I am solving a stochastic ODE with additive noise:
$\frac{dx(t)}{dt} = -\alpha x(t) + L(t)$
where L(t) is Gaussian noise meaning $<L(t)> = 0, <L(s)L(t')> = 2D\delta(s -t').$
I am trying to calculate $<x^2(t)>$ but I am getting confused by the resulting double integral:
$2D\int_0^tds e^{-\alpha(t-s)} \int_0^tdt'e^{-\alpha(t-t')}\delta(s-t')$ which should equate to $\frac{D}{\alpha}(1-e^{-2\alpha t})$.
Do you use the fact $\delta(s-t')$ = $\delta(t'-s)$.Then apply the following property: $\int_{\omega}dzf(z)\delta(z-z_0) = f(z_0)$ when $\omega$ includes the point $z_0$ to get $\int_0^tds e^{-2\alpha(t-s)}$ ? Our lecturer didn't cover these steps and they give very blunt replies so I just wanted to get some alternative response. Thanks in advance!