The dirac delta can be defined in such a way that

$$ \int_{-\infty}^{\infty}\delta(z)\,dz = 1 $$

and, for a suitable $f$, it holds that

$$ \int_{-\infty}^{\infty}f(z)\,\delta(z)\,dz = f(0). $$

What can be said about the integral (take $a>0$)

$$ \int_0^{a}f(z)\,\delta(z)\,dz\quad? $$

Is still true that $\int_0^{a}f(z)\,\delta(z)\,dz=f(0)$ ? I am pretty sure that, for all $\varepsilon>0$, then

$$ \int_{-\varepsilon}^{\varepsilon}f(z)\,\delta(z)\,dz=f(0), $$

but I am in trouble when the interval of integration has the zero at its boundary.

  • 3
    $\begingroup$ This has been asked several times before. The most coherent definition is that, for every $a>0$, $$\int_0^af(z)\delta(z)dz=\int_{-a}^0f(z)\delta(z)dz=\frac12f(0)$$ $\endgroup$ – Did Sep 7 '16 at 21:48

With 0 on the boundary it is a priori not defined. It corresponds to integrating $\delta$ against the function $f 1_{x>0}$ (or possibly $f 1_{x\geq 0}$) which is not continuous at $0$ where the support of $\delta$ lies. Note that the other case $(-\epsilon,\epsilon)$ presents no problem since $\delta$ has support in 0.

  • $\begingroup$ when extending the distributions outside $C^\infty_c$, I think it is important to say that the support of $\delta$ is a neighborhood of $0$ $\endgroup$ – reuns Sep 7 '16 at 21:55
  • $\begingroup$ It could be $\frac{1}{2} \mathbf{1}_{x>0} + \frac{1}{2} \mathbf{1}_{x \geq 0}$ as well! $\endgroup$ – Hurkyl Sep 7 '16 at 22:17
  • $\begingroup$ It depends on what the intended purpose is. You may e.g. want to say that when $\phi$ has compact support and integral 1 then $t\phi(tx)$ converges weakly (as $t\rightarrow \infty$) to a Dirac delta. But when $\phi$ is not symmetric you will get into trouble when integrating on $R_+$ $\endgroup$ – H. H. Rugh Sep 7 '16 at 22:23

$$\int_a^b f(x)\delta(x)\ \text{d}x = \begin{cases} f(0) & \text{if}\ 0\in\ [a, b] \\ 0 & \text{if}\ 0\notin\ [a, b]\end{cases}$$

  • 1
    $\begingroup$ Note that $[a,b)$, $(a,b]$, or $(a,b)$ could just as well be intended by the notation. The precise meaning isn't fixed when the notation is introduced, since all four possibilities give the same result when integrating actual functions. $\endgroup$ – Hurkyl Sep 7 '16 at 22:43
  • $\begingroup$ @Hurkyl Oh! You're right! Thank you for the add, I completely ignored the other cases! :D $\endgroup$ – Von Neumann Sep 7 '16 at 22:49
  • $\begingroup$ I suspect that the definition propose by @FourierTransform creates the problem $$ f(0)=\int_{-\infty}^{\infty}f(x)\delta(x)\neq \int_{-\infty}^{0}f(x)\delta(x)+\int_{0}^{\infty}f(x)\delta(x)=2\,f(0) $$ $\endgroup$ – AlmostSureUser Sep 8 '16 at 8:46
  • $\begingroup$ @AlmostSureUser I had some notes in which I read that property. But who knows, Delta always hides mysteries. $\endgroup$ – Von Neumann Sep 8 '16 at 9:48
  • $\begingroup$ @AlmostSureUser Intervals must be open $\ldots\to\ldots \left(\,a,b\,\right)$. $\endgroup$ – Felix Marin Sep 9 '16 at 5:02

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