Regarding Dirac Delta I'm having a little bit of trouble understanding Dirac Delta, or rather, finding a proper definition. I understand the way it is "found" by using Fourier transforms on a function, and that it isn't really a function but a distribution. I also get the idea that I shouldn't look at it as a common Riemann integral. But then, why are there limits of functions that represent the same mathematical object? Specifically I can't see how the following would be true: $$\int_{-\infty}^{\infty}\delta(x)dx = \int_{-\infty}^{\infty}\lim_{\sigma\to 0}\frac{1}{\sqrt{2\pi\sigma²}}e^{\frac{-x²}{2\sigma²}}dx = 1$$
If I think about the evolution of amplitude and width of the Gaussian as $\sigma$ approaches zero it kind of makes sense, but then, that's the "intuitive" and not rigorous definition of the Delta (being 0 everywhere except at $x=0$ where it is "$\infty$"). To summarize, is there a proper definition of the Delta that would allow me to show the relation above? Or any explanation as to "why the limit has the same properties"? Thanks!
 A: Generally, if $G_i$ is a sequence of distributions, their limit (if it exists) is given pointwise; i.e. by the formula
$$ \left( \lim_i G_i \right) [f] = \lim_i \left( G_i[f] \right) $$
for every test function $f$.
The problem begins when we introduce the faux-integral notation for evaluating a distribution. Recall that
$$ \int_{-\infty}^{\infty} F(x) f(x) \, \mathrm{d}x := F[f]$$
is how we define the meaning of the integral-like notation on the left. 
Using this, the limit above is written
$$ \int_{-\infty}^{\infty} \left( \lim_i G_i \right)(x) f(x) \, \mathrm{d} x =
\lim_i \int_{-\infty}^{\infty} G_i(x) f(x) \, \mathrm{d} x$$
Notice how the limit has magically transported from inside the integral to outside the integral.
The notation problem becomes serious when we use another form of shorthand. For any (sufficiently nice) function $h$, let me introduce the notation $\widehat{h}$ to mean the distribution defined by the formula
$$ \widehat{h}[f] := \int_{-\infty}^{\infty} h(x) f(x) \, \mathrm{d} x $$
Note that the integral on the right is an ordinary integral of functions. Suppose that each of the $G_i$ in the example above is of the form $G_i = \hat{g_i}$. Then the integral becomes
$$ \int_{-\infty}^{\infty} \left( \lim_i \hat{g}_i \right)(x) f(x) \, \mathrm{d} x =
\lim_i \int_{-\infty}^{\infty} g_i(x) f(x) \, \mathrm{d} x$$
The final abuse of notation is that people often don't decorate $g$ at all when they do this, and so we have the horribly notated that fact
$$ \int_{-\infty}^{\infty} \lim_i g_i (x) f(x) \, \mathrm{d} x =
\lim_i \int_{-\infty}^{\infty} g_i(x) f(x) \, \mathrm{d} x$$
where almost nothing on the left hand side means what it looks like it means.
A: On a certain class $F$ of real functions that is also  a vector space over $\Bbb R$ we have may have a class $G$ of  linear operators  mapping $F$ to $\Bbb R,$ where each $g\in G$ is identified with a real function $g^*$ such that $g(f)=\int_{\Bbb R}f(x)g^*(x)dx.$ And we  have the linear operator $D(f)=f(0),$ which (usually) is not in $G.$ But $D$ is (usually) a point-wise limit of members of $G,$ in that there is a sequence $(g_n)_n$ in $G$ such that for every $f\in F$ we have $f(0)=\lim_{n\to \infty}g_n(f)=\lim_{n\to \infty}\int_{\Bbb R}f(x)g_n^*(x)dx.$
There are advantages to writing $D$ as if it belonged to $G$ and as if there were some function $\delta=D^*$ such that  $f(0)=D(f)=\int_{\Bbb R}f(x)\delta (x)dx.$ Some of the formulas involving integration remain valid, and it easier to discuss $D$ and members of $G$ together. 
Reference topic: Heavide Calculus /Heaviside Operational Calculus. 
The $F$ and $G$ above  vary , depending on the topic, context, and application.
A: The way I was taught, the delta function was defined by the expression:
$$
\int_{-\infty}^\infty \delta(x)f(x)dx = f(0)
$$
(For continuous $f$.)
This then explains why $\int_{-\infty}^\infty \delta(x)dx = 1$, just let $f(x)=1$. To make your other integral work, it would only really make sense to have the limit on the outside of the integral, like so:
$$
\lim_{\sigma\to0} \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}} e^{-x^2/2\sigma^2} dx
$$
$$
=\lim_{\sigma\to0} 1 = 1
$$
EDIT: In general, it is probably best to put the limit on the very outside of whatever expression using the delta function you are computing. As you decrease $\sigma$, $\frac{1}{\sqrt{2\pi\sigma^2}} e^{-x^2/2\sigma^2}$ approximates the delta function better and better, so in theory, the expression should approach its true value.
A: The proper definition (and by proper I mean what we did in my class) is of a rectangle that gets thinner and thinner, but longer and longer. Say we define $d_c(x)$ as the following
$$
d_c(x)=\begin{cases} 
      0 & x < -\frac{1}{2c} \\
      c & -\frac{1}{2c} \leq x \leq \frac{1}{2c} \\
      0 & x > \frac{1}{2c}
   \end{cases}
$$
As you can see, this makes a rectangle. Solving for the area we get $(\frac{1}{2c}+\frac{1}{2c})c = 1$ no matter what $c$ value we say. We then define the Dirac Delta function as
$$
\delta(x) = \lim_{c\rightarrow\infty}d_c(x)
$$
