I'm trying to find the answer to this series: $$\sum_{p=1}^{\infty} \frac{(\sin(\frac{\pi}{4}p))(1-\cos(\alpha p))}{p}$$ where $\alpha$ is a constant. This is my attempt at solving it: $$\sum_{p=1}^{\infty} \frac{(\sin(\frac{\pi}{4}p))(1-\cos(\alpha p))}{p}= \sum_{p=1}^{\infty}\frac{(\sin(\frac{\pi}{4}p))}{p}-\sum_{p=1}^{\infty}\frac{\sin(\frac{\pi}{4}p)\cos(\alpha p)}{p}$$
I already know that the first term equals $\frac{\pi-\frac{\pi}{4}}{2}=\frac{3\pi}{8}$ but whatever I do to the second term it'll end up being something independent of $\alpha$ which doesn't make sense. For example: $$\begin{align}\sum_{p=1}^{\infty}\frac{\sin(\frac{\pi}{4}p)\cos(\alpha p)}{p}&=\frac{1}{2}\sum_{p=1}^{\infty}\frac{\sin((\frac{\pi}{4}+\alpha)p)}{p}+\frac{1}{2}\sum_{p=1}^{\infty}\frac{\sin((\frac{\pi}{4}-\alpha)p)}{p}\\ \\ &=\frac{1}{4}(\pi-(\frac{\pi}{4}+\alpha))+\frac{1}{4}(\pi-(\frac{\pi}{4}-\alpha))=\frac{3\pi}{8}\end{align}$$ Where am i making a mistake? I'd be glad if someone could help me out!
Note the series $$S(x)=\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}2,~~~x\in(0, 2\pi),~~~S(x+2\pi)=S(x)$$
hence, $$\lim_{x\to0^+}S(x)=\frac{\pi}2,~~S(0)=0,~~\lim_{x\to0^-}S(x)=-\frac{\pi}2$$ This tells us $S(x)$ is discontinuous at $x=0$. Actually, $S(x)$ is discontinuous at $x=2k\pi, k\in\mathbb{Z}$. Therefore, your derivation works only if
$$\frac{\pi}{4}\pm\alpha\in (2k\pi, ~2k\pi+2\pi)$$
However, at those discontinuous endpoints, your result doesn't hold, for example, as your claim:
If you set $\alpha=\frac{\pi}4$
$$T(\frac{\pi}4)=\frac{1}{2}\sum_{p=1}^{\infty}\frac{\sin(\frac{\pi}{2}p)}{p}+\frac{1}{2}\sum_{p=1}^{\infty}\frac{\sin(0p)}{p}=\frac{\pi}8$$
If you let $\alpha\to\frac{\pi}4^+$
$$\lim_{\alpha\to \frac{\pi}4^+}T(\alpha)=\frac{\pi}8-\frac{\pi}4=-\frac{\pi}8$$
If you let $\alpha\to\frac{\pi}4^-$
$$\lim_{\alpha\to \frac{\pi}4^-}T(\alpha)=\frac{\pi}8+\frac{\pi}4=\frac{3\pi}8$$