So say I do a fourier transform of a rectangular function with magnitude 1 from (0, NT). The fourier transform of this would be: [itex]f(jΩ) = \frac{1-e^{-jΩNT}}{jΩ} = NT\cdot{e^{-jΩNT/2}}\cdot{sinc(ΩNT/2)}[/itex] Now say if I sample this rectangle at time T producing N samples, the DTFT of this is: [itex]f(e^{jw}) = \frac{1-e^{-jwN}}{1-e^{-jw}} = e^{-jw(N-1)/2}\cdot\frac{sin(wN/2)}{sin(w/2)}[/itex] Since DTFT and Fourier Transform is related by Ω = wT where [itex]f(e^{jw}) = \frac{1}{T}\sum{f(jΩ + j2\pin)}[/itex] Now if I try this method I get to this point: [itex]f(e^{jw}) = e^{-jw(N-1)/2}\cdot{sin(wN/2)}\cdot{\sum\frac{1}{w+2{\pi}n}}[/itex] This is where I get stuck, because that last summation needs to somehow equal sin(w/2) or 1-e^{-jw}. The summation is from negative infinity to infinity. Was wondering if there is some math trick that gives the result of that summation. Thanks.