Consider the following double integral $$ \int_0^t \int_{x-s}^{x+s}\sum_i \delta(y-i)f(i,s)\,dyds $$ where $\delta$ is the Dirac delta function and $x,i\in \mathbb{Z}$. Is there a way to simplify this?
My attempt: Using the properties of the Dirac delta, we get $$ \int_0^t \sum_{i=\lceil x-s\rceil}^{\lfloor x+s \rfloor} f(i,s)\,ds $$ since $x\pm s$ are not necessarily integers. I feel there is not much else I can do, but I might be missing something. Any ideas?
The Dirac Delta is not a function. And the linear functionals are not integrals. In fact, the product of Heaviside function and the Dirac Delta is not defined.
If we were to proceed heuristically, we might write
$$\begin{align} \int_{x-s}^{x+s}\delta(y-i)\,dyf(i,s)\,dy&=\int_{-\infty}^\infty \left(H(y-x-s)-H(y-x+s)\right)\delta(y-i)\,dy\\\\ &=H(i-x-s)-H(i-x+s) \end{align}$$
Then, we would have
$$\begin{align} \int_0^t \int_{x-s}^{x+s}\sum_{i\in \mathbb{Z}}\delta(y-i)\,dy\,f(i,s)\,ds&=\int_0^t \int_{-\infty}^\infty \sum_{i\in \mathbb{Z}}\left(H(y-x-s)-H(y-x+s)\right)\delta(y-i)\,dy\,f(i,s)\,ds\\\\ &=\sum_{i\in\mathbb{Z}}\int_0^t \left(H(i-x-s)-H(i-x+s)\right)\,f(i,s)\,ds \end{align}$$
And you can finish now.
