KelvinKei[\[Nu], z] == (1/2) Sum[((I - 1)^k/(2^(3 (k/2)) k!)) (Sum[Binomial[k, 2 j] ((1 + I^k) (-2 I Pi (-1)^k Cos[Pi \[Nu]] Floor[Arg[z - x]/(2 Pi)] KelvinBei[-\[Nu] + k - 4 j, x] + E^(2 I Pi \[Nu] Floor[Arg[z - x]/(2 Pi)]) KelvinKei[ \[Nu] - k + 4 j, x]) - I (1 - I^k) ((-(-1)^k) 2 I Pi Cos[Pi \[Nu]] Floor[Arg[z - x]/(2 Pi)] KelvinBer[-\[Nu] + k - 4 j, x] + E^(2 I Pi \[Nu] Floor[Arg[z - x]/(2 Pi)]) KelvinKer[ \[Nu] - k + 4 j, x])), {j, 0, Floor[k/2]}] - Sum[Binomial[k, 2 j + 1] ((1 + I^k) ((-(-1)^k) 2 I Pi Cos[Pi \[Nu]] Floor[Arg[z - x]/(2 Pi)] KelvinBei[-\[Nu] + k - 4 j - 2, x] + E^(2 I Pi \[Nu] Floor[Arg[z - x]/(2 Pi)]) KelvinKei[ \[Nu] - k + 4 j + 2, x]) - I (1 - I^k) ((-(-1)^k) 2 I Pi Cos[Pi \[Nu]] Floor[Arg[z - x]/(2 Pi)] KelvinBer[-\[Nu] + k - 4 j - 2, x] + E^(2 I Pi \[Nu] Floor[Arg[z - x]/(2 Pi)]) KelvinKer[ \[Nu] - k + 4 j + 2, x])), {j, 0, Floor[(k - 1)/2]}]) (z - x)^k, {k, 0, Infinity}] /; Element[x, Reals] && x < 0

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2007-05-02