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Re: Definite Integration in Mathematica
*To*: mathgroup at smc.vnet.net
*Subject*: [mg74454] Re: Definite Integration in Mathematica
*From*: "dimitris" <dimmechan at yahoo.com>
*Date*: Thu, 22 Mar 2007 01:07:17 -0500 (EST)
*References*: <etqo3f$10i$1@smc.vnet.net>
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
>
>
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