# If unit step function is not absolutely integrable. then how does it have fourier transform?

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]

• 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>.
– MBaz
Feb 18 at 19:07
• .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. Feb 18 at 19:35
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.