Re: Why does this lead to an answer with complex numbers?
- To: mathgroup at smc.vnet.net
- Subject: [mg71494] Re: Why does this lead to an answer with complex numbers?
- From: "dimitris" <dimmechan at yahoo.com>
- Date: Mon, 20 Nov 2006 18:11:57 -0500 (EST)
- References: <ejosmm$n3k$1@smc.vnet.net><ejrmr9$97b$1@smc.vnet.net>
$Version
"5.2 for Microsoft Windows (June 20, 2005)"
FullSimplify[Integrate[Log[Sqrt[A^2 + x^2] - B*x], x], x > 0 && A > 0
&& B > 0 && B < 1]
(1/(2*Sqrt[-1 + B^2]))*(2*Sqrt[-1 + B^2]*x*(-1 + Log[(-B)*x + Sqrt[A^2
+ x^2]]) +
A*(2*ArcTanh[(Sqrt[-1 + B^2]*x)/A] - 4*Log[A] - 4*Log[B] +
Log[-((4*(-1 + B)*(1 + B)*(A*Sqrt[-1 + B^2] - x + B*Sqrt[A^2 +
x^2]))/(A*Sqrt[-1 + B^2] + (-1 + B^2)*x))] +
Log[((-1 + B^2)*(A*Sqrt[-1 + B^2] + x + B*Sqrt[A^2 +
x^2]))/(A*Sqrt[-1 + B^2] + x - B^2*x)]))
FullSimplify[Assuming[A > 0 && B > 0 && B < 1, Integrate[Log[Sqrt[A^2 +
x^2] - B*x], {x, 0, 3}]], A > 0 && B > 0 && B < 1]
(1/(2*Sqrt[1 - B^2]))*(-6*Sqrt[1 - B^2] + 2*A*Pi + 6*Sqrt[1 -
B^2]*Log[Sqrt[9 + A^2] - 3*B] -
I*A*(2*Log[1 - B^2] - Log[-A + 3*I*Sqrt[1 - B^2]] + Log[A +
3*I*Sqrt[1 - B^2]] +
Log[(3 + Sqrt[9 + A^2]*B + I*A*Sqrt[1 - B^2])/(-3 + 3*B^2 -
I*A*Sqrt[1 - B^2])] +
Log[(-3 + Sqrt[9 + A^2]*B + I*A*Sqrt[1 - B^2])/(-3 + 3*B^2 +
I*A*Sqrt[1 - B^2])] - 2*Log[-1 + B^2 + I*B*Sqrt[1 - B^2]]))
rul = {A -> Random[], B -> Random[]}
{A -> 0.4317060401451342, B -> 0.9756124768714767}
%% /. rul
-6.296006604829875 - 2.1835524462330213*^-15*I
Chop[%]
-6.296006604829875
NIntegrate[Evaluate[Log[Sqrt[A^2 + x^2] - B*x] /. rul], {x, 0, 3}]
-6.2960066048309775
aaronfude at gmail.com wrote:
> Hi,
>
> Thanks for all the answers. They were all very useful, even though I
> have done my best to confuse everyone by leaving a beta in there which
> had nothing to do with the problem.
>
> So I understand that the answer may be complex and the complex part is
> constant which is in a certain sense valid for a indefinite integral.
> But I very much need a real answer and I still can't quite extract.
> Consider the following:
>
> \!\(\(\(\ \)\(Assuming[x > 0 && A > 0 && B > 0 && \ B < 1, \
> FullSimplify[Integrate[Log[\@\(A^2 + x\^2\) - B*x\ ], \ x]]]\)\)\)
>
> The answer that I get is correct, but not very useful since it is
> appears complex and I could find a way to determine the real part. Do
> you have any suggestions?
>
>
> Thank you!
>
> Aaron Fude