While dealing with an autocorrelation function I ended up with the following:
$ R_{YY}(t+\tau, t)=\int_0^{t+\tau}{\int_0^{t}\delta(v-s)dsdv} , t\ge0 ,\tau\ge0 $
I do not know how to approach this, in $[0,t]\times[0,t+\tau]$ only the line $s=v$ will have contribution so inner integral gives $1$ until $v\le t$ (and $0$ otherwise) for some reason the following seems to be the answer:
$\int_0^{t+\tau}{\int_0^{t}\delta(v-s)dsdv}=t $
But then
$ R_{YY}(t+\tau, t)=t$ does not make any sense at all
(For those who are interested the original problem: "For $0$-mean WSS random process $X(t)$, $R_{XX}(t_1,t_2)=\delta(t_1-t_2)$, $Y(t)=\int_0^tX(v)dv$, is $Y$ WSS?". Integration is a linear operator though not time invariant as it starts from time $0$, so I am not sure ) Could you please help me?
Notice that$$\int_0^t\delta(\nu-s)ds = \begin{cases} 1\quad \nu\in (0,t) \\ 0\quad \text{otherwise}. \end{cases}$$ Thus $$\int_0^{t+\tau}{\int_0^{t}\delta(v-s)dsdv} = \int_0^t1d\nu+\int_t^{t+\tau}0d\nu = t$$