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Can someone explain why the formula below is a definition of the Dirac function?

$$\int_I y(0) \delta(t)\, dt =\begin{cases} y(0) \text{ if 0}\in \text{I} \\\\ 0 \text{ otherwise} \end{cases}$$

where $I$ is an interval.

I am used to the definition $\int \delta(t) \,dt = 1$ and $\delta(t) = \infty$ if $t = 0$, otherwise $0$.

Mark Viola
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NightRain23
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2 Answers2

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The definition you give ($\int \delta(t) dt = 1$ and $\delta(t) = \infty$ if $t = 0$, otherwise $0$) is more of a characterization its behavior, rather than a true definition.

Rewrite your expression as:

$$ \int_\mathbb{R} y(0) 1_I \delta(t)\, dt$$

where $1_I$ is the indicator function for I.

This integrand is a representation of $[(y\cdot 1_I) * \delta](0)$ where $*$ is a convolution. Because the dirac delta is the identity with respect to the convolution, this is equal to $[(y\cdot 1_I)](0)$; this is $y(0)$ if $I$ contains 0, or zero otherwise.

nanofarad
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  • The Dirac Delta is NOT a function. The operator is NOT an integral so there is NO integrand. And the the relationship in the OP is NOT true is $0$ is an end point of $I$. – Mark Viola Apr 20 '20 at 17:45
  • @MarkViola Notation fixed (it's a bit rusty as I've taken relevant classes a while back). Or is there a deeper issue with the content and logic itself? – nanofarad Apr 20 '20 at 17:48
  • The notation given in the OP means (as you indicated) $$\langle y 1_I, \delta \rangle=\begin{cases}y(0)&,\text{if},0, \text{ is contained in the open set},I\\0&,\text{otherwise}\end{cases} $$ – Mark Viola Apr 20 '20 at 17:55
  • Can it be shown from this definition that the dirac delta is the identity in convolution? – NightRain23 Apr 21 '20 at 11:42
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In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

The notation $\int_a^b f(t)\delta(t-c)\,dt$, $a\le b$, is interpreted to mean the functional $\langle fp_{ab},\delta_c\rangle$.

Here, $p_{ab}$ is the "rectangular pulse" function, $p_{ab}(t)=u(t-a)-u(t-b)$, and $u$ is the unit step (or Heaviside Function) where

$$u(t)=\begin{cases}1&,t>0\\\\0&,t<0\end{cases}$$

Note that there are various conventions for the value $u(0)$.

Therefore, we have

$$\begin{align} \int_a^b f(t)\delta(t-c)\,dt&=\langle fp_{ab},\delta_c\rangle\\\\ &=\begin{cases}f(c)&,c\in(a,b)\\\\0&,c\notin [a,b]\end{cases} \end{align}$$

Note that if $c=a$ or $c=b$, the functional is not defined.

Mark Viola
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  • @nightrain23 Please let me know how I can improve my answer. I really want to give you the best answer I can. If it was not useful, then I can simply delete this answer if you prefer. – Mark Viola Jul 30 '20 at 15:32