5
$\begingroup$

For the sine function we can do the following formal computation:

$$\mathcal{F} (\sin(2\pi kt))(x) = \int_{-\infty}^{\infty} e^{-2\pi i xt} \frac{e^{2\pi i kt}-e^{-2\pi i kt}}{2i}dt= \frac{i}{2} \int_{-\infty}^{\infty} (-e^{2\pi i (x-k)t}+e^{-2\pi i (x+k)t})dt$$ $$ = \frac{i}{2}[\delta(x+k)-\delta(x-k)]$$

Is there a formal way to make sense of this given the fact that $\sin(2\pi k t)\notin L_1$ and so the Fourier transform doesn't converge in the typical sense. I also noticed that it seems to defy the scaling property $\mathcal{F}(f(at))$=$\frac{1}{|a|} \mathcal{F(f(t))}(\frac{x}{a})$

  • 0
    Duplicate of [Scaling property of Fourier series and Fourier transform](http://math.stackexchange.com/questions/105877/scaling-property-of-fourier-series-and-fourier-transform)?2012-02-07
  • 0
    You can make this computation rigorous using the theory of distributions.2013-11-24

1 Answers 1