RE: Integrate

*To*: mathgroup at smc.vnet.net*Subject*: [mg74323] RE: Integrate*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Sun, 18 Mar 2007 00:53:21 -0500 (EST)

Hello again! Of course sometimes things work quite unexpectedly! Consider again the integral (no! I am not obsessed with it!) Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}] Integrate::idiv: Integral of Log[Sin[x]^2]*Tan[x] does not converge on \ {x,0,Pi}. Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}] As we saw Integrate[Log[Sin[x]^2]*Tan[x],{x,0,Pi/2,Pi}] 0 Note now that the following are evaluated well Integrate[Log[Sin[x]^2]*Tan[x],{x,0,2Pi}] 0 Integrate[Log[Sin[x]^2]*Tan[x],{x,0,11Pi}] 0 Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, 4Pi/3}] (1/6)*(-Pi^2 - 3*(3*Log[2]^2 - 4*Log[2]*Log[3] + Log[3]^2 + 2*(PolyLog[2, -(1/2)] + PolyLog[2, -(1/3)]))) However Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, 2Pi/3}] Integrate::idiv: Integral of Log[Sin[x]^2]*Tan[x] does not converge on \ {x,0,(2*Pi)/3}. Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, (2*Pi)/3}] whereas of course Integrate[Log[Sin[x]^2]*Tan[x],{x,0,Pi/2,2Pi/3}]//FullSimplify Chop@N@% (4*Pi^2)/9 - Log[2]*Log[4] + Log[3]*Log[(2*(-1 + Sqrt[3]))/3^(1/4)] - PolyLog[2, -(1/3)] + 2*(PolyLog[2, -(1/Sqrt[3])] + PolyLog[2, 1 - Sqrt[3]] - PolyLog[2, - ((1 + I)/(-I + Sqrt[3]))] - PolyLog[2, (1 - I)/(-I + Sqrt[3])] - PolyLog[2, -((1 - I)/(I + Sqrt[3]))] - PolyLog[2, (1 + I)/(I + Sqrt[3])]) -0.688640713882748 So integrating between 0 and a point left than or including Pi needs to specify the point Pi/2 while integrating between 0 and a point right than Pi needs no specifying! You want more fun? Try this Integrate[Log[Sin[x]^2]*Tan[x], {x, Pi/3, 2*Pi}] Integrate[Log[Sin[x]^2]*Tan[x], {x, Pi/3, Pi}] Integrate[Log[Sin[x]^2]*Tan[x], {x, Pi/3, Pi/2,Pi}]//FullSimplify As I mentioned the integral Integrate[Log[Sin[x]^2]*Tan[x], {x, 0, Pi}] is platform dependent. Somewhere the Integrate algorithm in the Windows version of Mathematica 5.2 fails. I describe the role of Pi above but I can't find any explanation! If it will return a complex value I showed in the previous post that it could have been an improper application of Newton-Leibniz formula but now it returns Infinity. Anyway... The explanation is for more specialized persons than me or for Mathematica developers. But what I have found is how to overcome the problem. As I have said, use Integrate with your mind and you can achieve results close enough to tell them outstanding! =CF/=C7 Michael Weyrauch =DD=E3=F1=E1=F8=E5: > Hello, > > Dimitris, this is a marvelous solution to my problem. I really apprecia= te > your help. I will now see if I can solve all my other (similar) integrals= using the same trick. > Timing is not really the big issue if I get results in a reasonable amoun= t of time. > > Also the references you cited are quite interesting, because they give so= me insight > what might go on inside Mathematica concerning integration. > > I also agree with you that it is my task as a programmer to "help" Mathem= atica > and guide it into the direction I want it to go. Sometimes this is an "ar= t"... > > Nevertheless, in the present situation I do not really understand why Mat= hematica wants > me to do that rather trivial variable transformation, which is at the hea= rt of your solution. > The integrand is still a rather complicated rational function of the same= order. The form > of the integrand did not really change > substantially as it is the case with some other ingenious substitutions o= ne uses in order to > do some complicated looking integrals "by hand". > > I think the fact that we are forced to such tricks shows that the Mathema= tica integrator > is still a bit "immature" in special cases, as also the very interesting = article by D. Lichtblau, > which you cite, seems to indicate. So all this is probably perfectly know= n to the > Mathematica devellopers. And I hope the next Mathematica version has all = this "ironed out"?? > > Many thanks again, Michael