Sum over Integers of Sine of n + alpha of theta over n + alpha
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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 | ||||||||||
| \(\text {(1)}: \quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle 2 \sin \alpha \theta \cos n \theta\) | 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 | ||||||||||
| \(\text {(2)}: \quad\) | \(\displaystyle \) | \(=\) | \(\displaystyle 2 \cos \alpha \theta \sin n \theta\) | 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 $(2)$ |
From Half-Range Fourier Cosine Series for $\cos \alpha x$ over $\openint 0 \pi$:
- $\displaystyle \cos \alpha \theta \sim \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \theta} {\alpha^2 - n^2} }$
From Half-Range Fourier Sine Series for $\sin \alpha x$ over $\openint 0 \pi$:
- $\displaystyle \sin \alpha \theta \sim \frac {2 \sin \alpha \pi} \pi \paren {\sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {n \sin n \theta} {\alpha^2 - n^2} }$
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Sources
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text {II}$: $1$.
