Prove that
$$ X_N(w) = \sum_{n=-N}^{N} \frac{\sin \omega_c n}{\pi n}e^{-j \omega n} $$
mayb be expressed as
$$ X_N(w) = \frac{1}{2 \pi} \int^{\omega_c} - \omega_c \frac{\sin[(2N+1)(\omega - \frac{\theta}{2})]}{\sin[\frac{(\omega - \theta)}{2}]}d\theta$$
I started by trying to build up the function x(n) and w(n):
$$ x(n) = \frac{\sin\omega_c n}{\pi n} $$
and $$ \omega_c (n) = 1 , \quad -N \le n \le N$$
Then I figured that I could use the property of the Fuorier Transform for the multiplication of two signals:
$$ x(n)w(n) = \frac{1}{2 \pi} \int_{-\pi}^{\pi} X(\Theta)W(\omega - \Theta)d\Theta$$
where capital $X(\Theta)$ and $W(\Theta)$ are the DTFT. Is this the correct way to go by it? Can anyone help me complete this? I honestly don't know how to continue to make it look like the requested form....\
Edit: I found the answer from the solution manual, but I feel like it's not complete and I am not sure it is correct. Here is an image of the solution:
Is it correct, and if so can you explain what it did exactly? Why is it missing the $ - \omega_c $ ?
$$2\sum_{n=-N}^N e^{i b n} \frac{\sin( an)}{n} = \sum_{n=-N}^N e^{i b n} \int_{-a}^a e^{i nt}dt= \int_{-a}^a \sum_{n=-N}^N e^{i (b+t) n}dt \\=\int_{-a}^a e^{-i (b+t)N} \sum_{n=0}^{2N} e^{i (b+t) n}dt=\int_{-a}^a e^{-i (b+t)N} \frac{1-e^{i (b+t) (2N+1)}}{1- e^{i (b+t) }}dt\\ =\int_{-a}^a \frac{e^{-i (b+t) (2N+1)/2}-e^{i (b+t) (2N+1)/2}}{e^{-i (b+t)/2}- e^{i (b+t)/2 }}dt=\int_{-a}^a \frac{\sin((b+t) (2N+1)/2)}{\sin((b+t) /2)}dt$$