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MathGroup Archive 2006

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verify the result of indefinite integration

  • To: mathgroup at smc.vnet.net
  • Subject: [mg71440] verify the result of indefinite integration
  • From: "dimitris" <dimmechan at yahoo.com>
  • Date: Sat, 18 Nov 2006 04:41:10 -0500 (EST)

I try to verify that the result of the following indefinite integration
is corect

f = ArcCosh[Sec[ArcSinh[x]]];

indef = Integrate[f, x]
x*ArcCosh[Sec[ArcSinh[x]]] - (1/2 + I/2)*E^((-1 +
I)*ArcSinh[x])*Cot[ArcSinh[x]/2]*
   ((-I)*E^(2*ArcSinh[x])*Hypergeometric2F1[1/2 - I/2, 1, 3/2 - I/2,
-E^(2*I*ArcSinh[x])] +
    Hypergeometric2F1[1/2 + I/2, 1, 3/2 + I/2,
-E^(2*I*ArcSinh[x])])*Sqrt[Tan[ArcSinh[x]/2]^2]

Here is one simple attempt

der = FullSimplify[D[indef, x]];

rad = Table[Random[Real, {-10, 10}], {10}];

Chop[(der /. x -> #1 & ) /@ rad]
Chop[(f /. x -> #1 & ) /@ rad]

% == %%
True

However, the following expression cannot be simplified directly to zero

FullSimplify[der - f]
-((x*Cot[ArcSinh[x]/2]*Sec[ArcSinh[x]]^2*(-1 + Cos[ArcSinh[x]]*(1 +
Sqrt[-1 + Sec[ArcSinh[x]]]*Sqrt[1 +
Sec[ArcSinh[x]]]*Sqrt[Tan[ArcSinh[x]/2]^2])))/(Sqrt[1 + x^2]*Sqrt[-1 +
Sec[ArcSinh[x]]]*Sqrt[1 + Sec[ArcSinh[x]]]))

Only with the following assumption we have the desired simplification

FullSimplify[%, Tan[ArcSinh[x]/2]^2 > 0]
0

Any comments?

Dimitris


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