Re: HypergeometricPFQ
- To: mathgroup at yoda.physics.unc.edu
- Subject: Re: HypergeometricPFQ
- From: paul at earwax.pd.uwa.oz.au (Paul C Abbott)
- Date: Fri, 20 Nov 1992 12:17:34 +0800
Sam Finn writes:
>The evaluation rules for HypergeometricPFQ are messed up:
>
>In[1]:= Integrate[1/(x^(7/3)(x^2+1.44^2)),{x,1,Infinity}]
>
> 5 8
>Out[1]= 0.3 HypergeometricPFQ[{1, -}, {-}, -2.0736]
> 3 3
Simply replacing the HypergeometricPFQ by Hypergeometric2F1 will do the
trick. This can be done automatically (and should be :-)
%1 /. HypergeometricPFQ[num:{a__},den:{b__}, z_] :>
Hypergeometric2F1[a, b, z] /;
Length[num] == 2 && Length[den] == 1
0.139291
Alternatively, Mathematica can handle the more general symbolic case
without a problem:
Integrate[1/(x^(7/3) (x^2+a^2)),{x,1,Infinity}]
2 1/3 2 1/3
3 (a ) Log[1 + (a ) ]
---- + ------------------------ +
2 4
4 a 2 a
2/3 2 1/3 2 1/3 -I/3 Pi
(-1) (a ) Log[1 - (a ) E ]
----------------------------------------- +
4
2 a
4/3 2 1/3 2 1/3 I/3 Pi
(-1) (a ) Log[1 - (a ) E ]
----------------------------------------
4
2 a
Substituting in your numerical value (and using Chop to remove the small
complex part) yields the same answer:
% /. a -> 1.44 // N // Chop
0.139291
>P.S. And also, will somebody please remind me why I even bother trying
>to make productive use of Mathematica?
Because it is a good system :-)
Cheers,
Paul C. Abbott
Department of Physics
University of Western Australia
Nedlands 6009