What's the variance of the following stochastic integral? The stochastic integral is defined as
$$u_t = \int_{t-1}^t e^{-\kappa(t-s)}\int_0^s e^{-c(s-r)} \, dW(r) \, ds.$$
where $W(t)$ is a standard Brownian motion, $\kappa$ and $c$ are both positive.
I know this integral can be viewed as
$$u_t = \int_{t-1}^t e^{-\kappa(t-s)}J_c(s) \, ds,$$
where $J_c(s) = \int_0^s e^{-c(s-r)} \, dW(r)$, is an Ornstein-Uhlenbeck process. But how do we handle this double integral and use Ito's isometry to get the variance of it? 
Further, does this integral admit a Wold representation, that is,
$$u_t = \sum_{j = 0}^\infty F_j \varepsilon_{t-j},$$
where $\varepsilon_t \sim \mathrm{iid}(0, \sigma^2)$
 A: Here is my calculation, which may be wrong or not, I am not quite sure...
$$\begin{align}
Var(u_t) &= E(u_t^2) = E\left(\int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) \,ds\right)^2 \\
&= E\left(\int_{0}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}\,ds\, dW(r) \right)^2 \\
&= \int_{0}^{t} \left(\int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\right)^2\, dr \\
&= \int_{0}^{t} e^{-2\kappa t + 2cr} \left(\int_{r}^{t}e^{(\kappa-c)s}\,ds\right)^2\, dr \\
&= \dfrac{1}{(\kappa-c)^2}\int_{0}^{t} e^{-2\kappa t + 2cr} \left(e^{(\kappa-c)t}- e^{(\kappa-c)r} \right)^2 \,dr \\
&= \dfrac{1}{(\kappa-c)^2}\left(\int_{0}^{t} e^{-2c(t-r)} \,dr + \int_{0}^{t} e^{-2\kappa(t-r)}\, dr - 2\int_{0}^{t} e^{-(\kappa+c)(t-r)} \,dr\right)\\
&= \dfrac{1}{(\kappa-c)^2}\left(\dfrac{1-e^{-2ct}}{2c} + \dfrac{1-e^{-2\kappa t}}{2\kappa} - 2\frac{1-e^{-(\kappa+c)t}}{\kappa+c}\right)
\end{align}$$
which indicates $u_t$ exhibits heteroskedasiticity. 
A: Your idea is fine, however, there are a few mistakes. Here is another solution. Note that,
\begin{align*}
u_t &= \int_{t-1}^{t} \int_{0}^{s}e^{-\kappa(t-s)-c(s-r)}dW(r) \,ds \\
&=  \int_{0}^{t-1} \int_{t-1}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r) + \int_{t-1}^{t} \int_{r}^{t}e^{-\kappa(t-s)-c(s-r)}ds\,dW(r)\\
&=  e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds \int_0^{t-1} e^{cr}dW(r) + \frac{e^{-\kappa t}}{\kappa -c}\int_{t-1}^t e^{cr} \left[e^{(\kappa -c)t}-e^{(\kappa-c)r} \right] dW(r)\\
&=  e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds \int_0^{t-1} e^{cr}dW(r) + \frac{e^{-c t}}{\kappa -c}\int_{t-1}^t e^{cr} dW(r) - \frac{e^{-\kappa t}}{\kappa -c}\int_{t-1}^t e^{\kappa r} dW(r).
\end{align*}
Then, 
\begin{align*}
E(u_t^2) &=\left(e^{-\kappa t} \int_{t-1}^t e^{(\kappa -c)s}ds\right)^2 \int_0^{t-1} e^{2cr} dr \\
&\quad + \frac{e^{-2c t}}{(\kappa -c)^2}\int_{t-1}^t e^{2cr} dr + \frac{e^{-2\kappa t}}{(\kappa -c)^2}\int_{t-1}^t e^{2\kappa r} dr - 2\frac{e^{-(\kappa+c) t}}{(\kappa -c)^2}\int_{t-1}^t e^{(\kappa + c)r} dr.
\end{align*}
