EllipticPi[n, m] == ((Pi (-1)^Floor[1/2 - Arg[1 - m]/(2 Pi) - Arg[(n - m)/(n (1 - m))]/(2 Pi)])/ (2 Sqrt[1 - m])) (1/Sqrt[(n - m)/(n (1 - m))]) (1/Sqrt[1 - n]) - ((Pi m)/4) Sum[(-1)^k Hypergeometric2F1[3/2, 3/2 + k, 2, m] (n - 1)^k, {k, 0, Infinity}]

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2007-05-02