# Integral of delta function over a small interval around zero

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?

• yes, it is correct, take also a look here – Masacroso Nov 29 '18 at 18:15

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:$$\int_{-\Delta x}^{\Delta x}\delta(x)dx=\int_{0^-}^{0^+}\delta(x)dx=1$$