Summation Formula over Half-Integers

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Theorem

Let $C_N$ be the square with vertices $N \paren {\pm 1 \pm i}$ for real $N \in \N$.

Let $f$ be a function meromorphic on $C_N$.

Let $\cmod {\map f z} < \dfrac M {\cmod z^k}$, for constants $k > 1$ and $M$ independent of $N$, for all $z \in \partial C_N$.

Let $X$ be the set of poles of $f$.

Let $Y$ be the set of poles of $\map f {\dfrac {2 z + 1} 2}$.

Then:

$\displaystyle \sum_{n \in \Z \setminus Y} \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$

If $Y \cap \Z = \O$, this becomes:

$\displaystyle \sum_{n \mathop = -\infty}^\infty \map f {\frac {2 n + 1} 2} = \sum_{z_0 \in X} \Res {\pi \tan \paren {\pi z} \map f z} {z_0}$


Proof


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