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Assuming we denote Fourier transforms as follows:

$\mathcal{F}[f(t)](\omega)=\tilde{f}(\omega)$, then we have the following identity:

$\mathcal{F}[\frac{df}{dt}](\omega)=i\omega \tilde{f}(\omega)$

In addition, we know that the fourier transforms of sines and cosines are given by

$\mathcal{F}[\sin(\omega_0 t)]=\frac{i}{2}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$

and

$\mathcal{F}[\cos(\omega_0 t)]=\frac{1}{2}\left(\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right)$

However, if we apply the first identity:

$\mathcal{F}[\cos(\omega_0 t)]=\mathcal{F}[\frac{1}{\omega_0}\frac{d}{dt}\sin(\omega_0 t)]=\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$

which looks nothing like the first formula for the fourier transform of the cosine. I realize it's zero everywhere except at $\pm \omega_0$, so the prefactor isn't an issue, but the signs are wrong.

Where is the error?

EDIT: corrected sign on initial identify as per first comment

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  • $\begingroup$ I'll correct that, but it still doesn't fix the issue - the extra negative signs switches the signs of both delta functions, but still leaves one negative and one positive on the last line, whereas the cosine transform has both positive signs. I think the choice of whether the initial identity is positive or negative is one of convention anyway, no? I can choose that sign as I please, as long as I have the opposite sign on the inverse transform. $\endgroup$ – KBriggs Apr 6 '16 at 20:40
  • $\begingroup$ I don't see how gets rid of the sign: $(\delta(t-t_0)+\delta(t+t_0))g(t)=\delta(t-t_0)g(t_0)+\delta(t+t_0)g(t_0)$, but $(\delta(t-t_0)-\delta(t+t_0))g(t)=\delta(t-t_0)g(t_0)-\delta(t+t_0)g(t_0)$ - the second one still has a negative prefactor. It's the negative sign between the two delta functions that is the issue, not the negative signs in the arguments. $\endgroup$ – KBriggs Apr 6 '16 at 20:47
  • $\begingroup$ forget the Fourier transform of $\sin t$ and $\cos t$, consider only the Fourier transform of $e^{i \omega_0 t}$ $\endgroup$ – reuns Apr 6 '16 at 20:51
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$$\mathcal{F}[\frac{df}{dt}](\omega)=i\omega \tilde{f}(\omega) $$

and

$$\mathcal{F}[\cos(\omega_0 t)]=\mathcal{F}[\frac{1}{\omega_0}\frac{d}{dt}\sin(\omega_0 t)]=\frac{1}{\omega_0}(i\omega)\frac{i}{2}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)=\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)$$

$$\frac{-\omega}{2\omega_0}\left(\delta(\omega+\omega_0)-\delta(\omega-\omega_0)\right)=\left(\frac{-\omega}{2\omega_0}\delta(\omega+\omega_0)-\frac{-\omega}{2\omega_0}\delta(\omega-\omega_0)\right)=\left(\frac{\omega_0}{2\omega_0}\delta(\omega+\omega_0)+\frac{\omega_0}{2\omega_0}\delta(\omega-\omega_0)\right)=\frac{1}{2}\left(\delta(\omega+\omega_0)+\delta(\omega-\omega_0)\right)$$

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  • $\begingroup$ And suddenly everything is clear. I was simply misunderstanding your explanation. Thanks. $\endgroup$ – KBriggs Apr 6 '16 at 20:50
  • $\begingroup$ sorry man. your first statement about $$\mathcal{F}[\frac{df}{dt}](\omega)=i\omega \tilde{f}(\omega)$$ was true. $\endgroup$ – K.K.McDonald Apr 6 '16 at 21:20

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