Integral of Stochastic Integral $ x(t) = \int_0^te^{as}dW_s $

Question

I'm working through an exercise where I have got the following object:

$$ x(t) = \int_0^te^{as}dW_s $$

for some constant $a$. Now, I need to find the distribution of $\int_0^t x(s) ds$. I'm struggling to do this and I don't know where to look for help.

So, I want to find the distribution of:

$$ \int_0^t \int_0^s e^{au} dW_uds $$

How do I go about solving this problem? In particular I need to find it's mean and variance. I assume that the order of integration can't be swapped, but I don't know for certain. Can anyone enlighten me?

Answer 1

Integrate by parts

\begin{align} \int_0^t x(s) ds&= tx(t) -\int_0^t sdx(s) \\ & =t\int_0^te^{as}dW_s-\int_0^t se^{as}dW_s \\ &= \int_0^t(t-s)e^{as}dW_s \end{align}

which is a normal distribution of zero-mean and variance

$$var=\int_0^t(t-s)e^{2as}ds=\frac{e^{2at}-2at-1}{4a^2}$$