Show that the partial sum in equation (3) may be written as:$$f_N(x)=\frac{2}{\pi}\int_{0}^x \frac{\sin(2Nt)}{\sin(t)}\,dt$$
Can someone please explain me how to show these 2 are equal? The first step I can only think of is to use property of an even function.
Equation (3): $$ f_N(x) = \frac{4}{\pi} \sum_{n=1}^N \frac{\sin((2n-1)x)}{2n-1} $$
Observe that both terms of the desired identity vanish at $x=0$, then by differentiating (a finite sum and a primitive), it is sufficient to prove that $$ \sum_{n=1}^{N} \cos ((2n-1)\theta)=\frac{\sin (2N\theta)}{2\sin\theta}.\tag1 $$
To prove $(1)$, here are some possible steps: $$ \begin{align} \sum_{n=1}^{N} \cos ((2n-1)\theta)&=\Re \sum_{n=1}^{N} e^{i(2n-1)\theta}\\\\ &=\Re\left( e^{i\theta}\frac{e^{2iN\theta}-1}{e^{2i\theta}-1}\right)\\\\ &=\Re\left( e^{i\theta}\frac{e^{iN\theta}\left(e^{iN\theta}-e^{-iN\theta}\right)}{e^{i\theta}\left(e^{i\theta}-e^{-i\theta}\right)}\right)\\\\ &=\Re\left( e^{iN\theta}\frac{2i\sin(N\theta)}{2i\sin\theta}\right)\\\\ &=\Re\left( \left(\cos (N\theta)+i\sin (N\theta)\right)\frac{\sin(N\theta)}{\sin\theta}\right)\\\\ &=\frac{\cos (N\theta)\sin (N\theta)}{\sin\theta}\\\\ &=\frac{\sin (2N\theta)}{2\sin\theta}. \end{align} $$
Hope this helps!
– user2543 Mar 11 '15 at 20:56