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

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Theorem

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

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


Proof

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

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


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


\(\ds \) \(\) \(\ds \sum_{n \mathop = -m}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\map \sin {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -m}^{-1} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^m \dfrac {\map \sin {-n + \alpha} \theta} {-n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\map \sin {\alpha + n} \theta} {\alpha + n} + \dfrac {\map \sin {\alpha - n} \theta} {\alpha - n} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\paren {\alpha - n} \map \sin {\alpha + n} \theta + \paren {\alpha + n} \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \map \sin {\alpha + n} \theta - n \map \sin {\alpha + n} \theta + \alpha \map \sin {\alpha - n} \theta - n \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {\alpha \paren {\map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta} - n \paren {\map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta} } {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^m \paren {\dfrac {2 \alpha \sin \alpha \theta \cos n \theta - 2 n \cos \alpha \theta \sin n \theta} {\alpha^2 - n^2} }\) from $(1)$ and $(2)$


From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$:

$\ds \cos \alpha \theta \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \theta} {\alpha^2 - n^2} }$

From Half-Range Fourier Sine Series for $\sin \alpha x$ over $\openint 0 \pi$:

$\ds \sin \alpha \theta \sim \frac {2 \sin \alpha \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n \theta} {\alpha^2 - n^2} }$



Sources