Let $X(t)$ be the following time integral of a Wiener process $W(t)$ $$ X(t) = \int_{t}^{t+1} W(s)\, ds. $$ This process is zero mean, and I want to compute its covariance $EX(t)X(s)$.
I am not sure that the following is correct. It is obvious that if $|t-s|>1$ then $EX(t)X(s) = 0$. In the case when, for definiteness, $t \leq s \leq t+1$ we can consider the following partition $$ \Delta_1: t=a_0<a_1<\ldots<a_n=s, $$ $$ \Delta_2: s=b_0<b_1<\ldots<b_k=t+1, $$ $$ \Delta_3: t+1=c_0<c_1<\ldots<c_l=s+1. $$ Each corresponding Riemann sum can be represented in the following way (here $a_{-1}=0$ by definition) \begin{align*} R_{\Delta_1} &= \sum_{i=0}^{n-1} W(a_i)(a_{i+1}-a_{i})\\ &= \sum_{i=0}^{n-1} (W(a_{i-1})-W(a_i))a_{i} + W(a_{n-1})a_n\\ &=\sum_{i=0}^{n-1} (W(a_{i-1})-W(a_i))(a_{i}-a_n).\\ \end{align*} And now using the independence of increments we have \begin{align*} E (R_{\Delta_1}+R_{\Delta_2})(R_{\Delta_2}+R_{\Delta_3}) &= E (R_{\Delta_2})^2\\ &= \sum_{i=0}^k (b_i-b_k)^2 E(W(b_{i-1})-W(b_i))^2 \\ &= \sum_{i=0}^k (b_i-b_k)^2 (b_{i}-b_{i-1}). \end{align*} And $$EX(t)X(s) = \lim_{k\to\infty} \sum_{i=0}^k (b_i-b_k)^2 (b_{i}-b_{i-1}) = \int_{s}^{t+1} (y-t-1)^2dy = \frac{(t+1-s)^3}{3}.$$ Am I right?
\begin{align} \mathsf{E}[X(t)X(s)]&=\mathsf{E}\left[\int_t^{t+1}W(r)dr\int_s^{s+1}W(q)dq\right] \\ &=\mathsf{E}\left[\int_t^{t+1}\int_s^{s+1}W(r)W(q)dqdr\right] \\ &=\int_t^{t+1}\int_s^{s+1}\mathsf{E}[W(r)W(q)]dqdr \\ &=\int_t^{t+1}\int_s^{s+1}\min\{r,q\}dqdr. \end{align}
If $s\ge t+1$, then $$ \mathsf{E}[X(t)X(s)]=\int_t^{t+1}rdr=t+\frac{1}{2}. $$