Fourier transformation of a step function can be taken because it is integrable. One should use the definition of limit.

In general, functions and their Fourier images are mapped as generalized functions; see: http://mathworld.wolfram.com/GeneralizedFunction.html ,  https://en.wikipedia.org/wiki/Generalized_function , http://www.encyclopediaofmath.org/index.php/Generalized_function

Hi Arul! The problem is that Fourier transforms are defined by means of integrals from - to + infinities and such integrals do not exist for the unit step and signum functions. Try to integrate them?

Fourier transformation can be of several types, also for functions defined on the finite interval.

But what is more important, it is mapping in Hilbert space, from L2 into L2 functions, see, for example, http://math.arizona.edu/~faris/methodsweb/ftrans.pdf . Obviously, step function defined at infinite interval is not L2, as its norm is not finite.

However, from application perspectives in radio-physics, etc it is important to represent Fourier spectrum of step function defined on finite interval (rectangular periodic signal). Here the problem is not L2, but non-continuity.

I think that the main problem is that you do not present your question in full details. Also you use slang 'not directly taken' (in what sense?) instead of rigorous mathematical language.

There are several common conventions for defining the Fourier transform of an integrable function (Kaiser 1994, p. 29), (Rahman 2011, p. 11)..

We also use  F[f]    instead of  \hat{f}.

If  we  denote step function by  u  we can   directly compute   the Fourier transform F[u](z)= \int_0^\ifty  e^{-2\pi i x z}  dx.

For a fixed real number z,  the function

G(x)= -e^{-2\pi i x z}  /(2\pi i  z)  tends to 0  if   x  tends infty.

Here x is real and     therefore  |e^{-2\pi i x z} |=1.

As  far as I know using  this  as motivation we define  F[delta]. But engineering people 

using impulse can   interpretate that  F[u]=1.

If this answer is in the direction of your question,  we can give more details.

Dear Arul,

as Jerzy said before, the reason is that the functions you mentioned are not integrable over R and the Fourier transform can be defined classically as an integral only for integrable functions. Non-integrable functions are called distributions (or generalized functions). For beginners it is instructive to see the Fourier transform of some distributions via some kind of limiting process (in my opinion).

For example, instead of considering the signum function itself, you can consider the function g_a(t) defined as e^{-at} for t>0 and -e^{at} for t 0+, you will get the signum function. On the other hand, you can find the Fourier transform of g_a(t) by a direct integration, since it is an integrable function for any a>0. By integration you will get F(f_a(t))(f)=-4*pi*i*f/(a^2+4*pi^2 f^2). Sending a->0+ you will get formally F(sign(t))=1/(i*pi*f). The reasoning can be made rigorous by using the distribution theory.