How to unterstand the condition of absolute integrability for Fourier transform?

Question

To apply the Fourier transform on a function, it has to be absolutely integrable, so the following must be true: $$\int_{-\infty}^\infty \Big | f(t)\Big | \, dt<\infty.$$

For example this is not true for a $\sin$ or $\cos$ function because they are periodic and with $|\cdot|$ this integral becomes greater $0$ (and $\infty$ for the function above).

In a network in the sense of electrical engineering, inductivities and capacities may differentiate input function. For example: $$u(t)=L*\frac{\mathrm{d}i(t)}{\mathrm{d}t},$$ $$\underline{U}=j\omega L*\underline{I}.$$ The first function relates real time functions and the second one describes it via complex $(\underline{\cdot})$ functions. In certain steps of calculation, both perspectives are related via the Fourier transform.

This is (more or less) equal to the equation $$\mathcal F\{f'(t)\}(\omega)=i\omega *\mathcal F\{f(t)\}$$ where $\mathcal F\{\cdot\}$ applies the Fourier transform on a function and $f'$ is the derivate of $f$.

In electrical engineering, you often use harmonic functions like $\sin$, $\cos$. The Fourier transform works very well to make complex networks easier.

But is it allowed (considering the absolute integrability ) to perform this transform on such harmonic functions? It is done very often!

Answer 1

The Fourier transform is generally defined on a much larger space than $L^1$, including all so-called tempered distributions. In particular the unitary Fourier transform (with a $\frac{1}{\sqrt{2\pi}}$ in front of both the forward and inverse transforms) obeys the identity

$$\langle \mathcal{F}(d),f \rangle = \langle d,\mathcal{F}^{-1}(f) \rangle$$

for any tempered distribution $d$ and Schwarz function $f$. This bracket notation is the pairing between a distribution and a test function, which for a pair of functions just consists of integrating the product of the pair of functions. A wide variety of things form tempered distributions, including all bounded* functions.

The catch is that the Fourier transform of a tempered distribution is in general not a function, even if the distribution itself is given by integration against a function. For example, the Fourier transform of $e^{ix}$ is a Dirac delta, which physically represents the fact that $e^{ix}$ consists of just one frequency.

Other normalizations of the Fourier transform obey the same identity but with some power of $2\pi$ on one side or the other.

* Technically all bounded measurable functions, but that is a distinction only mathematicians are likely to worry about.

Answer 2

If the function is locally integrable, and globally square-integrable, then you're in business. This is because of the Plancherel identity $\|f\|_{L^2}=\|\hat{f}\|_{L^2}$, which is to say $$ \int_{-\infty}^{\infty}|f(t)|^2dt = \int_{-\infty}^{\infty}|\hat{f}(s)|^2ds. $$ If $f$ is in $L^2$ (i.e., square integrable,) then $\int_{-R}^{R}f(t)e^{-ist}dt$ exists because $|f| \le \frac{1}{2}(|f|^2+1)$ forces $|f|$ to be locally integrable in that case. And, then, $\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(t)e^{-ist}dt$ converges in mean to a function generally denoted by $\hat{f}$ as $R\rightarrow \infty$, which follows from the Plancehrel identity. So square integrable is a context where the Fourier transform and its inverse are well-defined. That's a context that fits well with Electrical Engineering and is also semi-classical.