Sum over Integers of Cosine 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 $0 \le \theta < 2 \pi$:

$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \sum_{n \mathop \in \Z} \dfrac {\cos \paren {n + \alpha} \theta} {n + \alpha}$


Corollary

$\ds \dfrac 1 \alpha + \sum_{n \mathop \ge 1} \dfrac {2 \alpha} {\alpha^2 - n^2} = \pi \cot \pi \alpha$


Proof

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

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


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


We have:

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


Setting $\theta = 0$:

\(\ds \sum_{n \mathop = -\infty}^\infty \dfrac {\map \cos {\paren {n + \alpha} 0} } {n + \alpha}\) \(=\) \(\ds \dfrac {\cos \alpha 0} \alpha + 2 \alpha \cos \alpha 0 \sum_{n \mathop = 1}^\infty \paren {\dfrac {\cos n 0 } {\alpha^2 - n^2} } + 2 \sin \alpha 0 \sum_{n \mathop = 1}^\infty \paren {\dfrac {n \sin n 0} {\alpha^2 - n^2} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 \alpha + \sum_{n \mathop = 1}^\infty \dfrac {2 \alpha} {\alpha^2 - n^2} + 0\) Sine of Zero is Zero and Cosine of Zero is One
\(\ds \) \(=\) \(\ds \dfrac 1 \alpha + \sum_{n \mathop = 1}^\infty \dfrac {2 \alpha} {\alpha^2 - n^2}\)


To establish this identity for all other values of $\theta$ on the interval $0 \le \theta < 2\pi$, 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:

\(\ds \map f \theta\) \(=\) \(\ds \sum_{n \mathop \in \Z} \dfrac {\map \cos {n + \alpha} \theta} {n + \alpha}\)
\(\ds \map {f'} \theta\) \(=\) \(\ds \sum_{n \mathop \in \Z} -\map \sin {n + \alpha} \theta\) Derivative of Cosine Function/Corollary


Then:

\(\ds \map {f'} \theta\) \(=\) \(\ds -\sum_{n \mathop \in \Z} \paren {\map \sin {\alpha \theta} \map \cos {n \theta} + \map \cos {\alpha \theta } \map \sin {n \theta} }\) Sine of Sum
\(\ds \) \(=\) \(\ds -\map \sin {\alpha \theta } \sum_{n \mathop \in \Z} \map \cos {n \theta} - \map \cos {\alpha \theta} \sum_{n \mathop \in \Z} \map \sin {n \theta}\) 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:

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

$\blacksquare$


Also see


Sources

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