Find the inverse Fourier transform of the following:
$$\sin(2 \pi \nu T) \cos (10 \pi \nu T) / (\nu T)$$
Attempt:
I was told it was easier if we rewrite this in terms of a $sinc$ function. I think this would be useful since we already know that the inverse FT of the $sinc$ function is the rectangular pulse function ($rect(t)$).
So the inverse Fourier transforms for this function becomes:
$$\int^\infty_{-\infty} \left( \frac{\sin(2 \pi \nu T)}{\nu T} \ \cos (10 \pi \nu T) \right) \ e^{j 2 \pi \nu T} d \nu$$
I'm stuck here. It looks like it requires some form of integration by parts for three terms. How should I proceed from here?
And what can I do with the $2 \pi$ scaling parameter in the $sinc$ function?
Any help would be greatly appreciated.
It will be simpler if you express $\sin(2 \pi \nu T) \cos (10 \pi \nu T) $ as $$\sin(2 \pi \nu T) \cos (10 \pi \nu T) =\frac{1}{2}\left(\sin(12\pi\nu T)-\sin(8\pi\nu T)\right)$$ so that you can write the function as a difference between two sinc functions.
As for the scaling factor, just multiply numerator and denominator by $2\pi$, so that you have $$\DeclareMathOperator{\sinc}{sinc}\frac{\sin(2\pi\nu T)}{\nu T}=\frac{2\pi\sin(2\pi\nu T)}{2\pi\nu T}=2\pi \sinc(2\pi\nu T)$$