       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,

> 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.

>>

>> 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,#[],#[]}]&/@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, 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

>>

>> 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, ArcTan[Sqrt[2/3]]/Sqrt}, {-(Pi/

>> (2*Sqrt)), Pi/(2*Sqrt)},

>> {-(ArcTan[Sqrt[2/3]]/Sqrt), -(ArcTan[Sqrt[2/3]]/Sqrt)}, {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]}, {x -> 3*Pi - I*Log[5 - 2*Sqrt]},

>> {x -> Pi + I*Log[5 - 2*Sqrt]},

>> {x -> 3*Pi + I*Log[5 - 2*Sqrt]}}

>>

>> Of course F[4Pi]-F=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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