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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?

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  • $\begingroup$ yes, it is correct, take also a look here $\endgroup$ – Masacroso Nov 29 '18 at 18:15
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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:$$\int_{-\Delta x}^{\Delta x}\delta(x)dx=\int_{0^-}^{0^+}\delta(x)dx=1$$

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