Abs[ArcCoth[x + I y]] == (1/2) Sqrt[(ArcTan[1 - x/(x^2 + y^2), y/(x^2 + y^2)] - ArcTan[1 + x/(x^2 + y^2), -(y/(x^2 + y^2))])^2 + (1/4) (Log[(1 - 2 x + x^2 + y^2)/(x^2 + y^2)] - Log[(1 + 2 x + x^2 + y^2)/(x^2 + y^2)])^2]

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2001-10-29