$\varepsilon$ in $\int_{-\varepsilon}^\infty e^{-x}\delta(\sin{x})\;dx$

Question

Why is $\varepsilon$ in the following integral and not just $0$ instead? My teacher said he didn't have time to explain it; however, I am curious about that.

$$\int_{-\varepsilon}^\infty e^{-x}\delta(\sin{x})\;dx$$

where

$$0 \lt \varepsilon \ll 1$$

Answer 1

It depends on how the delta function $\delta(\cdot)$ is defined. To be sure to get "all" of the contribution at $x=0$, the support of the integral extends a little to the left of zero.

The integral will evaluate to $\sum_{k=0}^\infty e^{-\pi k x} = \frac{1}{1-e^{-\pi}}$ because $\sin x$ has zeros at $\pi k x$.