What's the point of Dirac delta function? I have heard that The main useful property of Dirac delta function is it's fundamental property that
$$
 \int_{-\infty}^{\infty}f(x)\delta(x-a)dx=f(a) 
$$
I don't understanding why this equation is useful. The left hand side $\int_{-\infty}^{\infty}f(x)\delta(x-a)dx$ , which is the inner product of two functions, seems much harder to compute than the left hand side $f(a)$, which just applies $f$ to $a$ directly. Besides, in order to compute $\int_{-\infty}^{\infty}f(x)\delta(x-a)dx$ we must know the value of $f(a)$ first , right?   Why do we need such a complicated process to get the value of $f(a)$ when we have already know the value of $f(a)$.
 A: Keep in mind that the left-side $$\int_{-\infty}^{\infty} f(x) \delta(x - a) dx$$ is not an inner product of two functions, since the Dirac delta function isn't a function. It's properly understood in the context of distribution theory, or regarded as a generalized function. That being said, you're right: The utility isn't in computing the value at $a$, but rather in knowing that the integral can be simplified. 
Another nice reason to study the delta function is that it can be regarded (again, in the sense of distributions) as the derivative of the Heaviside function. 
As an example of this that comes up in differential equations, suppose you have a mass hanging on a spring; the displacement $y$ from the equilibrium can then be modeled by
$$y'' + y = 0$$
which has oscillatory solutions. More generally, if we have a driving force $f(t)$ applied to the system, we have
$$y'' + y = f(t)$$
Now suppose we want to deliver a sharp impulse to the system to get it going initially; the delta function can be thought of as the limit of very short, but strong impulses acting on the system (and hence, as an instantaneous shock). This can be studied via the "ODE" 
$$y'' + y = \delta(t - 1)$$
subject to initial conditions $y(0) = y'(0) = 0$ (so our block is not moving, and not displaced at time $0$; then we hit it $1$ second later). A standard way to study such equations is via the Laplace transform; recall that
$$\mathcal{L}\Big(f(t)\Big)(s) = \int_0^{\infty} e^{-st} f(t) dt$$
Now notice that 
$$\mathcal{L}\Big(\delta(t-1)\Big)(s) = \int_0^{\infty} e^{-st} \delta(t - 1) dt = e^{-s}$$
The left-hand side of the equation can be transformed easily, and then standard techniques (e.g. a table of Laplace transforms) can give the final solution quite easily from here. 
The key point is that the niceness with which $\delta$ interacts with integrals allows us to extend some techniques to difficult situations.
A: In Digital Signal Processing we use the Dirac delta as a set of basis functions that spans the discrete space of signals with finite energy and are absolutely summable, hence it's used to model the process of sampling signals to convert them into discrete signals. 
So in that way we see that integral as an inner product to obtain the coefficients in the new basis. So instead of having an infinite continuous function now it is an infinitely countable function which can be stored and treated by computers.
