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Re: Definite Integration in Mathematica
 To: mathgroup at smc.vnet.net
 Subject: [mg74678] Re: Definite Integration in Mathematica
 From: "dimitris" <dimmechan at yahoo.com>
 Date: Sat, 31 Mar 2007 01:28:47 0500 (EST)
 References: <etqo3f$10i$1@smc.vnet.net><euigeg$ov1$1@smc.vnet.net>
David,
I really appreciate your assistance!
Thanks a lot both for your integration insight and as well your
wishes!
Kind Regards
Dimitris
=CF/=C7 dimitris =DD=E3=F1=E1=F8=E5:
> > 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])])]
>
> Could we show me how you get this antiderivative for the integrand
>
> f[x_] = (x^2 + 2*x + 4)/(x^4  7*x^2 + 2*x + 17);
>
> BTW,
>
> Using Mathematica I was able to get a continuous in the real axis
> antiderivative as follows
>
> Times @@ Apply[#1[[1]]  #1[[2]] & , Solve[Denominator[f[x]] == 0, x],
> 1];
> Apart[(4 + 2*x + x^2)/%];
> Map[FullSimplify, %, 1];
> (Integrate[#1, x] & ) /@ %;
> FF[x_] = FullSimplify[%]
> FullSimplify[D[FF[x], x] == f[x]]
>
> (1/4)*(2*ArcTan[(2*x)/(1 + Sqrt[15  4*I])] + 2*ArcTan[(2*x)/(1 +
> Sqrt[15  4*I])] 
> 2*ArcTan[(2*x)/(1 + Sqrt[15 + 4*I])] + 2*ArcTan[(2*x)/(1 +
> Sqrt[15 + 4*I])] 
> I*(Log[(7 + 2*I) + Sqrt[15  4*I]  2*x^2]  Log[(7  2*I) +
> Sqrt[15 + 4*I]  2*x^2] +
> Log[(7  2*I) + Sqrt[15  4*I] + 2*x^2]  Log[(7 + 2*I) +
> Sqrt[15 + 4*I] + 2*x^2]))
> True
>
> Plot[FF[x], {x, 0, 4}]
>
>
> Thanks a lot!
> Dimitris
>
>
>
> =CF/=C7 David W.Cantrell =DD=E3=F1=E1=F8=E5:
> > "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 be=
lo=
> w=2E
> >
> > > 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 calc=
ul=
> us
> > > 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 consid=
er
> > > > the
> > > > subject very fundamental I would like any kind of insight.
> > > >
> > > > In the section about Proper Integrals in his article, Adamchik
> > > > mentions that the NewtonLeibniz 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[{$DisplayFunction=Identity},
> > > > Plot[F[x],{x,#[[1]],#[[2]]}]&/@Partition[Range[0,4,2],2,1]];
> > > > {(Pi/2), Pi/2}
> > > > (*plot to be displayed*)
> > > >
> > > > The right way of applying the Fundamental theorem is the following
> > > >
> > > > (Limit[F[x], x > 4, Direction > 1]  Limit[F[x], x > 2, Directio=
n 
> > > >> 1]) +
> > > > (Limit[F[x], x > 2, Direction > 1]  Limit[F[x], x > 0, Directi=
on
> > > > > 1])
> > > > N[%]
> > > > Pi  ArcTan[1/4]  ArcTan[5/12]
> > > > 2.501822870763167
> > > >
> > > > Integrate works in precisely this way
> > > >
> > > > Integrate[f[x], {x, 0, 4}]
> > > > N[%]
> > > >
> > > > Pi  ArcTan[1/4]  ArcTan[5/12]
> > > > 2.501822870763167
> > > >
> > > > A little later, he (i.e. Adamchik) says "The origin of
> > > > discontinuities
> > > > along the path of integration is not in the method of indefinite
> > > > integration but rather in the integrand."
> > > >
> > > > Adamchik mentions next that the four zeros of the integrand's
> > > > denominator
> > > > are two complexconjugate pairs having real parts +/ 1.95334. It
> > > > then
> > > > seems that he is saying that, connecting these conjugate pairs by
> > > > vertical line segments in the complex plane, we get two branch
> > > > cuts...
> > > >
> > > > BUT didn't the relevant branch cuts for his int cross the real axis
> > > > at x = +/ 2, rather than at x = +/ 1.95334?
> > > >
> > > > (NOTE: The difference between 1.95334 and 2 is not due to numerical
> > > > error).
> > > >
> > > > Exactly what's going on here?
> > > >
> > > > Show[GraphicsArray[Block[{$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]}, {=
(P=
> i/
> > > > (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
> > > >
> > > >
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