Probability distribution of $\int_0^t \frac{W_s}{s} \,ds$ I am currently working on an exercise that requires the knowledge of the distribution of $\int_0^t \frac{W_s}{s} \,ds$, where $W$ is a Brownian motion. 
I can compute the distribution of $\int_{0}^T W_t \,dt$ easily by using the stochastic Fubini's theorem:
\begin{eqnarray}
\int_{0}^T W_t \,dt & = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dW_s \,dt \\
& = & \int_0^T \int_0^T \mathbf{1}_{[0,t]} (s) \,dt  \,dW_s\\
& = &  \int_0^T T-s \,dW_s\\
& \sim & N \bigg( 0, \int_{0}^T (T-s)^2 \,ds \bigg).
\end{eqnarray}
However, such method seems to not work in this case, as I do not know how to express $\frac{W_t}{t}$ as a stochastic integral. Any suggestions?
 A: From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma,
$$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$
so
$$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$
as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent Gaussian variables, it is Gaussian as well. Taking the expectation, it is zero. 
The variance of $I$ can be computed at least two ways. One is using the expression above and compute
$$\mathbf E[I^2] = \int_0^t \Big(\ln\frac{t}{s}\Big)^2ds = t\int_0^1 (\ln x)^2dx = t\int_0^\infty y^2 e^{-y}dy=2t.$$
The last equality comes from differentiation (to the second order) under the integral. The second way is to compute directly from the original expression
$$\mathbf E[I^2] = \int_0^tdu\frac{1}{u}\int_0^tdv\frac{\mathbf E[W_uW_v]}{v}=\int_0^tdu\frac{1}{u}\int_0^t dv \frac{\min(u,v)}{v}=2\int_0^tdu\frac{1}{u}\int_0^u dv \frac{v}{v} = 2t.$$
So $I\sim N(0,2t).$
