can you integrate the dirac delta over any domain?

Question

At eq 28 the following author takes it from 1 to infinity. I've never seen this before and i'm not sure his equation is correct.

http://bado-shanai.net/The%20Table%20of%20Integrals/intResiduesofDiracDelta.htm

Answer 1

An integral of the form \begin{equation} \int_1^∞\delta(x)f(x)dx \end{equation} is just $0$. Intuitively, the spike of the Dirac delta is not in the domain, so the integral only picks up $0$.

If you look closer at the integral you mention: \begin{equation} \int_1^∞x^{-4}\delta[\sin (\pi x)]\cos(\pi x)\pi dx \end{equation} you will see that the argument in the delta is in fact oscillating between $-1$ and $1$. The value of the integral is just the sum of $x^{-4}\cos(\pi x)\pi$ divided by $\pi$ whenever $\sin (\pi x)=0$ that is for $x=1,2,3...$ \begin{equation} \sum_{x=1}^∞\frac{1}{x^4}\cos(\pi x)=\sum_{x=1}^∞\frac{(-1)^x}{x^4} \end{equation} which is in fact different from what your source states. In fact I would have: \begin{equation} \int_1^∞x^{-4}\delta[\sin (\pi x)] \pi dx = \sum_{x=1}^∞\frac{1}{x^4} \end{equation} but the author justifies the cosine by saying it "comes from the derivative in the functional differential". Perhaps you know what it means, or somebody can correct my mistakes.