Re: Definite Integration in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74656] Re: Definite Integration in Mathematica*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Fri, 30 Mar 2007 02:55:22 -0500 (EST)*References*: <eufq77$7hj$1@smc.vnet.net>

"dimitris" <dimmechan at yahoo.com> wrote: > Hi. > > I apologise for the delay. > > I will reply to your questions in this post; but I want the current > post to serve as a sum up of this beautiful thread. So I apologise also > for its considerable length and not anwering at once to your queries. There is certainly no need to apologize. Congratulations that you are to be a father! Due to the length of your message, I've removed (snipped) a great deal below, leaving only the material essentially related to my response. > f[x_] :== 1/(2 + Cos[x]) > > The indefinite integral by Mathematica > > F[x_] == Integrate[f[x], x] > (2*ArcTan[Tan[x/2]/Sqrt[3]])/Sqrt[3] > >From an old book about Mathematica (Mathematica for Scientists and > Enginners by R. Gass we got (but without further explanation...) > > FF[x_] == (-2*3^(1/2)*ArcTan[Sin[x]/(Cos[x] + 1)])/3 + > (2*3^(1/2)*ArcTan[(3^(1/2)*Sin[x])/(3*(Cos[x] + 1))])/3 + > (3^(1/2)*x)/3 > as a continuous antiderivative in the real axis. > > I will be very glad if you could show me how exactly the author get > previous antiderivative or what method in general would you follow > in order to get a continuous antiderivative in the real axis. Below, I'll show what I suspect that Gass did and also show what I would then do in this problem. > It is time to answer your first second question > > > Unfortunately, I don't have Trott's book. Since it's very easy to > > obtain, could you also please show us the continuous > > antiderivative for > > 1/(5 + Cos[x]) obtained by Trott's method? > > As you see I chose another example (similar to Trott's) because I > want to show also the continuous antiderivative by Gass' book. > > I don't know what you have understood but stating that > > > > 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!) > > I really mean something like > > Fc[x_] :== Piecewise[{{F[x], Inequality[-4*Pi, LessEqual, x, Less, > -3*Pi]}, {Pi/Sqrt[3], x ==== -3*Pi}, > {F[x] + 2*(Pi/Sqrt[3]), -3*Pi < x < -Pi}, {4*(Pi/Sqrt[3]), x > ==== - > Pi}, {F[x] + 4*(Pi/Sqrt[3]), -Pi < x < Pi}, > {6*(Pi/Sqrt[3]), x ==== Pi}, {F[x] + 6*(Pi/Sqrt[3]), Pi < x < > 3*Pi}, > {8*(Pi/Sqrt[3]), x ==== 3*Pi}, > {F[x] + 8*(Pi/Sqrt[3]), Inequality[3*Pi, Less, x, LessEqual, > 4*Pi]}}] > > I hope you are not dissapointed! To be honest, yes, that is disappointing because it is messy and, more importantly, it is not an antiderivative valid over the _whole_ real line. -------------------------------------- For 1/(2 + Cos[x]), we wish to get an antiderivative which is valid on R. As you showed (in some of what I snipped), the standard Weierstrass substitution, u = Tan[x/2], leads to the same antiderivative as that returned by Mathematica, (2*ArcTan[Tan[x/2]/Sqrt[3]])/Sqrt[3]. The only type of singularity occurs when x = (2n + 1)Pi, n integer. By looking at the difference between limits from either side of, say, Pi, we see that these discontinuities are caused by jumps of 2 Pi/Sqrt[3]. To remove those jumps, we may simply add 2 Pi/Sqrt[3] Floor[(x + Pi)/(2 Pi)] to Mathematica's antiderivative. Let's call this result F1[x]: (2 ArcTan[Tan[x/2]/Sqrt[3]] + 2 Pi Floor[(x + Pi)/(2 Pi)])/Sqrt[3] Although F1[x] has no jumps, it is still Indeterminate at x = (2n + 1)Pi. But these singularities are removable. The only question remaining is how to remove them "nicely". Note that 2 Pi Floor[(x + Pi)/(2 Pi)] is the same as x - 2 ArcTan[Tan[x/2]], except that the latter is Indeterminate at x = (2n + 1)Pi. Substituting accordingly in F1[x], we get F2[x]: (2 ArcTan[Tan[x/2]/Sqrt[3]] + x - 2 ArcTan[Tan[x/2]])/Sqrt[3] This seems no better, and perhaps even worse, than F1[x]. But, simply noting that Tan[x/2] and Sin[x]/(1 + Cos[x]) are identical, F2[x] is the result given by Gass. (Of course, whether he obtained his result by my method, I have no idea.) We still have removable singularities to get rid of. The trick (which can perhaps be done with Mathematica, although I don't know how) is to combine the two arctangents of F2[x] into a single arctangent. Using just basic trig to do so, F2[x] can be "rewritten" as F3[x]: (x - 2 ArcTan[Sin[x]/(Cos[x] + 2 + Sqrt[3])])/Sqrt[3] which is my suggested antiderivative on R. The nice thing is that, having combined the two former arctangents, each of which was Indeterminate at x = (2n + 1)Pi, we have gotten rid of the removable discontinuities, the new single arctangent being defined for all reals. - - - - - More generally, for 1/(a + Cos[x]) with a > 1 or a < -1, the above process can be applied to give an antiderivative valid on R: (x - 2 ArcTan[Sin[x]/(Cos[x] + a + Sign[a] Sqrt[a^2 - 1])])/(Sign[a] Sqrt[a^2 - 1]). Kind regards, David W. Cantrell P.S. There is a way that F1[x] could be _made_ valid throughout R. One might suggest that the reason that F1[x] is Indeterminate at x = (2n + 1)Pi in Mathematica is the presence of Tan[x/2], but I do not agree. Tan[x/2] is defined for _all_ real x in Mathematica; in particular, at x = (2n + 1)Pi, its value is ComplexInfinity. The reason that F1[x] is Indeterminate is that ArcTan[ComplexInfinity] is Indeterminate. That, in my opinion, need not be the case! ArcTan is, after all, just the principal branch of the multivalued inverse tangent relation. What we choose to call the principal branch is, in essence, _arbitrary_. That being the case, what harm would be done by assigning a value to ArcTan[ComplexInfinity]? Granted, ArcTan[Infinity] is +Pi/2, while ArcTan[-Infinity] is -Pi/2. But I see no reason why we could not reasonably choose to say that ArcTan[ComplexInfinity] is one of those. If we chose ArcTan[ComplexInfinity] = -Pi/2, then F1[x], as originally given, would be valid throughout R. If we instead chose ArcTan[ComplexInfinity] = +Pi/2, then the Floor in F1 would need to be replaced by Ceiling to give us an antiderivative valid throughout R. My point is that Indeterminate is a "dead end" which should be avoided when there is a reasonable, even if arbitrary, way to do so.

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