Re: Boundary cond. at Infinity

```f[x_, a_] :=
Hypergeometric2F1[(a + 1)/2, 1, (a + 1)/2 + 1, -Exp[2*x]]

There is a discontinuity at

Solve[(a + 1)/2 + 1 == 0, a]

{{a -> -3}}

Plot3D[Hypergeometric2F1[(a + 1)/2, 1, (a + 1)/2 + 1,
-Exp[2*x]], {a, -4, 4}, {x, 0, 4}, PlotPoints -> 21,
AxesLabel -> {"a", "x", None}];

Linear transformation (Abramowitz and Stegun, 15.3.5):

trans = Hypergeometric2F1[a_, b_, c_, z_] ->
Hypergeometric2F1[b, c - a, c, z/(z - 1)]/(1 - z)^b;

Simplify[f[x, a] /. trans]

Hypergeometric2F1[1, 1, (a + 3)/2, Exp[2*x]/(1 + Exp[2*x])]/
(1 + Exp[2*x])

As x approaches Infinity, the numerator will tend towards

Hypergeometric2F1[1, 1, (a + 3)/2, 1]

(a + 1)/(a - 1)

And the denominator tends to Infinity.  The limit is then zero.

Bob Hanlon

In a message dated 4/25/98 6:18:19 AM, gold@dresden.isi.kfa-juelich.de
wrote:

>And Math.3.0 give me an answer in terms of PolyGamma's and
>Hypergeometric2F1's with one constant C[2] which I want to find by
>myself. To find it I need to know the behavior of e.g.
>Hypergeometric2F1[(1 + a)/2, 1, 1 + (1 + a)/2, -E^(2*x)] at infinity.
>How can I ask Mathematica to show me the behavior of arbitrary function
>at infinity? I expect to get something like x*Exp[-2*x] i.e. Taylor is
>of no use in this case since the behavior is exponential with unknown
>power.

```

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