Show that $\sum_\limits{k=1}^{\infty} (-1)^{k-1}\frac{\sin((2k-1)y)}{2k-1} = \frac{1}{2}\ln\tan{(\frac{\pi}{4}+\frac{y}{2})}$

Question

Show that $$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{\sin((2k-1)y)}{2k-1} = \frac{1}{2}\ln\tan{(\frac{\pi}{4}+\frac{y}{2})}$$ for $$-\frac{\pi}{2}<y<\frac{\pi}{2}$$

What I tried is $$\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$$ $$\frac{1}{1-(-z^2)}=\sum_{n=0}^{\infty}(-z^2)^n$$ $$\frac{1}{1+z^2}=\sum_{n=0}^{\infty}(-1)^nz^{2n}$$ $$\int\frac{1}{1+z^2}=\int\sum_{n=0}^{\infty}(-1)^nz^{2n}$$ $$\arctan z=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n+1}}{2n+1}$$

Let $k-1=n$, $$\arctan z=\sum_{k=1}^{\infty}(-1)^{k-1}\frac{z^{2k-1}}{2k-1}$$ Let $z=e^{iy}$ $$\Im \arctan ({e^{iy}})=\Im \sum_{k=1}^{\infty}(-1)^{k-1}\frac{{e^{iy(2k-1)}}}{2k-1}={\sum_{k=1}^{\infty} (-1)^{k-1}\frac{\sin((2k-1)y)}{2k-1}}$$

What is $\Im \arctan ({e^{iy}})$?

Answer 1

Let's calculate the l.h.s. It's the antiderivative of $\;\displaystyle\sum_\limits{k=1}^{\infty} (-1)^k\cos(2k-1)y=-\cos y+\cos 3y-\cos 5 y+\dotsm$

Now $\cos y-\cos 3y+\cos 5 y-+\dotsm\;$ is the real part of $$\text{e}^{iy}-\text{e}^{3iy}+\text{e}^{5iy}-\dotsm=\frac{\text{e}^{iy}}{1+\text{e}^{2iy}}=\frac{1}{\text{e}^{-iy}+\text{e}^{iy}}=\frac1{2\cos y}.$$ Thus \begin{align} \sum_{k=1}^{\infty}(-1)^{k-1}\frac{\sin(2k-1)y}{2k-1}&=-\int\frac{\text{d}\mkern1muy}{2\cos y}=-\frac12\int\frac{\cos y\,\text{d}\mkern1muy}{\cos^2 y} \cr & =-\frac12\int\frac{\cos y\,\text{d}\mkern1muy}{1-\sin^2 y} =-\frac12\int\frac{\text{d}\mkern1mu u}{1-u^2}\qquad(u=\sin y)\cr &=-\frac14\int\frac{\text{d}\mkern1mu u}{1-u}-\frac14\int\frac{\text{d}\mkern1mu u}{1+u}\cr &=\frac14\,\ln\frac{1-u}{1+u}=\frac14\,\ln\frac{1-u}{1+u}=\frac14\,\ln\frac{1-\sin y}{1+\sin y} \end{align} We'll rewrite this last fraction, doing some trigonometry: we'll use the formula $\;\sin y =\dfrac{2t}{1+t^2}$, where $t=\tan\dfrac y2$: $$\frac{1-\sin y}{1+\sin y}=\frac{1+\cfrac{2t}{1+t^2}}{1-\cfrac{2t}{1+t^2}}=\frac{1+t^2+2t}{1+t^2-2t}=\biggl(\frac{1+t}{1-t}\biggr)^2=\biggl(\tan\Bigr(\dfrac \pi4+\dfrac y2\Bigr)\biggr)^2.$$ by the addition formula for the tangent. Finally we have $$\sum_{k=1}^{\infty}(-1)^{k-1}\frac{\sin(2k-1)y}{2k-1}=\frac12\,\ln\,\biggl\lvert\tan\Bigr(\dfrac \pi4+\dfrac y2\Bigr)\biggr\rvert.$$