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

```

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