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