I am trying to verify that the identity $$ \int_{\delta\leq |t|<\frac{1}{2}}f(x-t)\frac{\sin(\pi(2N+1)t)}{\sin(\pi t)}dt=\widehat{(ge^{\pi i\cdot})}(N)+\widehat{(ge^{-\pi i\cdot})}(-N) $$ where $$ g(t)=\frac{f(x-t)}{2i\sin(\pi t)}\chi_{\{ \delta\leq|t|<\frac{1}{2} \}}(t) $$ where $g$ is integrable holds. My question is how to go about attacking this integral? Would integration by parts by a good first step?
Thank you in advance for your comments and help.
I show you some steps. Your left-hand side can be written $$ \int_{-\infty}^{+\infty} g(t)2i\sin(\pi(2N+1)t)\,dt $$ Then, also use that $$ 2i\sin(\pi(2N+1)t)=e^{i\pi(2N+1)t}-e^{-i\pi(2N+1)t}, $$ and hence $$ g(t)2i\sin(\pi(2N+1)t)=\bigl(e^{i\pi t}g(t)\bigr)e^{2\pi iNt}+\bigl(e^{-i\pi t}g(t)\bigr)e^{-2\pi iNt}. $$ Can you take it from here? (The final result depends a bit on where one put the constant $2\pi$ in the definition of the Fourier transform.)