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
- 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text {II}$: $1$.