Integral of delta function over a small interval around zero

Question

I know that the dirac delta function, $\delta(x)$, satisfies the following properties $\int_{-\infty}^{\infty}\delta(x)=1$ and $\delta(0)=\infty$.

However if I integrate the delta function over a $\Delta x$ rather than over an infinite interval around $0$ what do I get?

I would say the following holds: $\int_{0-\Delta x}^{0+\Delta x} \delta(x) = 1$, for arbitrarily small $\Delta x$. Is this correct?

Answer 1

Yes it is correct since the delta function is $zero$ outside of $(0^-,0^+)$ and the integral with $\Delta x$ reduces to the same form as before:$$\int_{-\Delta x}^{\Delta x}\delta(x)dx=\int_{0^-}^{0^+}\delta(x)dx=1$$