Fourier transform of exponential

Question

When I read wikipedia I see that Fourier transform is defined for functions in $L^1.$ And then I see that the Fourier transform of the complex exponential function $f(t)=e^{it},$ is given by $\hat{f}(\omega)=\delta(\omega-1).$ But $f(t)\notin L^1.$ Can someone clarify this discrepancy?

Answer 1

This is most likely done in the sense of distributions. You can define the Fourier transform of a so-called tempered distribution.

A distribution is a continuous linear functional on the space $C_c^{\infty}(\mathbb{R})$ of smooth functions with compact support. These include locally integrable functions, measures such as the $\delta$ "function", and many other things. A tempered distribution doesn't grow too fast at infinity in some sense, so we exclude things like $e^{t^2}$ but any bounded function such as $e^{it}$ is tempered.

This is a fascinating topic, and this comment can not possibly do justice to it.

The Fourier transform of tempered distributions obeys the same algebraic rules as the usual one, so that for instance: $$\widehat{e^{it}g(t)}=\widehat{g}(\cdot\,\,-1)$$ (which we know is valid if $g$ is nice enough). So we only need to show that $$\widehat{1}=\delta$$ To do so we approach the constant function $1$ by $$f_{\varepsilon}(t)=e^{-\varepsilon t^2}$$ which tends to $1$ sufficiently well so that $f_{\varepsilon}\rightarrow 1$ in the sense of distributions. The Fourier transform of $f_{\varepsilon}$ is of course well known and is $$\widehat{f_{\varepsilon}}(\omega)=\sqrt{\dfrac{\pi}{\varepsilon}}e^{-\pi^2\omega^2/\varepsilon}$$ and as $\varepsilon\rightarrow 0$ you can see that this approximates the delta "function". The functions $f_{\varepsilon}$ concentrate near $t=0$ while getting higher and higher. So this shows that $e^{it}$ can be approximated by integrable functions whose Fourier transform approaches $\delta(\omega-1)$ in some way. To be able to write $\widehat{e^{it}}=\delta(\omega-1)$, you need however to look at these objects as distributions.