Re: Definite Integration in Mathematica

*To*: mathgroup at smc.vnet.net*Subject*: [mg74488] Re: Definite Integration in Mathematica*From*: "David W.Cantrell" <DWCantrell at sigmaxi.net>*Date*: Fri, 23 Mar 2007 19:01:42 -0500 (EST)*References*: <etqo3f$10i$1@smc.vnet.net> <ett73k$h3g$1@smc.vnet.net>

"Michael Weyrauch" <michael.weyrauch at gmx.de> wrote: > Hello, > > another nice example, where the result for the integral given by > Mathematica just cannot be right. I understand your point, but it's impossible to have a result here which is "right" throughout the complex plane. > The indefinite integral of a continuuos function cannot have a jump. Again, I understand your point. You're thinking just about integrating along the real line; the integrand (x^2 + 2x + 4)/(x^4 - 7x^2 + 2x + 17) is indeed continuous there. But computer algebra systems sometimes give antiderivatives which are "right" only piecewise. You're welcome to consider that unfortunate in this case because we can do better; see below. > That -- to my opinion -- is mathematical > nonsens. We all learned that integration "smoothens", i.e. if the > integrand is somewhat "ugly" the integral is less "ugly". (Never mind my > English!). > > So the result should actually be presented as > > F[x_]=ArcTan[(1 + x)/(4 - x^2)]*UnitStep[2 - x] + > (Pi + ArcTan[(1 + x)/(4 - x^2)])*UnitStep[-2 + x] No, not if you're wanting an antiderivative valid along the whole real line. Your F has a jump discontinuity at x = -2. Furthermore, it is Indeterminate at x = +2, although that singularity is removable. Back in December, in the thread which led to this one, I gave an antiderivative which is continuous along the whole real line: ArcTan[(4x + Sqrt[2(15 + Sqrt[241])])/(2 - Sqrt[2(-15 + Sqrt[241])])] + ArcTan[(4x - Sqrt[2(15 + Sqrt[241])])/(2 + Sqrt[2(-15 + Sqrt[241])])] David W. Cantrell > This is a perfectly nice function without jumps and it is the > antiderivative of your integrand, and the fundamental theorem of calculus > works with it... > > So, despite my love for Mathematica, here it fools me.... > > Regards Michael > > "dimitris" <dimmechan at yahoo.com> schrieb im Newsbeitrag > news:etqo3f$10i$1 at smc.vnet.net... > > Hello to all of you! > > > > Firstly, I apologize for the lengthy post! > > Secondly, this post has a close connection with a recent (and well > > active!) > > thread titled "Integrate" and one old post of mine which was based on > > a older > > post of David Cantrell. Since there was no response and I do consider > > the > > subject very fundamental I would like any kind of insight. > > > > In the section about Proper Integrals in his article, Adamchik > > mentions that the Newton-Leibniz formula (i.e. the Fundamental > > Theorem of Integral Calculus: Integrate[f[x],{x,a,b}]=F[b]-F[a], > > F[x]: an antiderivative), does not hold any longer if the > > antiderivative F(x) has singularities in the integration interval > > (a,b). > > > > To demonstrate this, he considers the integral of the function: > > > > f[x_] = (x^2 + 2*x + 4)/(x^4 - 7*x^2 + 2*x + 17); > > > > over the interval (0,4). > > > > Plot[f[x], {x, 0, 4}]; > > (*plot to be displayed*) > > > > The integrand posseses no singularities on the interval (0,4). > > > > Here is the corresponding indefinite integral > > > > F[x_] = Simplify[Integrate[f[x], x]] > > ArcTan[(1 + x)/(4 - x^2)] > > > > Substituting limits of integration into F[x] yields an incorrect > > result > > > > Limit[F[x], x -> 4, Direction -> 1] - Limit[F[x], x -> 0, Direction - > >>1] > > N[%] > > NIntegrate[f[x], {x, 0, 4}] > > > > -ArcTan[1/4] - ArcTan[5/12] > > -0.6397697828266257 > > 2.501822870767894 > > > > This is because the antiderivative has a jump discontinuity at x=2 > > (also at x = -2), so that the Fundamental theorem cannot be used. > > > > Indeed > > > > Limit[F[x], x -> 2, Direction -> #1]&/@{-1, 1} > > Show@Block[{$DisplayFunction=Identity}, > > Plot[F[x],{x,#[[1]],#[[2]]}]&/@Partition[Range[0,4,2],2,1]]; > > {-(Pi/2), Pi/2} > > (*plot to be displayed*) > > > > The right way of