I would like to find the distribution of $\int_0 ^T uW_u du$ where $(W_u)_{u\geq0}$ is the Brownian motion.
What I have tried:
$$\int_0 ^T uW_u du = \int_0 ^T B_udu - \int_0^T \int_0^tB_sdsdt$$ by integration by parts. I know each term on the RHS is normal as any integration of a Gaussian process is again a Gaussian process.
However, I cannot conclude RHS is normals since I don't have independence of $\int_0 ^T B_udu $ and $\int_0^T \int_0^tB_sdsdt$ since $Cov(\int_0 ^T B_udu, \int_0^T \int_0^tB_sdsdt)=T^4 /8 \neq 0$
Any help is appreciated.
Provided that $$ {\rm d}\left(u^2W_u\right)=2uW_u{\rm d}u+u^2{\rm d}W_u+{\rm d}\left<t^2,W_t\right>_u=2uW_u{\rm d}u+u^2{\rm d}W_u, $$ we have \begin{align} \int_0^TuW_u{\rm d}u&=\frac{1}{2}\int_0^T{\rm d}\left(u^2W_u\right)-u^2{\rm d}W_u\\ &=\frac{1}{2}\left(T^2W_T-\int_0^Tu^2{\rm d}W_u\right)\\ &=\frac{1}{2}\left(T^2\int_0^T{\rm d}W_u-\int_0^Tu^2{\rm d}W_u\right)\\ &=\frac{1}{2}\int_0^T\left(T^2-u^2\right){\rm d}W_u, \end{align} which is obviously normal.