# Fourier transform of sine and cosine function

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 $\mathcal{F}(f(at))$=$\frac{1}{|a|} \mathcal{F(f(t))}(\frac{x}{a})$

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– Henning Makholm Feb 7 '12 at 0:32
You can make this computation rigorous using the theory of distributions. – Potato Nov 24 '13 at 20:24

1. The fact that the function is not in $L^1$ does not imply that it does not have a Fourier transform, as this example shows. What the classical theory tells us is that if the function is in $L^1$ then its Fourier transform is well defined.
2. It does not defy the scaling property: Without loss of generality forget about all the $2\pi$ factors in your derivations (both in the Fourier transform and in the function). Then $$\mathcal{F}[sin(kt)](x)=\frac{i}{2}[\delta(x+k)-\delta(x-k)],$$ so $$\frac{1}{|k|}\mathcal{F}[sin(t)](x/k)=\frac{i}{2}\Big[\frac{\delta(x/k+1)}{|k|}-\frac{\delta(x/k-1)}{|k|}\Big].$$ The function $\frac{\delta(x/k+1)}{|k|}$ is zero everywhere except at -k and $$\int_{\mathbb{R}}\frac{\delta(x/k+1)}{|k|}dx=\int_{\mathbb{R}}\delta(y+1)dy=1,$$ so it is equal to $\delta(x+k)$. The same happens with $\frac{\delta(x/k-1)}{|k|}$.