From: Matteo Fasiello on 3 Dec 2009 06:19 Hi, first time I use the mailing list so I hope I am doing this properly. I am trying to do the following indefinite integral with mathematica: --- \[Integral](1/( 64 (a0 + Sqrt[b0])^3 Gamma[5/4 + a0/(4 a0 - 4 I Sqrt[b0])]^4) E^( 1/2 I Sqrt[b0] k1^2 \[Tau]^2 + 1/2 I Sqrt[b0] k2^2 \[Tau]^2 + 1/2 I Sqrt[b0] k3^2 \[Tau]^2 + 1/2 I Sqrt[b0] k4^2 \[Tau]^2) H^6 Sqrt[k1] Sqrt[k2] Sqrt[k3] Sqrt[ k4] \[Tau]^6 Gamma[ 5/4 - (I a0)/( 4 Sqrt[b0])]^4 (-2 Sqrt[b0] HypergeometricU[-(1/4) - (I a0)/(4 Sqrt[b0]), -(1/2), -I Sqrt[ b0] k1^2 \[Tau]^2] + (I a0 + Sqrt[b0]) HypergeometricU[ 3/4 - (I a0)/(4 Sqrt[b0]), 1/ 2, -I Sqrt[b0] k1^2 \[Tau]^2]) (-2 Sqrt[b0] HypergeometricU[-(1/4) - (I a0)/(4 Sqrt[b0]), -(1/2), -I Sqrt[ b0] k2^2 \[Tau]^2] + (I a0 + Sqrt[b0]) HypergeometricU[ 3/4 - (I a0)/(4 Sqrt[b0]), 1/ 2, -I Sqrt[b0] k2^2 \[Tau]^2]) (-2 Sqrt[b0] HypergeometricU[-(1/4) - (I a0)/(4 Sqrt[b0]), -(1/2), -I Sqrt[ b0] k3^2 \[Tau]^2] + (I a0 + Sqrt[b0]) HypergeometricU[ 3/4 - (I a0)/(4 Sqrt[b0]), 1/ 2, -I Sqrt[b0] k3^2 \[Tau]^2]) (-2 Sqrt[b0] HypergeometricU[-(1/4) - (I a0)/(4 Sqrt[b0]), -(1/2), -I Sqrt[ b0] k4^2 \[Tau]^2] + (I a0 + Sqrt[b0]) HypergeometricU[ 3/4 - (I a0)/(4 Sqrt[b0]), 1/ 2, -I Sqrt[b0] k4^2 \[Tau]^2])) \[DifferentialD]\[Tau] --- Mathematica can't do it apparently.. It does it if one uses HankelH1 functions instead of the Hyper but what I really need is the Hyper. I also need to keep all these k1,2,3,4 a0, b0 parameters alive. I tried to use the fact a0 > 0 && b0 > 0 && k1 > 0 && k2 > 0 && k3 > 0 && k4 > 0 as assumptions, to no avail. Thanks a lot, Matteo
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