Re: Oh, those complex values!!!

*To*: mathgroup at smc.vnet.net*Subject*: [mg72206] Re: Oh, those complex values!!!*From*: "David W. Cantrell" <DWCantrell at sigmaxi.net>*Date*: Thu, 14 Dec 2006 05:49:31 -0500 (EST)*Organization*: NewsReader.Com Subscriber*References*: <elja1i$76b$1@smc.vnet.net>

aaronfude at gmail.com wrote: > Integrating a real function, getting complex values again. How to get > the correct answer here? Sorry I don't have time for a detailed answer, but maybe this will help. > How to systematically avoid this pitfall? > > f=x Log[a+b x+Sqrt[c+2d x+x^2]]; > t = Assuming[c>d^2&&a>0&&b>0,Integrate[f, x]]; > u=t/.a\[Rule]2/.b\[Rule]2/.c\[Rule]10/.d\[Rule]1; > > N[u/.x\[Rule]1-u/.x\[Rule]0] > 9.50705\[InvisibleSpace]-29.7626 \[ImaginaryI] I don't know what you did above, but surely it's not what you intended. Note the parentheses below. In[32]:= N[(u /. x -> 1) - (u /. x -> 0)] Out[32]= 0.954442 + 6.28319 I The real part is correct, and the imaginary part is exactly 2 Pi: In[33]:= FullSimplify[Im[(u /. x -> 1) - (u /. x -> 0)]] Out[33]= 2 Pi which I suppose is due to some branch cut. Anyway, we don't want that, so to get what you want from (u /. x -> 1) - (u /. x -> 0), you could either just subtract 2 Pi I from it or take its real part: In[34]:= FullSimplify[Re[(u /. x -> 1) - (u /. x -> 0)]] Out[34]= (1/4)*(1 + 4*Sqrt[3]*ArcCoth[Sqrt[3]] - 4*Log[3] + 4*Log[6] - 12*Log[1 + Sqrt[10]] + 12*Log[2 + Sqrt[13]] + 2*Log[4 + Sqrt[13]] + 2*(2 + Sqrt[3])* Log[(18*(5 + 3*Sqrt[3]))/(9 - Sqrt[3] + 2*Sqrt[30])] + 2*(-2 + Sqrt[3])*Log[(18*(-5 + 3*Sqrt[3]))/(9 + Sqrt[3] + 2*Sqrt[30])] + 2*(2 + Sqrt[3])*Log[(1/18)*(7 - 4*Sqrt[3])*(9 - 2*Sqrt[3] + 2*Sqrt[39])] + 2*(-2 + Sqrt[3])*Log[(1/18)*(7 + 4*Sqrt[3])*(9 + 2*Sqrt[3] + 2*Sqrt[39])] - 4*Sqrt[3]*Re[ArcTanh[2/Sqrt[3]]]) In[35]:= N[%] Out[35]= 0.954442 David W. Cantrell > NIntegrate[f/.a\[Rule]2/.b\[Rule]2/.c\[Rule]10/.d\[Rule]1, {x, 0, 1}] > 0.954442 > > Many thanks in advance! > > Aaron Fude