if$\newcommand{\Cov}{\operatorname{Cov}}\newcommand{\Var}{\operatorname{Var}}$if $X(t)$ is the Ornstein-Uhlenbeck Ornstein–Uhlenbeck process and $Y(t)$ the time integrated OU process I am trying to calculate the autocovariance $cov(Y_t, Y_s)$$\Cov(Y_t, Y_s)$.
I have a bunch of results but I don't know how to connect them. First in this post they treat the OU process and it's also the result of the variance: $$ Var(Y_t) = \mathbb{E} Y_t^2 = \mathbb{E} \int_0^t \int_0^t X_s X_u ds du = \int_0^t \int_0^t Cov(X_s, X_u) ds du = 2 \int_0^t \int_0^u Cov(X_s, X_u) ds du.$$\begin{align} \Var(Y_t) & = \mathbb{E} Y_t^2 = \mathbb{E} \int_0^t \int_0^t X_s X_u \,ds \,du \\ &= \int_0^t \int_0^t \Cov(X_s, X_u) \,ds\, du \\ & = 2 \int_0^t \int_0^u \Cov(X_s, X_u) \,ds\, du. \end{align}
and they use $Cov(X_s, X_u) = \frac{\sigma^2}{2\theta}\left( e^{-\theta(u-s)} - e^{-\theta(u+s)} \right)$$\Cov(X_s, X_u) = \frac{\sigma^2}{2\theta} \left( e^{-\theta(u-s)} - e^{-\theta(u+s)} \right)$ for $s\leq u$. And in that case the result is $$ Var(Y_t)=\frac{\sigma^2 t}{\theta^2}-\frac{3\sigma^2}{2 \theta^3}+\frac{\sigma^2}{2 \theta^3}(4 e^{-\theta t}-e^{-2 \theta t}) $$$$ \Var(Y_t)=\frac{\sigma^2 t}{\theta^2}-\frac{3\sigma^2}{2 \theta^3}+\frac{\sigma^2}{2 \theta^3}(4 e^{-\theta t}-e^{-2 \theta t}) $$
The thing is that (if I am not wrong) the way to calculate the autocovariance is using a similar integral but with different limits. I am not sure about this but I was thinking in $ Cov(Y_t,Y_v)=\mathbb{E} \int_0^t \int_0^v X_s X_u ds du $$ \Cov(Y_t,Y_v)=\mathbb{E} \int_0^t \int_0^v X_s X_u \,ds \,du $ but I am not completely sure.
And in that case I don't know how to solve the integral because I have to respect the validity of the formula of the covariance for $s \leq u$
On the other hand I also found the following result[Bhattacharya] : $$ Cov(Y_t, Y_{t+v}-Y_t)=\frac{\sigma^2}{2 \theta³}(1-e^{-\theta v}) (1-e^{-\theta t})^2 $$$$ \Cov(Y_t, Y_{t+v}-Y_t)=\frac{\sigma^2}{2 \theta^3}(1-e^{-\theta v}) (1-e^{-\theta t})^2 $$
And also I was thinking in using that $$ Cov(Y_t, Y_{t+v}-Y_t) = Cov(Y_t, Y_{v+t}) - Var(Y_t) $$$$ \Cov(Y_t, Y_{t+v}-Y_t) = \Cov(Y_t, Y_{v+t}) - \Var(Y_t) $$ But again, I am not completely sure about this, and if this is true, there is a way to solve the integral and have the same result.
Any help would be really appreciated.
REF: Bhattacharya, Stochastic Processes with Applications
