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

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Theorem

Let $\alpha \in \R$ be a real number which is specifically not an integer.


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.

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


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


\(\ds \) \(\) \(\ds \sum_{n \mathop = -\infty}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\map \sin {0 + \alpha} \theta} {0 + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = -\infty}^{-1} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha} + \sum_{n \mathop = 1}^\infty \dfrac {\map \sin {-n + \alpha} \theta} {-n + \alpha}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\map \sin {\alpha + n} \theta} {\alpha + n} + \dfrac {\map \sin {\alpha - n} \theta} {\alpha - n} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\paren {\alpha - n} \map \sin {\alpha + n} \theta + \paren {\alpha + n} \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {\alpha \map \sin {\alpha + n} \theta - n \map \sin {\alpha + n} \theta + \alpha \map \sin {\alpha - n} \theta + n \map \sin {\alpha - n} \theta} {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty\paren {\dfrac {\alpha \paren {\map \sin {\alpha + n} \theta + \map \sin {\alpha - n} \theta} - n \paren {\map \sin {\alpha + n} \theta - \map \sin {\alpha - n} \theta} } {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + \sum_{n \mathop = 1}^\infty \paren {\dfrac {2 \alpha \sin \alpha \theta \cos n \theta - 2 n \cos \alpha \theta \sin n \theta} {\alpha^2 - n^2} }\) from $(1)$ and $(2)$
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \theta} \alpha + 2 \sin \alpha \theta \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \theta } {\alpha^2 - n^2} - 2 \cos \alpha \theta \sum_{n \mathop = 1}^\infty \dfrac {n \sin n \theta} {\alpha^2 - n^2}\) Linear Combination of Indexed Summations


Setting $\theta = \pi$:

\(\ds \sum_{n \mathop = -\infty}^\infty \dfrac {\map \sin {n + \alpha} \pi} {n + \alpha}\) \(=\) \(\ds \dfrac {\sin \alpha \pi} \alpha + 2 \sin \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \pi } {\alpha^2 - n^2} - 2 \cos \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {n \sin n \pi} {\alpha^2 - n^2}\)
\(\ds \) \(=\) \(\ds \dfrac {\sin \alpha \pi} \alpha + 2 \sin \alpha \pi \sum_{n \mathop = 1}^\infty \dfrac {\alpha \cos n \pi } {\alpha^2 - n^2} - 0\) $\sin n \pi = 0$
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\sin \pi \alpha} \alpha + 2 \sin \pi \alpha \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac \alpha {\alpha^2 - n^2}\) $\cos n \pi = \paren {-1}^n$


Then:

\(\ds \pi \cosec \pi \alpha\) \(=\) \(\ds \dfrac 1 \alpha + 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}\) Mittag-Leffler Expansion for the Cosecant Function
\(\ds \leadsto \ \ \) \(\ds \pi\) \(=\) \(\ds \sin \pi \alpha \paren{\dfrac 1 \alpha + 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2} }\) multiplying both sides by $\sin \pi \alpha$
\(\text {(4)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\sin \pi \alpha} \alpha + 2 \sin \pi \alpha \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}\)
\(\ds \leadsto \ \ \) \(\ds \pi\) \(=\) \(\ds \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \pi} {n + \alpha}\) equating $(3)$ and $(4)$


To establish this identity for all other values of $\theta$ on the interval $0 < \theta < 2 \pi$, we will demonstrate that the sum is a constant function.

We will do this by showing that the derivative of the function is zero everywhere which by Zero Derivative implies Constant Function will complete the proof.


We have:

\(\ds \map f \theta\) \(=\) \(\ds \sum_{n \mathop \in \Z} \dfrac {\map \sin {n + \alpha} \theta} {n + \alpha}\)
\(\ds \map {f'} \theta\) \(=\) \(\ds \sum_{n \mathop \in \Z} \map \cos {n + \alpha} \theta\) Derivative of Sine Function/Corollary
\(\ds \leadsto \ \ \) \(\ds \map {f'} \theta\) \(=\) \(\ds \sum_{n \mathop \in \Z} \paren {\map \cos {\alpha \theta } \map \cos {n \theta} - \map \sin {\alpha \theta } \map \sin {n \theta} }\) Cosine of Sum
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \sum_{n \mathop \in \Z} \map \cos {n \theta} - \map \sin {\alpha \theta} \sum_{n \mathop \in \Z} \map \sin {n \theta}\) Linear Combination of Indexed Summations


From Cosine Function is Even and Sine Function is Odd, we have:

$\map \cos {-n \theta} = \map \cos {n \theta}$

and:

$\map \sin {-n \theta} = -\map \sin {n \theta}$


Therefore:

\(\ds \map {f'} \theta\) \(=\) \(\ds \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} } - \map \sin {\alpha \theta} \times 0\)
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \map \cos {n \theta} }\)
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {1 + 2 \sum_{n \mathop = 1}^\infty \paren {\dfrac {e^{i n \theta} + e^{-i n \theta} } 2} }\) Euler's Cosine Identity
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {1 + \sum_{n \mathop = 1}^\infty e^{i n \theta} + \sum_{n \mathop = 1}^\infty e^{-i n \theta} }\) Linear Combination of Indexed Summations
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {\sum_{n \mathop = 0}^\infty e^{i n \theta} + \sum_{n \mathop = 0}^\infty e^{-i n \theta} - 1 }\) Re-index the sum
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {\dfrac 1 {1 - e^{i \theta} } + \dfrac 1 {1 - e^{-i \theta} } - 1}\) Sum of Infinite Geometric Sequence
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }\)
\(\ds \) \(=\) \(\ds \map \cos {\alpha \theta} \paren {\dfrac {\paren {1 - e^{-i \theta} } + \paren {1 - e^{i \theta} } - \paren {1 - e^{i \theta} - e^{-i \theta} + 1} } {\paren {1 - e^{-i \theta} } \paren {1 - e^{i \theta} } } }\)
\(\ds \) \(=\) \(\ds 0\)

$\blacksquare$


Also see


Sources

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