Mittag-Leffler Expansion for Cosecant Function/Real Domain

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Theorem

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

Then:

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


Proof

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

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


Setting $x = 0$:

\(\ds \cos 0\) \(=\) \(\ds \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\alpha^2 - n^2} }\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \frac {2 \alpha \sin \alpha \pi} \pi \paren {\frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {\alpha^2 - n^2} }\) Cosine of Zero is One
\(\ds \leadsto \ \ \) \(\ds \frac \pi {2 \alpha \sin \alpha \pi}\) \(=\) \(\ds \frac 1 {2 \alpha^2} + \sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n} {\alpha^2 - n^2}\)
\(\ds \leadsto \ \ \) \(\ds \pi \cosec \pi \alpha\) \(=\) \(\ds \dfrac 1 \alpha + 2 \sum_{n \mathop \ge 1} \paren {-1}^n \dfrac {\alpha} {\alpha^2 - n^2}\) Definition of Cosecant and rearranging

$\blacksquare$


Sources