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Let $\delta(x)$ denote a Dirac Delta function on $R$.

Then, I know that $$\delta(x) = 0 \text{ if } x \neq 0$$ and

$$\int_R \delta(x) f(x) dx = f(0)$$

Then, since $\delta(x)$ is zero everywhere except at $0$, it seems to me that

$$\int_{(-\epsilon, \epsilon)} \delta(x) f(x) dx = f(0)$$

for any $\epsilon >0$, but since $\delta$ is not a "function", how does one jusfity this?

nan
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    Treat "$\delta(x) dx$" as a measure that measures a function by taking the value at a point (in this case $0$), then this makes sense. – Sarvesh Ravichandran Iyer Feb 18 '17 at 00:03
  • @астонвіллаолофмэллбэрг If I treat it as a measure, what is the measurable space that it acts on? It is the set of ALL functions on $R$? – nan Feb 18 '17 at 00:06
  • The way to make this rigorous is using the theory of distributions, if you think delta function as a limit of some distributions you can prove that these are equivalent. – R.W Feb 18 '17 at 00:12
  • @RafaelWagner True, using an approximation via convolution with mollifiers.Also the measure space is the set of all subsets of $\mathbb R$, the measure is : zero if the set doesn't contain $1$, and one otherwise. Using the definition of function integration, one arrives at the formula that you have given. – Sarvesh Ravichandran Iyer Feb 18 '17 at 00:12
  • EDIT: I meant zero if the set doesn't contain $0$ above. – Sarvesh Ravichandran Iyer Feb 18 '17 at 00:32
  • Have you read https://en.wikipedia.org/wiki/Dirac_delta_function ? –  Feb 18 '17 at 01:14

2 Answers2

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In THIS ANSWER and THIS ONE, I provided primers on the Dirac Delta.

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

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

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

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

Therefore, with $a=\epsilon$, $b=-\epsilon$, and $c=0$, we have

$$\begin{align} \int_{-\epsilon}^\epsilon f(x)\delta(x)\,dx&=\langle fp_{-\epsilon\,\epsilon},\delta_0\rangle\\\\ &=\int_{-\infty}^\infty p_{-\epsilon\,\epsilon}(x)f(x)\delta(x)\,dx \\\\ &=f(0)p_{-\epsilon\,\epsilon}(0)\\\\ &=f(0) \end{align}$$

as was to be shown!

Community
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Mark Viola
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The "function" that you speak of is not actually a function, but a mathematical object called a "distribution" or a generalized function.

To remain brief, we can talk about linear functionals over a function space, i.e. the set of all functions $T$ such that $T : C^\infty_0 \to \mathbb{C},$ for example. Here, $C^\infty_0$ is the set of infinitely differentiable functions with compact support. The delta function is one such linear functional. To be called a distribution only requires a few more properties (such as continuity). Using the suggestive notation $$ \delta(f(x)) = f(0) \text{ as } \langle \delta, f \rangle = f(0)$$ then perhaps this bracketed notation behaves like an inner product, which would mean $$ \langle \delta, f \rangle = \int \delta(x)f(x)\; dx = f(0).$$

However, as you pointed out, this integral doesn't make sense in the genuine function sense, and hence we stick with $\delta$ being a function that maps a given test function to its value at the origin.

Merkh
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