Derivative[m, 0][KelvinKei][\[Nu], z] == (1/2) Pi (Sum[(1/k!) (z/2)^(2 k) D[((z/2)^\[Nu] Cos[(1/4) Pi (2 k + 3 \[Nu])])/Gamma[k + \[Nu] + 1], {\[Nu], m}], {k, 0, Infinity}] + Sum[Binomial[m, j] (Pi^(m - j) (-I)^(m - j + 1) Sum[(1/2^i) (((-1)^i i!) StirlingS2[m - j, i] ((I Cot[(Pi \[Nu])/2] + 1)^i (I Cot[(Pi \[Nu])/2] - 1) - 2^(m - j) (I Cot[Pi \[Nu]] + 1)^i (I Cot[Pi \[Nu]] - 1))), {i, 0, m - j}]) Sum[(1/k!) (z/2)^(2 k) D[Sin[(1/4) Pi (2 k - 3 \[Nu])]/((z/2)^\[Nu] Gamma[k - \[Nu] + 1]), {\[Nu], j}], {k, 0, Infinity}], {j, 0, m}] - Sum[Binomial[m, j] Pi^(m - j) ((-KroneckerDelta[m - j]) I + (-I)^(m - j + 1) 2^(m - j) (I Cot[Pi \[Nu]] - 1) Sum[((-1)^i i! StirlingS2[m - j, i] (I Cot[Pi \[Nu]] + 1)^i)/2^i, {i, 0, m - j}]) Sum[(1/k!) (z/2)^(2 k) D[((z/2)^\[Nu] Sin[(1/4) Pi (2 k + 3 \[Nu])])/Gamma[k + \[Nu] + 1], {\[Nu], j}], {k, 0, Infinity}], {j, 0, m}]) /; !Element[\[Nu], Integers]

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2007-05-02