applying the Fundamental theorem is the following > > > > (Limit[F[x], x -> 4, Direction -> 1] - Limit[F[x], x -> 2, Direction - > >> -1]) + > > (Limit[F[x], x -> 2, Direction -> 1] - Limit[F[x], x -> 0, Direction > > -> -1]) > > N[%] > > Pi - ArcTan[1/4] - ArcTan[5/12] > > 2.501822870763167 > > > > Integrate works in precisely this way > > > > Integrate[f[x], {x, 0, 4}] > > N[%] > > > > Pi - ArcTan[1/4] - ArcTan[5/12] > > 2.501822870763167 > > > > A little later, he (i.e. Adamchik) says "The origin of > > discontinuities > > along the path of integration is not in the method of indefinite > > integration but rather in the integrand." > > > > Adamchik mentions next that the four zeros of the integrand's > > denominator > > are two complex-conjugate pairs having real parts +/- 1.95334. It > > then > > seems that he is saying that, connecting these conjugate pairs by > > vertical line segments in the complex plane, we get two branch > > cuts... > > > > BUT didn't the relevant branch cuts for his int cross the real axis > > at x = +/- 2, rather than at x = +/- 1.95334? > > > > (NOTE: The difference between 1.95334 and 2 is not due to numerical > > error). > > > > Exactly what's going on here? > > > > Show[GraphicsArray[Block[{$DisplayFunction = Identity}, > > (ContourPlot[#1[F[x + I*y]], {x, -4, 4}, {y, -4, 4}, Contours -> 50, > > PlotPoints -> 50, ContourShading -> False, Epilog -> {Blue, > > AbsoluteThickness[2], Line[{{0, 0}, {4, 0}}]}, > > PlotLabel -> #1[HoldForm[F[x]]]] & ) /@ {Re, Im}]], ImageSize > > - > 500]; > > (*contour plots to be displayed*) > > > > > > Consider next the following function > > > > f[x_] = 1/(5 + Cos[x]); > > > > Then > > > > Integrate[f[x], {x, 0, 4*Pi}] > > N[%] > > NIntegrate[f[x], {x, 0, 4*Pi}] > > > > Sqrt[2/3]*Pi > > 2.565099660323728 > > 2.5650996603270704 > > > > F[x_] = Integrate[f[x], x] > > ArcTan[Sqrt[2/3]*Tan[x/2]]/Sqrt[6] > > > > D[F[x], x]==f[x]//Simplify > > True > > > > Plot[f[x], {x, 0, 4*Pi}, Ticks -> {Range[0, 4*Pi, Pi/2], Automatic}] > > Plot[F[x], {x, 0, 4*Pi}, Ticks -> {Range[0, 4*Pi, Pi/2], Automatic}] > > > > The antiderivative has jump discontinuities at Pi and 3Pi inside the > > integration range > > > > Table[(Limit[F[x], x -> n*(Pi/2), Direction -> #1] & ) /@ {-1, 1}, {n, > > 0, 4}] > > {{0, 0}, {ArcTan[Sqrt[2/3]]/Sqrt[6], ArcTan[Sqrt[2/3]]/Sqrt[6]}, {-(Pi/ > > (2*Sqrt[6])), Pi/(2*Sqrt[6])}, > > {-(ArcTan[Sqrt[2/3]]/Sqrt[6]), -(ArcTan[Sqrt[2/3]]/Sqrt[6])}, {0, > > 0}} > > > > Reduce[5 + Cos[x] == 0 && 0 <= Re[x] <= 4*Pi, x] > > {ToRules[%]} /. (x_ -> b_) :> x -> ComplexExpand[b] > > x /. %; > > ({Re[#1], Im[#1]} & ) /@ %; > > poi = Point /@ %; > > > > x == 2*Pi - ArcCos[-5] || x == 4*Pi - ArcCos[-5] || x == ArcCos[-5] || > > x == 2*Pi + ArcCos[-5] > > {{x -> Pi - I*Log[5 - 2*Sqrt[6]]}, {x -> 3*Pi - I*Log[5 - 2*Sqrt[6]]}, > > {x -> Pi + I*Log[5 - 2*Sqrt[6]]}, > > {x -> 3*Pi + I*Log[5 - 2*Sqrt[6]]}} > > > > Of course F[4Pi]-F[0]=0 incorrectly. > > > > The reason for the discrepancy in the above result is not because of > > any problem with the fundamental theorem of calculus, of course; it is > > caused by the multivalued nature of the indefinite integral arctan. > > > > > > Show[GraphicsArray[Block[{$DisplayFunction = Identity}, > > (ContourPlot[#1[F[x + I*y]], {x, 0, 4*Pi}, {y, -4, 4}, Contours -> > > 50, PlotPoints -> 50, ContourShading -> False, > > FrameTicks -> {Range[0, 4*Pi, Pi], Automatic, None, None}, > > Epilog -> {{PointSize[0.03], Red, poi}, > > {Blue, Line[{{0, 0}, {4*Pi, 0}}]}}] & ) /@ {Re, Im}]], > > ImageSize -> 500] > > > > So, in this example the discontinuities are indeed from the branch > > cuts that start and end from the simple poles of the integrand which > > is in agreement with V.A. paper! > > > > > > I think I am not aware of something fundamental! > > Can someone point out what I miss? > > > > > > Regards > > Dimitris > > > >

**Re: Definite Integration in Mathematica**

**Re: Re: Definite Integration in Mathematica**

**Re: Definite Integration in Mathematica**

**Re: Definite Integration in Mathematica**