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Compute the Fourier transform of $f(t) = \sin t$. Does that converge?

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The transform exists in the sense of distributions only. Are you familiar with distributions? – Hans Lundmark Oct 22 '10 at 8:54
distributions are not familiar to me but I know the answer made by Maple: I*Pi*(-Dirac(omega-1)+Dirac(omega+1)) – alvoutila Oct 22 '10 at 9:53
Dirac δ is a distribution... – J. M. Oct 22 '10 at 10:22
Did we finally get our name changed, to – Aryabhata Oct 22 '10 at 13:51
The function must be in $L^1$ or (by density) $L^2$ or in $L^p$ $1 < p < 2$ (by interpolation) for the most common form of the Fourier transform to be defined, I believe. But, $f$ is not in any of these spaces. – Jonas Teuwen Oct 22 '10 at 14:41
up vote 3 down vote accepted

If you consider $f:=\sin$ as a function on $\mathbb R$ then it has a Fourier transform only in the distribution sense, as remarked in H.L.'s comment. If, however, you consider $\sin$ as a function on $\mathbb R/(2\pi)$ then its Fourier transform $\hat f$ is a function on $\mathbb Z$, namely the doubly infinite sequence $(c_k)_{k\in\mathbb Z}$ obtained by integration of $\sin$ against the functions $t\mapsto e^{-ikt}$: $$c_k={1\over 2\pi}\int_{-\pi}^\pi \sin(t)e^{-ikt}dt\qquad (k\in\mathbb Z).\qquad(*)$$ As $\sin$ is a nice function Fourier theory guarantees that we can get $\sin (t)$ back using the $c_k$ as follows: $$\sin (t)=\sum_{k=-\infty}^\infty c_k e^{ikt}.$$ Now, by Euler's formula, $\sin(t) =(e^{it}-e^{-it})/(2i)$, so we can immediately read off $\hat f(1)=c_1=-i/2$, $\hat f(-1)=c_{-1}= i/2$ and $\hat f(k)=c_k=0$ otherwise, without doing the integration (*).

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It is obvious that the Fourier series for $\sin t$ is $\sin t$!!! – Mariano Suárez-Alvarez Oct 22 '10 at 12:54
$\mathbb R/(2\pi)$ should be $\mathbb R/(2\pi \mathbb Z)$? – anon Oct 23 '10 at 0:03

$\sin(t)=\frac{e^{it}-e^{-it}}{2i}$. Since, $e^{it}$ and $e^{-it}$ are basis functions of the Fourier transform. From this easily can be computed using duality property and there will be no convergence issue.

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