On the square, ∣z∣≥N. The identity ∣sinπ(x+α+iy)∣2=sin2π(x+α)+sinh2πy bounds the sine away from zero on the vertical sides because x=±N; on horizontal sides it grows exponentially. Therefore the whole contour integral is O(N−1) or smaller and vanishes. The residue theorem then giveswhich proves n∈Z∑(−1)n/(n−α)2=π2cscπαcotπα. Absolute convergence justifies the limiting residue sum.
0=2πi[−πcscπαcotπα+∑n∈Zπ(n−α)2(−1)n],
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