Question about dirac Delta function 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?
 A: 
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!
A: 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.
