Some questions Regarding Fourier Transform I have some questions regarding Fourier Transform.


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*I have studied a lot of books for Fourier Transform and currently, I am confused weather I should use Riemann integral or Lebesgue integral for the Fourier Transform.

*How would you differentiate between Fourier Transform of a function or a distribution, what I got is that you will use Lebesgue integral for a distribution whereas for a function you will use Riemann integral. May be, I would be wrong.

*The integral we use in Laplace Transform is Riemann integral so if you use Lebesgue integral over here for Fourier so how can you approach Fourier Transform from the Laplace Transform, as we say Fourier as a special case of Laplace Transform.

*There is a condition of absolute integrability for the existence of Fourier Transform which is a sufficient condition. Isn't there a necessary condition for it.

*Some of the functions such as cosine and sine do not meet the integrability condition but still they have the Fourier Transform. How would we take it.
I checked these question thoroughly on this forum but could not find any.
Please response to all of my above question. Thanks in advance.
 A: 
I have studied a lot of books for Fourier Transform and currently, I am confused weather I should use Riemann integral or Lebesgue integral for the Fourier Transform.

I'd suggest you use Lebesgue integral, it gives a more complete theory. But both are defined they would yield the same result result.

How would you differentiate between Fourier Transform of a function or a distribution, what I got is that you will use Lebesgue integral for a distribution whereas for a function you will use Riemann integral. May be, I would be wrong.

When transforming distributions you should be aware how they differ from functions. You define distributions as something that acts on a test function - or put in layman terms you define a distribution $u$ in a way in terms of what value the $\int u\phi dx$ takes depending on the test function $\phi$. The test function is supposed to be very well behaved and operations on distributions are often transferred to operation on the test function. The Fourier transform of a distribution is defined by using the Fourier transform of test functions. Since test functions are so well behaved it shouldn't matter if we use Lebesgue or Riemann integration.

The integral we use in Laplace Transform is Riemann integral so if you use Lebesgue integral over here for Fourier so how can you approach Fourier Transform from the Laplace Transform, as we say Fourier as a special case of Laplace Transform.

That's a bit problematic yes, but normally one would use the same type of integral. As pointed out it would only be a matter of making the theory more complete by using Lebesgue integral - with Riemann integral you only suffer from less integrands being integrable, but otherwise they don't differ.

There is a condition of absolute integrability for the existence of Fourier Transform which is a sufficient condition. Isn't there a necessary condition for it.

This sounds to be due to you're using Riemann integration. The Riemann integral and Lebesgue integral uses different approach when integrating over unbounded sets. While the Riemann integral approaches this by restricting the set with upper and lower bound and taking limit as the bounds goes to infinity the Lebesgue integral has it built into the definition from start.
The result is that Riemann integral has a well defined order of summing up the integral while Lebesgue integral doesn't. This means that a integrand that isn't absolutely integrable wouldn't be Lebesgue integrable. 
For Lebesgue integral it's required that the function is integrable (and therefore absolutely integrable), but that's also a sufficient condition.

Some of the functions such as cosine and sine do not meet the integrability condition but still they have the Fourier Transform. How would we take it.

Even though $\sin$ and $\cos$ need not be considered distribution you would need to do it in this case in order to allow for transformation as the result would need to be a distribution. For distributions the requirement is not the same, what we talk about here is called tempered distributions (where we allow a wider set of test function than normal distributions need to accept). By that one could accept a wider range of distributions that would be transformable - which is a bit of the point in using distributions.
To be clear how fourier transform on distribution work, recall that a distribution is a functional (ie function that takes function) that maps test functions to real numbers. Normally one denotes this mapping as $\langle u,\phi\rangle$ where $u$ is a distribution and $\phi$ is a test function (in laymans term this is thought as $\int u\phi dx$). 
When defining the Fourier transform of distributions we don't actually integrate the distribution, instead we throw the problem at the test function. So:
$$\langle \mathscr Fu, \phi\rangle = \langle u,\mathscr F\phi\rangle$$
which will mean that the only integral that is involved is when Fourier transforming test functions, but these are restricted to be very well behaved and can be integrated by using Riemann or Lebesgue integral.
To be concrete assume that we consider the distribution $u = e^{ix}$ we see that this is in fact the inverse fourier transform at $1$ as it's defined as $\langle u,\phi\rangle = \int e^{ix}\phi(x) dx  = (\mathscr F^{-1} \phi)(1)$, now look whats happen if we Fourier transforms $u$:
$$\langle\mathscr u, \phi\rangle = \langle u,\mathscr F\phi\rangle = (\mathscr F^{-1}\mathscr F\phi)(1) = \phi(1)$$
but that distribution is known as $\delta_1$ (since the dirac distribution is defined as $\langle \delta_a, \phi\rangle = \phi(a)$).
Note that you couldn't pull this of if you used the normal definition of the Fourier transform (for $\omega\ne1$:
$${1\over 2\pi}\int_a^b e^{ix} e^{-ix\omega} dx = {1\over 2\pi}\int_a^b e^{ix(1-\omega)} dx = {e^{ib(1-\omega)}-e^{ia(1-\omega)}\over 2\pi (1-i\omega)}$$ 
and the limit of that when $a,b\to\infty$ do not exists for any $\omega\ne1$, and it doesn't get better if $\omega=1$
$${1\over 2\pi}\int_a^b e^{ix} e^{-ix\omega} dx = {1\over 2\pi}\int_a^b e^{ix(1-\omega)} dx = {1\over 2\pi}\int_a^b 1dx = b-a$$
and using Lebesgue integration doesn't help.
