Sum from -m to m of Sine of n + alpha of theta over n + alpha

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Theorem

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

$\ds \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} = \int_0^\theta \map \cos {\alpha \theta} \dfrac {\sin \paren {m + \frac 1 2} \theta \rd \theta} {\sin \frac 1 2 \theta}$


Proof

We have:

\(\ds \sum_{n \mathop = -m}^m e^{i \paren {n + \alpha} \theta}\) \(=\) \(\ds \sum_{n \mathop = -m}^m e^{i n \theta} e^{i \alpha \theta}\)
\(\ds \) \(=\) \(\ds e^{i \alpha \theta} e^{-i m \theta} \sum_{n \mathop = 0}^{2 m} e^{i n \theta}\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta} - 1} {e^{i \theta} - 1} }\) Sum of Geometric Sequence
\(\ds \) \(=\) \(\ds e^{i \alpha \theta} e^{-i m \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} \paren {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } } {e^{i \theta / 2} \paren {e^{i \theta / 2} - e^{i \theta / 2} } } }\) extracting factors
\(\ds \) \(=\) \(\ds e^{i \alpha \theta} \paren {\dfrac {e^{i \paren {2 m + 1} \theta / 2} - e^{-i \paren {2 m + 1} \theta / 2} } {e^{i \theta / 2} - e^{i \theta / 2} } }\) Exponential of Sum and some algebra
\(\ds \) \(=\) \(\ds e^{i \alpha \theta} \frac {\map \sin {\paren {2 m + 2} \theta / 2} } {\map \sin {\theta / 2} }\) Euler's Sine Identity
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = -m}^m \paren {\cos \paren {n + \alpha} \theta + i \sin \paren {n + \alpha} \theta}\) \(=\) \(\ds \paren {\map \cos {\alpha \theta} + i \map \sin {\alpha \theta} } \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }\) Euler's Formula and simplifying
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = -m}^m \cos \paren {n + \alpha} \theta\) \(=\) \(\ds \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} }\) equating real parts


Note that the right hand side at $(1)$ is not defined when $e^{i \theta} = 1$.

This happens when $\theta = 2 k \pi$ for $k \in \Z$.

For the given range of $0 < \theta < 2 \pi$ it is therefore seen that $(1)$ does indeed hold.


Then:

\(\ds \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta\) \(=\) \(\ds \intlimits {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } {\theta \mathop = 0} {\theta \mathop = \theta}\) Primitive of $\cos a x$
\(\ds \) \(=\) \(\ds \paren {\dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha} } - \paren {\dfrac {\sin \paren {n + \alpha} 0} {n + \alpha} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}\) Sine of Zero is Zero
\(\ds \leadsto \ \ \) \(\ds \sum_{n \mathop = -m}^m \dfrac {\sin \paren {n + \alpha} \theta} {n + \alpha}\) \(=\) \(\ds \sum_{n \mathop = -m}^m \int_0^\theta \cos \paren {\alpha + n} \theta \rd \theta\)
\(\ds \) \(=\) \(\ds \int_0^\theta \sum_{n \mathop = -m}^m \cos \paren {\alpha + n} \theta \rd \theta\) Linear Combination of Definite Integrals
\(\ds \) \(=\) \(\ds \int_0^\theta \map \cos {\alpha \theta} \frac {\map \sin {\paren {m + \frac 1 2} \theta } } {\map \sin {\theta / 2} } \rd \theta\) from $(2)$

Hence the result.

$\blacksquare$


Sources

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