Why is the Fourier transform valid only for absolutely integrable signals?

For example, why can't we do the Fourier transform of exponential order functions?

  • 1
    $\begingroup$ The Fourier transform can be also defined for energy signals that are not integrable (for example, the sinc function). See Chapter 6 of "A Foundation in Digital Communication" by A. Lapidoth, available for free online. $\endgroup$
    – MBaz
    Apr 20, 2018 at 22:27
  • $\begingroup$ So the Fourier Transform for $e^{at}$, for $a$ a real positive number, is ....? $\endgroup$
    – Andy Walls
    Apr 21, 2018 at 3:57

2 Answers 2


Fourier on exponential functions can indeed make sense. In theory, under specific conditions. And in practical DSP, not likely.

There are so many conditions under which Fourier transformations exist that whole books are devoted to them, like D. C. Champeney, A Handbook of Fourier Theorems, Cambridge University Press, 1987. And the domain remains an open topic, in other words not all conditions are known, under which Fourier series are unique for instance.

Simply put: if $x(t)$ is absolutely integrable ($x\in L_1$ space), so is $x(t)e^{-jwt}$, by the virtue of the dominated convergence theorem, since $|x(t)e^{-jwt}|\le | x(t)|$. As such, this is a very direct condition to satisfy. By the way, note that $e^{-jwt}$ IS NOT absolutely integrable, an interesting paradox.

In signal processing applications and in statistics, as orthogonality or energy preservation is important, it is common to further restrict the domain to $x\in L_1 \cap L_2$. And this is not necessary for discrete signals, as $\ell_2 \subset \ell_1$.

As exponential functions $\exp(-at)$, $a\neq 0$, are not integrable on the whole $\mathbb{R}$-line, they should be truncated or windowed to be used in practice.

And if you REALLY want to define it, you can go to some distribution theory, see for instance Fourier transform of (real) exponential, where you can restrict the space of tests functions in the spirit of windowing.


Note that Fourier transform is defined as $X(j\omega) = \int_{-\infty}^{\infty}x(t) e^{-j\omega t} dt$. For this to have closed form expression, $x(t)$ should be absolutely integrable. But this is just a sufficient condition but not a necessary condition. That is, even though signal is not absolutely integrable, still Fourier transform exist.

For exponential function, the term $\int_{-\infty}^{\infty}e^{at} e^{-j\omega t} dt$ cannot have closed form expression.


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