I have a sine series given by
$\sum^\infty_{n=0}{\frac{\sin(n\theta)}{2n-1}}$,
and I would like to find the sum assuming that $0 < \theta < \pi$. Using some similar posts on this site I was able to express the sum as
$\text{Im} \left( \sqrt{e^{i\theta}} \space \text{arctanh}(\sqrt{e^{i\theta}}) - 1\right)$.
I'm not well-versed with these functions aside from their definitions, so how can I simplify this expression further?
You did a good job showing that $$\sum_{n=0}^\infty \frac{e^{i n t}}{2 n-1}=e^{\frac{i t}{2}} \tanh ^{-1}\left(e^{\frac{i t}{2}}\right)-1$$ Do the same $$\sum_{n=0}^\infty \frac{e^{-i n t}}{2 n-1}=e^{-\frac{i t}{2}} \tanh ^{-1}\left(e^{-\frac{i t}{2}}\right)-1$$ Combine them to get $$\sum_{n=0}^\infty \frac{\sin(n t)}{2 n-1}=\frac i 2 \left(e^{-\frac{i t}{2}} \tanh ^{-1}\left(e^{-\frac{i t}{2}}\right)-e^{\frac{i t}{2}} \tanh ^{-1}\left(e^{\frac{i t}{2}}\right) \right)$$ Now, using the hint given by metamorphy, $$\tanh ^{-1}\left(e^{-\frac{i t}{2}}\right)=\frac 12 \log\left( \coth \left(\frac{it}{4}\right)\right)=\frac{1}{2} \log \left(-i \cot \left(\frac{t}{4}\right)\right)$$ $$\tanh ^{-1}\left(e^{\frac{i t}{2}}\right)=\frac 12 \log\left( \coth \left(-\frac{it}{4}\right)\right)=\frac{1}{2} \log \left(i \cot \left(\frac{t}{4}\right)\right)$$ After a bunch of simplifications, you should get $$\sum_{n=0}^\infty \frac{\sin(n t)}{2 n-1}=\frac{\pi}{4} \cos \left(\frac{t}{2}\right)+\frac{1}{2} \sin \left(\frac{t}{2}\right) \log \left(\cot \left(\frac{t}{4}\right)\right)$$