[Date Index]
[Thread Index]
[Author Index]
Re: Definite Integration in Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74501] Re: Definite Integration in Mathematica
*From*: dh <dh at metrohm.ch>
*Date*: Fri, 23 Mar 2007 19:08:46 -0500 (EST)
*References*: <etqo3f$10i$1@smc.vnet.net> <ett6i6$gd2$1@smc.vnet.net>
Hello Dimitris,
Yes, you are right, Mathematica chooses an antiderivative function that
has branch cuts at approx. +/-2. However, this is not the only possible
antiderivative, there are others with different branch cuts. But it is
not too easy to find them.
Daniel
dimitris wrote:
> Hello Daniel!
>
> Of course I am aware of branch cuts and the stuff!
>
> May be I was not too clear in order someone to understand me!
>
> Read the first 4 pages of the paper of V Adamichik's Definite
> Integration
> in Mathematica v. 3.0. available from here:
>
> http://library.wolfram.com/infocenter/Articles/3157/
>
> I will try to set my question as clear as I can!
>
> The integral that appears in V.A. paper is
>
> f = HoldForm[Integrate[(4 + 2*x + x^2)/(17 + 2*x - 7*x^2 + x^4), {x,
> 0, 4}]]
>
> and the integrand is a smooth integrable function in the whole real
> axis.
>
> Mathematica's answer of (symbolic) definite integration is
>
> ReleaseHold@%
> Pi - ArcTan[1/4] - ArcTan[5/12]
>
> which agrees with the result obtained by numerical integration
>
> {(N[#1, 20] & )[%], ReleaseHold[f /. Integrate[x___] :> NIntegrate[x,
> WorkingPrecision -> 40,
> AccuracyGoal -> 20]]}
> {2.5018228707631675676, 2.50182287076316756755}
>
> Everything is ok up to now! And as I have shown in an old post you can
> using Mathematica to get a continuus
> antiderivative in the integration range but this is not my goal! I am
> very happy by Mathematica's antiderivative which is rather simple
> contrary to my expression.
>
> Mathematica uses (an extended version of) Risch algorithm in order to
> get an antiderivative
> for [(x^2 + 2*x + 4)/(x^4 - 7*x^2 + 2*x + 17)
>
> Simplify[Integrate[(x^2 + 2*x + 4)/(x^4 - 7*x^2 + 2*x + 17), x]]
> ArcTan[(1 + x)/(4 - x^2)]
>
> and due to the jump discontinuity of ArcTan at x=2 the correct way in
> order to apply the Newton-Leibniz formula
> is
>
> (% /. x -> 4) - Limit[%, x -> 2, Direction -> -1] + Limit[%, x -> 2,
> Direction -> 1] - (% /. x -> 0)
> Pi - ArcTan[1/4] - ArcTan[5/12]
>
> Again everything is clear and absolutely easy to understand!
>
> My query appears exactly after this point!
>
> Victor Adamchik tries next (in the same article) to explain the
> situation.
>
> He states that discontinuities of antiderivatives along the path of
> integration are due to the nature of the integrand-not the method of
> indefinite integration. In the above example, the integrand has four
> singular poles that become branch points for the antiderivative.
> Connected in pairs these points make two branch cuts. The path of
> integration crosses one of them
>
> The poles of the integrand are at the following points in the complex
> plain
>
> NSolve[x^4 - 7*x^2 + 2*x + 17 == 0, x]
> {{x -> -1.9533360788744223 - 0.2440276635405635*I}, {x ->
> -1.9533360788744223 + 0.2440276635405635*I},
> {x -> 1.9533360788744223 - 0.7559723364594371*I}, {x ->
> 1.9533360788744223 + 0.7559723364594371*I}}
>
> (ContourPlot[Evaluate[#1[ArcTan[(1 + x)/(4 - x^2)] /. x -> x + I*y]],
> {x, -3, 4}, {y, -3, 4}, ContourShading -> False,
> Contours -> 20, PlotPoints -> 40, Epilog -> {Hue[0],
> Thickness[0.005], Line[{{0, 0}, {4, 0}}]}] & ) /@ {Re, Im}
>
> What I don't understand is exactly his statement "... Connected in
> pairs these points make two branch cuts..."
> and how is connected with the jump discontinuity at x=2.
>
> These pairs of chosen branch cuts by Mathematica for the
> antiderivative cross the horizontal axis at x=+/-2?
>
> If the answer is yes at last I have understand it!
>
>
> dh <dh at metrohm.ch> wrote:
>
> Hi Dimitris,
> if an anylytic function is multivalued with singularities, people
> often
> try nevertheless to define a continuous single valued function. This
> is
> not possible globally, there are always places where we will have a
> misfit. One therefore cuts the plane and defines "branch cuts" where
> the
> new function is not continuous. Away from these cuts the function is
> continuous. Note that these cuts are not clear cut. As long as we get
> a
> continuos function, we can choose the cuts as we please. Further, as
> the
> original function is multivalued, we can define even more different
> continuous functions.
> Now concerning the fundmental theorem, it seems clear that if we
> cross
> the branch cut the theorem can not be applied. It only works if the
> antiderivative is continuous in the integration range. In your case:
> ArcTan[(1 + x)/(4 - x^2)], the branch cut was chosen so that we have
> a
> discontinuity at x=2. If we had chosen the branch cuts differently,
> we
> may have been able to find a function that is continuous in 0..4. But
> this is not easily done! At least not analytically. However, you may
> easily do so numerically in the real domain (although the function
> may
> not be non-analytical (no power series)).
> Daniel
>
> dimitris wrote:
>> 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
>>
>>
>
>
Prev by Date:
**Re: Definite Integration in Mathematica**
Next by Date:
**Re: Definite Integration in Mathematica**
Previous by thread:
**Re: Definite Integration in Mathematica**
Next by thread:
**Re: Definite Integration in Mathematica**
| |