Sum over Integers of Sine of n + alpha of theta over n + alpha

Theorem

For $0 < \theta < 2 \pi$:

$\displaystyle \sum_{n \mathop \in \Z} \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} = \pi$

Proof

First we establish the following, as they will be needed later.

 $\displaystyle$  $\displaystyle \sin \paren {\alpha + n} \theta + \sin \paren {\alpha - n} \theta$ $\displaystyle$ $=$ $\displaystyle 2 \map \sin {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \cos {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}$ Prosthaphaeresis Formula for Sine plus Sine $(1):\quad$ $\displaystyle$ $=$ $\displaystyle 2 \sin \alpha \theta \cos n \theta$ simplification

 $\displaystyle$  $\displaystyle \sin \paren {\alpha + n} \theta - \sin \paren {\alpha - n} \theta$ $\displaystyle$ $=$ $\displaystyle 2 \map \cos {\dfrac {\paren {\alpha + n} \theta + \paren {\alpha - n} \theta} 2} \map \sin {\dfrac {\paren {\alpha + n} \theta - \paren {\alpha - n} \theta} 2}$ Prosthaphaeresis Formula for Sine minus Sine $(2):\quad$ $\displaystyle$ $=$ $\displaystyle 2 \cos \alpha \theta \sin n \theta$ simplification

 $\displaystyle$  $\displaystyle \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \paren {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -m}^{-1} \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\sin \paren {-n + \alpha} \theta} {-n + \alpha}$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\sin \paren {\alpha + n} \theta} {\alpha + n} + \dfrac {\sin \paren {\alpha - n} \theta} {\alpha - n} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\paren {\alpha - n} \sin \paren {\alpha + n} \theta + \paren {\alpha + n} \sin \paren {\alpha - n} \theta} {\alpha^2 - n^2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \sin \paren {\alpha + n} \theta - n \sin \paren {\alpha + n} \theta + \alpha \sin \paren {\alpha - n} \theta - n \sin \paren {\alpha - n} \theta} {\alpha^2 - n^2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \paren {\sin \paren {\alpha + n} \theta + \sin \paren {\alpha - n} \theta} - n \paren {\sin \paren {\alpha + n} \theta - \sin \paren {\alpha - n} \theta} } {\alpha^2 - n^2} }$ $\displaystyle$ $=$ $\displaystyle \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {2 \alpha \sin \alpha \theta \cos n \theta - 2 n \cos \alpha \theta \sin n \theta} {\alpha^2 - n^2} }$ from $(1)$ and $(2)$
$\displaystyle \cos \alpha \theta \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \theta} {\alpha^2 - n^2} }$
$\displaystyle \sin \alpha \theta \sim \frac {2 \sin \alpha \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n \theta} {\alpha^2 - n^2} }$