Tricky integral of a rational function

Heirot · 2010-08-08 04:29:36

Hi all!

I have a very hard time verifying this integral:

None of the standard methods work because of the irreducibility of both the numerator and the denominator (the denominator can be factored out as a product of two polynomials of sixth degree, but that's not very helpful). On the other hand, Mathematica seems to have no problem evaluating the integral. So I must be missing something. It seems that the coefficients in the denominator are carefully chosen because changing any one of them gives Mathematica a hard time and it doesn't evaluate the integral at all.

Any suggestions will be greatly appreciated

bobbym · 2010-08-08 09:37:50

Hi Heirot;

Just because Mathematica can do it and you can't is not meaningful. Mathematica perhaps starts with the Risch algorithm and then uses many more, all of them designed for computers. Also it has many integrals stored in tables. If it reduces a problem to one it has in the tables it then can just spit it out. You will be happy to know that Maple gags on this one.

Incidentally in the real world no one is likely to spend a lot of time trying to get that integral analytically. They would see those limits of integration and run to numerical methods.

Heirot · 2010-08-08 19:11:15

I doubt that Mathematica runs any algorithm because it takes as much time to do this integral as it takes to calculate integral of x. Perhaps this integral if from the tables. How could I check that?

Btw, I know nobody ever would spend a lot of time thinking about this problem, but it's like a puzzle for me, especially when the solution is so elegant.

bobbym · 2010-08-08 20:07:11

Hi;

If you wrap the timing command around the indefinite integral you will see there is a very great time difference between this integrand and x. At least on this machine.

You can ask the people at wolfram what method they are using to evaluate that integral. They might tell you but most likely not. They are very secretive.

It all comes down to your meaning of the word verify. If you require mathematical certainty that may not be possible here. Would numerics satisfy you?

Their is one remaining oddity of this integral that the numerical methods would reveal.

Heirot · 2010-08-08 21:49:05

I agree with you that the indefinite integral would be more challenging to calculate. Because of the limits of integration one can use methods of contour integration or something of the sort that would greatly simplify the calculation. I was hoping that the special choice of the coefficients in the denominator makes that choice possible. E.g. expand the lower limit to minus infinity and then use the semicircle contour. It would pick up singularities in the upper half-plane (which are zeros of the denominator with positive imaginary part), summing their residues to give the integral. Now, since all of the poles are simple ones, I was hoping to express the sum of their residues in the terms of the coefficients of the denominator via Viete formulas. Do you think that could work?

bobbym · 2010-08-09 06:54:06

Hi Heirot;

I don't know try it! Let me know how you do.

Bob · 2010-08-14 00:55:53

Hi

Because of the way this question is phrased, I'm assuming there's an analytical solution in there somewhere.

You can sub X = x^2 but then what? The 'answer' seems to point towards arctan but still then what?

I tried to find factors (by inspection!) and got nowhere. I haven't got Mathematica which I guess would help.

If you've got a factorisation please post it.

Thanks,

Bob

bobbym · 2010-08-14 02:16:38

Hi bob;

This is the factorization he is talking about. He is right it is not much help.

Bob · 2010-08-15 08:37:54

Hi

Thanks for the factorisation. The symmetry makes me think this is the way to go but, so far, I've not got anywhere. On Wednesday 18th I'm off to Germany to visit my son, David. He's much better at maths than me so I'll ask him to have a try. Let you know if he comes up with anything.

Question to Heirot.

Where did this problem come from ... maybe there's a clue in that?

Bob

Last edited by Bob (2010-08-15 08:39:36)

Bob · 2010-08-17 23:45:25

My son wants the complete complex factorisation for each x^6 polynomial. If mathematica can do this, please would you put it to work for us.

Thanks in anticipation.

The rational function is 'even' so the problem can be re-expressed as the integral from minus infinity to plus infinity and then halved.

You'll be pleased to know this makes the problem much easier!

Bob

Last edited by Bob (2010-08-17 23:50:09)

bobbym · 2010-08-18 01:35:26

Hi bob bundy;

That is what I have been unable to do satisfactorily. If I could analytically factor those 2 sixth degree polynomials I would have the poles. Then I could use the method of residues to evaluate that integral. Hope I am explaining it right.

Unfortunately, no package, neither mathematica, maple or derive will factor them into exact forms like (x - 2 + 3i)^2 etc. As you know passed the fourth degree there are no algebraic methods to get roots of a polynomial. The packages will only get the roots (factors) numerically. This is not sufficient as far as I understand to get π / 2 exactly.

Using algebraic extensions for factoring I can coax the packages to give me exact radical forms for the factorization. But the answers are to large to even fit into a post. I will post if I make any kind of progress. I am sorry for this.

