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Integrating with HypergeometricU

• To: mathgroup at smc.vnet.net
• Subject: [mg105445] Integrating with HypergeometricU
• From: Matteo Fasiello <matteorf at gmail.com>
• Date: Thu, 3 Dec 2009 06:13:22 -0500 (EST)
• References: <200912010912.EAA18696@smc.vnet.net> <200912021128.GAA27056@smc.vnet.net>
• Reply-to: matteorf at gmail.com

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

• References:
  • Plotting a table of piecewise functions
    • From: michael partensky <partensky@gmail.com>
  • Re: Plotting a table of piecewise functions
    • From: michael partensky <partensky@gmail.com>