I have a question that is similar to the one here: covariance of integral of Brownian, but the answer that I come up with does not match what the book claims the answer is.
Given that $$X_t = \int_0^t \frac{B_u}{u}du,$$ where $B$ is standard Brownian motion, and that $t\leq s$, what is $E[X_tX_s]$?
I have that $$E[X_tX_s] = \int_0^t\int_0^s\frac{u\wedge v}{uv}\,dv\,du.$$ This I can justify by Fubini, Holder's inequality and the fact that $E[B_uB_v] = u\wedge v$. Then,
$$\int_0^t\int_0^s\frac{u\wedge v}{uv}\,dv\,du = \int_0^t\int_0^u\frac{ v}{uv}\,dv\,du + \int_0^t\int_u^s\frac{u}{uv}\,dv\,du$$
I compute the sum of the two integrals on the RHS as $t + t\ln{s} - \int_0^t\ln{u}\,du$. The last integral here does not even converge. I must have done something wrong obviously because I have just shown that this expression was integrable before applying Fubini (furthermore, there is no logarithmic term in the given answer) but I don't see where I am going wrong.
Edit: As Tom-Tom pointed out the RHS converges to $t\ln{\frac{s}{t}}$ but the resulting answer is still not what is given in the book, which is $2t-t^2+st$.