limits of an integral with delta

Question

Having this integral:

$$\int_{x-t}^{x+t}\delta(z)dz$$

the solution is:

$$ \begin{cases} 1& \text{for $|x| < t$} \\ 0 & \text{else} \end{cases}$$

Namely, all the domain of $x-t<z<x+t$ will have $1$ in the integral, but why? Shouldn't I check when $z=x+t=0$ and $z=x-t=0$ so the solution should be: $1$ when$|x|=t$ else zero? why in the solution all the domain $x-t<z<x+t$ gets 1?

score 0 · Accepted Answer · answered Mar 25 '21 at 16:43

0

The delta function, $\delta(x)$, is defined by $\int_A \delta(x)dx= 1$ if $0\in A$ and 0 otherwise.

answered Mar 25 '21 at 16:43

user247327

18,710

But why the boundaries are not included in the solution? Namely abs(x)<1 but not abs(x)<=1 – Danny Dan Mar 25 '21 at 17:12
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@DannyDan the boundary of the integration domain is a null set with respect to the Lebesgue measure, so it doesn't make a difference: $$\int_{[a,b]} f(x) dx = \int_{(a, b)} f(x) dx.$$ – ViktorStein Mar 25 '21 at 18:04
thank a lot! @Ramanujan – Danny Dan Mar 25 '21 at 18:56

limits of an integral with delta

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