# Fourier Transform - Conditions

I've been looking at the topic in the Farlow book on PDEs, and it says:

"The major drawback of the Fourier Transform is that not all the functions can be transformed; for example even simple functions like $f(x)=e^x,f(x)=c\in\mathbb{R},f(x)=sin(x)$ cannot be transformed, since the integral involved does not exist. Only functions that damp to zero sufficiently fast have transforms."

I've looked also in the Tolstov book on Fourier Series that this condition of damping to zero is more like the function is absolutely integrable in the real axis.

But looking in a table of transforms (Or even looking at the development of this table) I can see that the transform of $sin(x)$ is $i\sqrt{\pi/2}(\delta(\omega+a)-\delta(\omega-a))$. And the transform of a constant is of the form $c\delta(t)$ for $c$ some other constant.

So I don't fully understand this restriction. I think that even if the function is not absolutely integrable, the function can still be transformed using these kind of weird "functions" like the $\delta$. But the transform won't be a function in the usual sense of the word. Am I right?

The Fourier transform as defined by the integral $\int_{-\infty}^{\infty}f(x) e^{-iux}dx$ exists if and only if $f$ is absolutely integrable.
However, the Fourier transform can be defined in a sensible way for functions not meeting this requirement. For example, the Fourier transform can be extended to functions which are in $L^2$ but not $L^1$ using a limiting argument.
And, as applicable to this question, the Fourier transform can also be extended to the class of "tempered distributions," also sometimes called generalized functions. Indeed, the Fourier transform is a bijective operator on this space, meaning that the Fourier transform of a tempered distribution is another tempered distribution, and the transform can be inverted. Objects such as the $\delta$ distribution are examples of tempered distributions which are not functions.