[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**RE: mathematica & fortran ??**
Next by Date:
**Re: will someone explain the behavior of Unevaluated in this example?**
Previous by thread:
**Integrate fails revised**
Next by thread:
**Re: will someone explain the behavior of Unevaluated in this example?**
| |