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Just for fun I was reading about the Heaviside step function on Wikipedia. The definition in terms of the Dirac delta function makes sense:

$$ H(x) = \int_{-\infty}^x \delta(s)\ ds $$

$\delta(s) = 0$ for all $s \in [-\infty, 0)$. For other values of $s$, $\delta(s) \not= 0$ only at one point, so the integral (area) equals the value at that point, $1$.

However, later on in the article it reads (emphasis mine):

Since $H$ is usually used in integration, and the value of a function at a single point does not affect its integral, it rarely matters what particular value is chosen of $H(0)$.

If the value at a single point does not affect the integral for $H$, why does it for $\delta$? In other words, why does the following equation not hold, assuming the above quote?

$$ \int_{-\infty}^\infty \delta(s)\ ds = 0 $$

I must be misunderstanding something about either integrals or the Dirac delta function.

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    $\begingroup$ What you are misunderstanding is about the Dirac delta "function." It is not a "function" in the classic sense of $f:X\to Y$ which assigns to each element in $X$ an element $f(x)$ in $Y$. It is instead a "distribution" which is a generalization of the notion of a function. $\endgroup$ – Math1000 Nov 28 '19 at 23:31
  • $\begingroup$ The way I learned it, the reverse is true. Dirac delta is defined to be the function that satisfies that property. It just so happens that it ends up being the thing you’re asking about. $\endgroup$ – user64742 Nov 29 '19 at 7:41
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Formally, the dirac delta is not a function in the same sense as the heaviside function.

Rather, it is a distribution.

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The best way to think of the dirac delta function is: $$\delta_n(x)=\begin{cases} 0 & x<0 \\ n & 0<x<\frac1n \\ 0 & x>\frac1n \end{cases}$$ $$\delta(x)=\lim_{n\to\infty}\delta_n(x)$$ Now try and integrate $\delta_n(x)$ first then take the limit: $$\int_{-\infty}^\infty\delta_n(x)dx=\int_{-\infty}^0\delta_n(x)dx+\int_0^{\frac1n}\delta_n(x)dx+\int_{\frac1n}^\infty\delta_n(x)dx$$ $$=\int_{-\infty}^00.dx+\int_0^{\frac1n}ndx+\int_{\frac1n}^\infty0.dx$$ $$=n\left[x\right]_0^{1/n}=1$$ and now: $$\int_{-\infty}^\infty\delta(x)dx=\lim_{n\to\infty}\int_{-\infty}^\infty\delta_n(x)dx=1$$

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