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As far as I know that one of the sufficient conditions of fourier transform to exist is that the function must be absolutely integrable(I learnt this from "Continuous and Discrete Signals and Systems" by Samir S. Soliman, 2nd edition). But so I'm concerned that the unit step function is not absolutely integrable yet it has fourier transform. I know that condition isn't necessary condition so there are fourier transforms for those functions which do not obey that condition. However if transforms of any function cannot be determined by the fundamental approach then how come we determine those using the properties that we do in case of unit step function. Moreover, I came to know through ChatGPT that fourier transform of distributional functions exist. What are these distributional class of functions? [My knowledge so far is limited within the scopes of chapter 1 to 4 of the book I mentioned earlier]

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    $\begingroup$ It works out because the math allows it, at the cost of introducing Dirac deltas. Also: personally, I wouldn't put any trust on chatbots; stick to textbooks </rant>. $\endgroup$
    – MBaz
    Feb 18 at 19:07
  • $\begingroup$ Please read the answers to this and this question. Then come back and edit your question to address any remaining doubts. $\endgroup$
    – Matt L.
    Feb 18 at 19:30
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    $\begingroup$ One very common theme in mathematics is that you start with a definition that is intuitive in one way or another but turns out to be limited. Then we explore possible unique ways to generalise the definition. Some of the established methods for doing that are analytic continuation, Cesaro summation, weak derivatives or topologies, etc. This is one such case and in fact, all of the methods I've just listed can be applied. And they all get you to the same construct. It's important to understand that sometimes, stretching some of the interesting properties can lose other nice-to-have ones. $\endgroup$
    – Jazzmaniac
    Feb 18 at 19:31
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    $\begingroup$ .cont'd: In that case, you can establish several different sets of frameworks with different rules and different scopes. For example, if you look at the Fourier transform in $L^2$, you can get a lot of nice theorems, but sometimes it still makes sense to work in $L^\infty$, because that's where some other things make more sense. Your specific question is also related to the generalisation of an inner-product space to a space with a dual of linear forms. This is also a very common theme and closely related to the concept of distributions and weak derivatives. $\endgroup$
    – Jazzmaniac
    Feb 18 at 19:35
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    $\begingroup$ Lastly, if you would like to understand all of this better, then I can recommend picking up a textbook on functional analysis. The most important concepts that you seem to be asking for are usually introduced right at the beginning. $\endgroup$
    – Jazzmaniac
    Feb 18 at 19:45

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In a sense, the Heaviside unit step function doesn't have a Fourier Transform, even if it has a Laplace Transform. Just as the non-zero constant function has no Fourier Transform. Not at every frequency. Not at $f=0$. If $u(t)$ is the unit step function, then $U(0)$ doesn't exist.

But, if you throw some limits at it, and stay away from $f=0$, you can get an expression for $U(f)$ for $f \ne 0$.

This stuff they're saying about including the dirac delta function is true, but $\delta(0)$ doesn't exist. If we're sloppy we might say $\delta(0)=\infty$. But that's a little sloppy.

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