathbb{E}\left[ f'(X) \right] = \mathbb{E}\left[ X f(X) \right],
$$
so let $f(x) = x^3$ and agreeing that $\mathbb{E}\left[ X^2 \right] = t$ then we have
$$
\begin{align*}
\mathbb{E}\left[ X^4\right] &= 3t\mathbb{E}\left[X^2 \right] =3t^2.
\end{align*}
$$
A: A fun method is to consider two independent $N(0,t)$ variables $Z_1$ and $Z_2$. If you know that $Z_1+Z_2\sim N(0,2t)\sim \sqrt{2t}\cdot N(0,1)$, then you can deduce that
$$(2t)^2 \mu_4 = \mathbb E(Z_1+Z_2)^4 = \mathbb EZ_1^4+6\mathbb EZ_1^2Z_2^2+\mathbb EZ_2^4=t^2\mu_4+6(t\mu_2)^2+t^2\mu_4$$
where $\mu_4$ and $\mu_2$, respectively, denote the fourth and second moment of a $N(0,1)$ random variable, and I've used the fact that the odd moments are zero. Plugging in $\mu_2=1$ gives $\mu_4=3$ as required.
A: Write it as twice the integral from $0$ to $\infty$. 
Then simply substitute $u=x^2/2t$, pull constants all out the front, recognize the gamma integral, use known facts (like $\Gamma(a+1)=a\Gamma(a)$) and simplify.
It drops out in a few lines.
