expectation of integral of power of Brownian motion Consider the process
$Z_t =\int_0^t W_s^n ds$
with $n\in \mathbb{N}$.  What is $\mathbb{E}[Z_t]$?  
It is easy to compute for small $n$, but is there a general formula?  What about if $n\in \mathbb{R}^+$?
In addition, is there a formula for $\mathbb{E}[|Z_t|^2]$?
 A: Recall that if $X$ is a $\mathcal{N}(0, \sigma^2)$ random variable then its moments are given by 
$$\mathbb{E}[X^n] = \begin{cases} 0 \qquad & n \text{ odd} \\
\sigma^n (n-1)!! \qquad & n \text{ even} \end{cases}$$
where $n \in \mathbb{N}$ and $!!$ is the double factorial. Since $W_s \sim \mathcal{N}(0,s)$ we have, by an application of Fubini's theorem,
$$\mathbb{E}\bigg[\int_0^t W_s^n ds\bigg] = \begin{cases} 0 \qquad & n \text{ odd} \\
(n-1)!! \int_0^t s^{\frac{n}{2}} ds \qquad & n \text{ even}\end{cases} $$
It is then easy to compute the integral to see that if $n$ is even then the expectation is given by
$2\frac{(n-1)!!}{n+2} t^{\frac{n}{2} + 1}$. 
For $n \not \in \mathbb{N}$, I'd expect to need to know the non-integer moments of a centered Gaussian random variable. I am not aware of such a closed form formula in this case.

Edit: You shouldn't really edit your question to ask something else once you receive an answer since it's not really fair to move the goal posts for whoever answered. Having said that, here is a (partial) answer to your extra question. 
Again, what we really want to know is $\mathbb{E}[X^n Y^n]$ where $X \sim \mathcal{N}(0, s), Y \sim \mathcal{N}(0,u)$. The general method to compute expectations of products of (joint) Gaussians is Wick's theorem (also known as Isserlis' theorem). 
This says that if $X_1, \dots X_{2n}$ are jointly centered Gaussian then 
$$\mathbb{E}[X_1 \dots X_{2n}] = \sum \prod \mathbb{E}[X_iX_j]$$
where the sum runs over all ways of partitioning $\{1, \dots, 2n\}$ into pairs and the product runs over pairs $(i,j)$ in the current partition. 
You should expect from this that any formula will have an ugly combinatorial factor. For a fixed $n$ you could in principle compute this (though for large $n$ it will be ugly). What is obvious though is that $\mathbb{E}[Z_t^2] = ct^{n+2}$ for some constant $c$ depending only on $n$. 
Indeed, 
$$\mathbb{E}[Z_t^2] = \int_0^t \int_0^t \mathbb{E}[W_s^n W_u^n] du ds$$
so we apply Wick's theorem with $X_i = W_s$ if $i \leq n$ and $X_i = W_u$ otherwise. Therefore
$$\mathbb{E}[Z_t^2] = \sum \int_0^t \int_0^t \prod \mathbb{E}[X_iX_j] du ds.$$
Now,
$$\mathbb{E}[X_iX_j] = \begin{cases} s \qquad& i,j \leq n \\
u \qquad& i,j > n \\
s \wedge u \qquad& \text{otherwise} \end{cases}$$
so the integrals are of the form
$$\int_0^t \int_0^t s^a u^b (s \wedge u)^c du ds$$
where $a+b+c = n$.
This integral we can compute. We get
\begin{align}
\int_0^t \int_0^t s^a u^b (s \wedge u)^c du ds =& \int_0^t \int_0^s s^a u^{b+c} du ds + \int_0^t \int_s^t s^{a+c} u^b du ds \\
=& \int_0^t \frac{1}{b+c+1} s^{n+1} + \frac{1}{b+1}s^{a+c} (t^{b+1} - s^{b+1}) ds 
\\=& \tilde{c}t^{n+2}
\end{align}
for some constant $\tilde{c}$.
This gives us that $\mathbb{E}[Z_t^2] = ct^{n+2}$, as claimed.
