Re: Definite Integration in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74589] Re: Definite Integration in Mathematica*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Tue, 27 Mar 2007 04:04:09 -0500 (EST)*References*: <etqo3f$10i$1@smc.vnet.net><eu7rtg$bgr$1@smc.vnet.net>

>(By the way, as I just realized, Michael Trott > in his book on symbolics has quite some interesting things to say related= to this issue.) There are a great many books about Mathematica. Bu definetely if you want a book that combines Mathematica with advanced mathematics for a various scientific fields (even form current research) this book is Trott's Guidebooks. The whole series deserves the money. > However, in many practical applications people want to have an integral i= n the > former sense, and it is very disconcerting to fall down a discontinuity s= tep along > an integral of a continous integrand on the real line due to a singularit= y of the > integrand somewhere in the complex plane. Therefore, I guess, the design= ers of "the other > CAS", which Dimitris uses for comparison, obviously opted to return an in= tegral, which is continuous along > the real axis (if such an integral is available in the set of possible so= lutions). It's just happen in this case the antiderivative returned by the other CAS to be continuous in the real axis. For other integrals usually discontinuous antiderivatives are returned. For example (*other CAS*) int(1/(5+cos(x)),x); 1/6*6^(1/2)*arctan(1/3*tan(1/2*x)*6^(1/2)) (*mathematica*) Integrate[1/(5 + Cos[x]), x] ArcTan[Sqrt[2/3]*Tan[x/2]]/Sqrt[6] both have jump discontinuity at x=+/- n*Pi, n=odd. See this thread for a clever method by Peter Pein in order to get a continuous antiderivative with Mathematica. But in case of recognizing the jump discontinuity and its position, it is very easy to obtain a continous antiderivative adding the piecewise constant (see the relevant example from Trott's book!) Also based on my experience Mathematica is quantum steps in front of other CAS (no I don't take any money from WRI!) regarding integration capabilities. For an example let f = HoldForm[Integrate[Cos[Log[z]]/(1 + z^2)^Pi, {z, 0, Infinity}]] The indefinite integral first ReleaseHold[f /. Integrate[h_, {x_, __}] :> Integrate[h, x]] FullSimplify[D[%, z] == Cos[Log[z]]/(1 + z^2)^Pi] (1/4 + I/4)*z^(1 - I)*Hypergeometric2F1[1/2 - I/2, Pi, 3/2 - I/2, - z^2] + (1/4 - I/4)*z^(1 + I)*Hypergeometric2F1[1/2 + I/2, Pi, 3/2 + I/2, - z^2] True The definite integral along with check... Timing[g=ReleaseHold[f]] {145.422*Second, (Pi*(-((Gamma[1/2 - I/2]*Sech[(1/2)*(1 - 2*I*Pi)*Pi])/ Gamma[(3/2 - I/2) - Pi]) - (Gamma[1/2 + I/2]*Sech[(1/2)*(1 + 2*I*Pi)*Pi])/Gamma[(3/2 + I/2) - Pi]))/(4*Gamma[Pi])} {Chop[N[F, 21]], ReleaseHold[f /. Integrate[x___] :> NIntegrate[x, WorkingPrecision -> 50, PrecisionGoal -> 20]]} {0.142291882821508343786,0.142291882821508343786} In my point of view this is a very tough integral (appeared in another forum). And getting an analytic result clearly demonstrates Mathematica's integration capabilities! Let compare it with the other CAS I used in order to compare sometimes its results with Mathematica and vice versa... The definite integral first int(cos(ln(z))/(1+z^2)^Pi, z= 0..infinity); int(cos(ln(z))/((1+z^2)^Pi),z = 0 .. infinity) and now the indefinite int(cos(ln(z))/(1+z^2)^Pi, z); int(cos(ln(z))/((1+z^2)^Pi),z) No results! A numerical integration is possibly of course evalf(Int(cos(ln(z))/(1+z^2)^Pi, z= 0..infinity);); 0=2E1422918828 Best Regards Dimitris =CF/=C7 Michael Weyrauch =DD=E3=F1=E1=F8=E5: > Hello, > > thanks, Dimitris, for bringing up this issue. (By the way, as I just r= ealized, Michael Trott > in his book on symbolics has quite some interesting things to say related= to this issue.) > > My remark in this thread clearly was written in the spirit of "functions = of one real > variable". As I see my statements are clearly incorrect in the sense of = "functions of a complex > variable". > > However, in many practical applications people want to have an integral i= n the > former sense, and it is very disconcerting to fall down a discontinuity s= tep along > an integral of a continous integrand on the real line due to a singularit= y of the > integrand somewhere in the complex plane. Therefore, I guess, the design= ers of "the other > CAS", which Dimitris uses for comparison, obviously opted to return an in= tegral, which is continuous along > the real axis (if such an integral is available in the set of possible so= lutions). > > This is not just a matter of aestetics or simplicity as Dimitris remark i= n this thread may suggest > > >Do you prefer the extend antiderivative over the compact one obtained > >directly > >by Mathematica only because is real in the real axis? > >As regards myself, no! > > In applications we have to use that solution which makes practical sense,= and in many > applications it's the integral which is continouus (and differentiable) f= or a continouus integrand > on the real axis. Of course, it's my task as a physicist or engineer to s= ee if a mathematical > solution serves my purposes or not, however, in such rather intriguing ca= ses, it would be > desirable that the designers of Mathematica would help me by providing a= n Option > to Integrate, in which I could ask for an integral which is continuous on= the real axis if such an integral > exists for a particular integrand. > (As I understand the remark of Daniel Lichtblau in some future version of= Mathematica > they may provide such an Option). > > Thanks again > > Michael Weyrauch

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