Series Expansion for Pi Cosecant of Pi Lambda

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Theorem

Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.


Then:

$\displaystyle \pi \csc \pi \lambda = \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {n + \lambda} + \frac 1 {n - 1 - \lambda} }$


Proof

Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:

$\map f x = \cos \lambda x$


From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as:

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


Because of the nature of this expansion, we have that:

$\map f {0^+} = \map f {0^-}$

and so the expansion holds for $x = 0$.


So, setting $x = 0$:

\(\displaystyle \cos 0\) \(=\) \(\displaystyle \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos 0} {\lambda^2 - n^2} }\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac \pi {\sin \lambda \pi}\) \(=\) \(\displaystyle 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac 1 {\lambda^2 - n^2} }\) Cosine of Zero is One and rearrangement
\(\displaystyle \leadsto \ \ \) \(\displaystyle \pi \csc \lambda \pi\) \(=\) \(\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda} {\lambda^2 - n^2}\) Definition of Real Cosecant Function
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\lambda + n + \lambda - n} {\paren {\lambda + n} \paren {\lambda - n} }\) Difference of Two Squares
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac {\lambda - n} {\paren {\lambda + n} \paren {\lambda - n} } + \frac {\lambda + n} {\paren {\lambda + n} \paren {\lambda - n} } }\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \paren {\frac 1 {\lambda + n} + \frac 1 {\lambda - n} }\)



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