Distribution of $\| W_t \|^2_{L^2([0,T])}$ Motivation: consider the SDE
$$dX_t = b(X_t) dt + \sqrt{\varepsilon} dW_t. \tag1$$
Consider the action, defined by
$$S(\phi)=\int_0^T |\phi'(t)-b(\phi(t))|^2 dt$$
if $\phi \in H^1([0,T])$ and $+\infty$ otherwise.
If $\phi$ is a sample path of $(1)$, then we can rearrange to see that $S(\phi)=\varepsilon \| W_t \|^2_{L^2([0,T])}$, where $W_t$ is the particular Brownian motion which drives $\phi$. I think this implies that the probability density of the paths is $\phi \mapsto f_\varepsilon(S(\phi))$ where $f_\varepsilon$ is the probability density for $\varepsilon \| W_t \|^2_{L^2([0,T])}$. 
So I get the question: what are the properties of the distribution of $\| W_t \|^2_{L^2([0,T])}$?
The one thing that I think I have correctly derived is:
$$\mathbb{E} \left ( \| W_t \|^2_{L^2([0,T])} \right ) = \mathbb{E} \left ( \int_0^T W_t^2 dt \right )= \int_0^T \mathbb{E} \left ( W_t^2 \right ) dt = \int_0^T t dt = T^2/2.$$
Any information about the distribution of this quantity, for instance how to compute its variance, would be appreciated.
Edit: Nate Eldredge has pointed out that my motivation, while intuitive, does not really work when treated formally as written, because $S(\phi)=+\infty$ with probability $1$. So I would also be interested in some description, formal or informal, of the extent to which $S$ and the distribution of paths of (1) are connected.
 A: You can get the variance by the following trick:
$$\begin{align*} \mathbb{E}\left[\left(\int_0^T W_t^2\,dt\right)^2\right] &= \mathbb{E}\left[\int_0^T W_t^2\,dt \int_0^T W_s^2\,ds\right] \\
&= \mathbb{E}\left[\int_0^T \int_0^T W_t^2 W_s^2 \,dt\,ds\right] \\
&= \int_0^T \int_0^T \mathbb{E}[W_t^2 W_s^2]\,dt\,ds.
\end{align*}$$
Now for $s \le t$, we can write $W_t = W_s + (W_t - W_s)$ and so we have $$W_t^2 W_s^2 = (W_t - W_s)^2 W_s^2 + 2 (W_t - W_s) W_s^3 + W_s^4.$$
Taking expectations and using independence of increments (and noting that the case $s \ge t$ is symmetric),
$$\mathbb{E}[W_t^2 W_s^2] = \begin{cases} (t-s) s + 3 s^2, & s \le t \\ (s-t) t + 3 t^2, & s \ge t \end{cases}$$
So we end up with
$$\begin{align*}\mathbb{E}\left[\left(\int_0^T W_t^2\,dt\right)^2\right]  &= \int_0^T \int_0^t ((t-s) s + 3 s^2)\,ds\,dt + \int_0^T \int_t^T ((s-t) t + 3 t^2)\,ds\,dt \\ &= \frac{7}{12} T^4\end{align*}$$
This means the variance is $\frac{7}{12} T^4 - \frac{1}{4} T^4 = \frac{1}{3} T^4$.  I think you could get higher moments with a similar approach.
I don't know of a closed form for the density function but there are certainly asymptotics for the probability of being very large or very small.  A phrase to look for is "quadratic functionals of Brownian motion" or "Wiener chaos of order 2".  
For example, in 

Wenbo V. Li, Small ball probabilities for Gaussian Markov processes under the $L^p$-norm,
  Stochastic Process. Appl. 92 (2001), no. 1, 87–102.

Lemma 2.3 implies the statement:
$$\lim_{\epsilon \to 0^+} \epsilon \log \mathbb{P}\left(\int_0^1 W_t^2\,dt < \epsilon\right) = -\frac{1}{8}.$$
