I was reading a journal and came to below equality which I couldn't figure out how to prove. Highly appreciate if anybody can help.
$$\int_{-\infty}^t\int_{-\infty}^{t+\tau}E[s(\theta)s(\theta')]\delta(\theta-\theta'+\tau)e^{\Delta(\theta+\theta'-\tau)}\,\mathrm d\theta\,\mathrm d\theta'=\int_{-\infty}^tE[s(\theta)s(\theta+\tau)]e^{2\Delta\theta}\,\mathrm d\theta$$
This is basically the definition of the delta-function. When you do the integral over $\theta'$ then everywhere you had $\theta'$ you will have $\theta+\tau$, since that's what makes the argument of the delta function equal 0.
The definition of the delta function is basically that it obeys $$ \int_a^b \delta(x) dx = \left\{ \begin{array}{ll} 1 & 0 \in (a,b) \\ 0 & \mathtt{otherwise} \end{array} \right. $$ and $$ \int_a^b f(x) \delta(x) dx = \left\{ \begin{array}{ll} f(0) & 0 \in (a,b) \\ 0 & \mathtt{otherwise} \end{array} \right. $$