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The finite Fourier sine transform of $f(x),0<x<l$ is defined as, $$f_s(n)=\int_0^lf(x)\sin\frac{n\pi x}{l}\:dx$$ And the inverse finite Fourier sine transform of $f_s(n)$ is defined as, $$f(x)=\frac{2}{l}\sum_{n=l}^\infty f_s(n)\sin\frac{n\pi x}{l}$$

Okay, I have some intuition about Fourier transform and inverse Fourier transform. Even their formula seems very similar, $$F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\:dt$$ $$f(t)=\frac{1}{2\pi}\int_{-\infty}^\infty F(\omega)e^{i\omega t}\:d\omega$$

But, when I see finite Fourier sine transform, and it's inverse, I couldn't understand why the transform use integral and the inverse use summation. Is there any interpretation of that?

WhyMeasureTheory
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  • The "Fourier sine transform" that you're referring to is a version of the Fourier series. It might be better to compare the Fourier transform to the complex version of the Fourier series, whereby $$ f_s(n) = \int_0^l f(x) e^{2 \pi i x/l},dx $$ – Ben Grossmann Nov 29 '21 at 14:53
  • One very coarse interpretation of the discrepancy is that a function over a finite interval requires only a countable "amount of information", while a function over $\Bbb R$ requires an uncountable amount. A more precise version of this might be a presentation of the Fourier transform as a limit of Fourier series. You might find this post to be helpful – Ben Grossmann Nov 29 '21 at 15:06
  • A more sophisticated explanation is that the Pontryagin dual of $\Bbb R$ is $\Bbb R$, whereas the Pontryagin dual of the circle $\Bbb T$ is $\Bbb Z$. – Ben Grossmann Nov 29 '21 at 15:15
  • How the Pontryagin duality explain the discrepancy of my question? @BenGrossmann I couldn't understand that concretely. Could you explain it a little more? – WhyMeasureTheory Nov 29 '21 at 17:01
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    Let $\Bbb T$ denote the set of complex numbers with magnitude $1$. The fact that we need a coefficient for every element of $\Bbb Z$ in order to represent a periodic function over $[0,l]$ corresponds to the fact that the set of group homomorphisms from $[0,l]/0\sim l$ to $\Bbb T$ is isomorphic to $\Bbb Z$. Similarly, the fact that we need a whole function over $\Bbb R$ in order to represent a function over $\Bbb R$ corresponds to the fact that the set of group homomorphisms from $\Bbb R$ to $\Bbb T$ is isomorphic to $\Bbb R$. – Ben Grossmann Nov 29 '21 at 18:39

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