In derivation of fourier transform, we start with the fourier series coefficients. If we let $T \to \infty$, it's common to say the spacing between consecutive fourier coefficient will approach $0$, and we get a continuous spectrum rather than distinct values.

So basically by letting $\omega_0 \to 0$, $n\omega_0$ becomes $\omega$. My question is - why? Isn't it an intederminate form? Maths is not about pure manipulation of symbols, we cannot magically interpret infinitely closely spaced coefficient as a continuous variable $\omega$. No matter how infinitely close to each other these coefficient will be, there will be places "without" them. We'd like to create a function with the domain of all real numbers, right? This is what we mean by "continuous spectrum". But putting these coefficients infinitely close to each other doesn't mean it's a continuous spectrum...

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Source and derivation of fourier transform from fourier series

  • $\begingroup$ No, you can just define a Fourier transform directly as $$\hat{f}(x)=\int_{-\infty}^\infty e^{itx}f(t)\,dt$$ for any $f$ such that $\int_{-\infty}^\infty |f(x)|\,dx<\infty$. See en.wikipedia.org/wiki/Fourier_transform for (a lot) more details on how this is done rigorously in general. $\endgroup$ Feb 8, 2015 at 10:38
  • $\begingroup$ I don't see your point here. Fourier transform is a generalization of fourier series, right? I'm just asking if the transition from FS to FT is rigorous enough, i.e. how to justify the transformation from distinct values of fourier coefficients into a continuous function. It's done by noting that if we let $\omega_0 \to 0$ then we can treat $n \omega_0$ as a continuous variable. My question is - why? Because we can express any real number by $n \omega_0$ as $\omega_0 \to 0$? If so, how can we actually PROVE we really can? $\endgroup$ Feb 8, 2015 at 11:46
  • $\begingroup$ You asked if that was the most rigorous definition we could come up with, so I told you "no, this is." You also asked why that change was justified, but I didn't answer that because it's basically symbol pushing and definitions of how integration works on continuous functions, and it was too tedious for me to post. Though, if you're interested, the link I posted addresses that as well in the definition section if you read it. $\endgroup$ Feb 8, 2015 at 11:50
  • $\begingroup$ OK, could answer this one briefly. Can we express every real number as $n \Delta\omega$, where $n$ is an integer and $\Delta \omega \to 0$? If yes, why? $\endgroup$ Feb 8, 2015 at 12:03
  • $\begingroup$ That's not a well-defined question: if $\Delta\omega\to 0$ then it isn't fixed, so neither is $n\Delta\omega$ hence it cannot be equal to a number since numbers don't change. $\endgroup$ Feb 8, 2015 at 12:06

1 Answer 1


Yes, you can make the transformation to the continuous transform case rigorous under restricted conditions, but it's rigorous development is more trouble than it's worth. However, the intuition in this argument is worth mentioning when trying to motivate the introduction of the Fourier transform.

The reason the argument has hung around so long is that Fourier came up his transform by using this argument. It's about the only natural and compelling derivation leading the Fourier transform; it motivated Fourier, and it's Fourier's argument. There are reasons to consider the discrete series, but Fourier's reasoning remains the only natural motivation for considering "continuous" integral versions of Fourier expansions, at least at an any reasonable elementary level.


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