Fourier integral form of the delta function?

I'm learning basic maths for physicist and was wondering what do we use the Dirac delta function for? What is the difference between "the Fourier integral form" and the usual way of expressing the delta function?

I know how to use it from elementary quantum mechanics classes so for instance $\delta _{ij}$ is equal to one if $i=j$ and zero if not.

Answer

Your question seems to confuse a number of things. Let me try to clear everything up:

1. 
   

   There is a difference between the Kronecker delta $\delta_{ij}$ where $i,j$ are discrete indices, and the Dirac delta $\delta(x-y)$, where $x,y$ are continuous variables. Superficially, the latter can be thought of as the continuous version of the former. There is something of a similarity in the following expressions, which should clarify the relation between the two: \begin{align} \sum_if_{i}\delta_{ij}=f_j\hspace{1cm}&\Leftrightarrow\hspace{1cm}\int dx f(x)\delta(x-y)=f(y)\\ \sum_{i}f_if_j\delta_{ij}=f_jf_j\hspace{1cm}&\Leftrightarrow\hspace{1cm}\int dx\ f(x)f(y)\delta(x-y)=f(y)f(y) \end{align}

   
   


2. 
   

   If you know what the Fourier transform is, you'll know that its definition is (depending on your conventions, some factors of $\sqrt{2\pi}$ may or may not appear) $$\mathscr{F}(f(x))(k)=\int f(x)e^{-ikx}dx$$ Moreover, you should know that $$f(x)=\mathscr{F}^{-1}\bigl(\mathscr{F}(f(x))(k)\bigr)(x)$$ Apply this identity to $\delta(x)$ and you'll find that the Dirac delta function (it's really a distribution!) can be written in a different way as well. That's all that's going on here, and the two ways of writing it are equivalent. The integral form is often just handy since it can be used to greatly simplify many integrals involving complex exponential functions.