Bob · 2010-08-22 22:21:18

Hi bobbym

I said I'd ask my son whilst in Germany, and he immediately suggested Cauchy's Integral Formula (see Wiki). From the way you have replied, it sounds like you've thought this too.

The real coefficients, plus the even function mean there must be 3 roots in each quadrant of the complex plane. He substituted y = x + 1 into one of the order 6 polys and got the coefficients of the other order 6 poly, but in the reverse order. From this he concluded that if z is one root, then the other two in that quadrant must be 1/(1-z) and -1 + 1/z. I've still got to get my head around how he reached this conclusion but I'm sure he's right.

He then thinks you can express the integral in terms of z, and conjectures that the zs will go out without having to be evaluated and leaving an imaginary value which, together with Cauchy, will lead to the required result.

But, he also thinks this problem may require one more twist to achieve all I have just stated. He will probably keep thinking now he's got a taste for the problem and may add his own posts subsequently.

All for now, as I've got some catching up to do on-line.

Bob

Bob · 2011-07-15 23:42:19

hi bobbym, gAr and anyone else who can maybe help.

(with apologies for the long post)

It was almost a year ago that this post first appeared. No one has posted a solution yet.

The proposal was:

Factorise the order 12 polynomial into 12 complex factors and use Cauchy.

But how to factorise?

After puzzling on this for some time, I devised an alternative strategy.

Factorise into 3 quartic factors, split into partial fractions with numeric numerators, and integrate each (without Cauchy).

As the polynomial has real and even coefficients each quartic must have four complex factors of the form (a + bi), (a bi), (-a + bi) and ( -a bi) and so each quartic will have the form:

where P and Q are real.

Lacking any other way to proceed I set up three such quartics

and multiplied this out to obtain 7 simultaneous equations in 6 unknowns.

I set up the equations in Excel and used goal seek to try and number crunch a solution.

To show if I was approaching a solution I took the 6 trial values for P, Q ...U, and computed how far distant from the target coefficients my solution lay.

My initial distance was over 100 units away, but by goal seeking each variable in turn I got down to about 18. But I couldnt get closer. Eventually, I worked out this is because of the way goal seek works. It kept hopping over the desired solution because its steps were to large.

So I switched to decimal search and got to within 0.6 of target.

After a pause I wrote a computer program to do the decimal searching for me and, within a few minutes, got to within 10^-16 of target. I dont think I can number crunch to a closer solution.

The factorisation is then:

with

P = -0.776376326
Q = 0.382818292
R = -0.608405391
S = 0.108362067
T = -8.61521826
U = 24.1062714

It would be better if I could arrive at exact values by some analytic method but I cannot.

Feeling I was getting close, I decided to go all out with the rest of the solution, planning to post it as my 1000th post.

But Ive hit a snag.:(

Firstly, I dont know if mathematica can test out that factorisation by multiplying it out, back to the power 12 poly.
If someone could test that, Id be grateful.

Then I started the partial fractions.

I used the coefficients for x^6, x^4 and x^2 to make three simultaneous equations and solved them:
I have
A = -1.23654966
B = 1.045292468
C = 4.0356581263

But these are not consistent with the x^8 and x^0 equations.

Furthermore, I cannot see how to integrate the 1/quartic expressions anyway.

So Im stuck.

Any help would be gratefully received.

Bob (post 987)

Last edited by Bob (2011-07-15 23:42:44)

bobbym · 2011-07-16 00:04:48

Hi;

I will try to do what you ask. In the meantime you should know that I have working with this integrand using the Ostrogradsky algorithm. This should have worked but it did not for some reason.

Bob · 2011-07-16 00:14:57

hi bobbym,

Many thanks.

It has just occurred to me that the partial fractions may not have numeric numerators. That would account for the discrepancy. And it might make it easier to integrate if the numerators are Ax + B

Bob
988

Last edited by Bob (2011-07-16 00:15:17)

gAr · 2011-07-16 00:20:57

Hi all,

I too tried this a few days ago using cauchy's.
And also tried whether I can rewrite it into some other form.

Let me give it another try.

Ostrogradsky's algorithm is new to me, I'll try that too!

bobbym · 2011-07-16 00:57:47

Hi;

P = -0.776376326
Q = 0.382818292
R = -0.608405391
S = 0.108362067
T = -8.61521826
U = 24.1062714

Those coefficients are quite close. But the exact, closed form for them Mathematica can not find.

Bob · 2011-07-16 01:06:57

hi bobbym

How close, exactly. Can you post the polynomial that you got?

Thanks,

Bob
989

Last edited by Bob (2011-07-16 01:07:17)

bobbym · 2011-07-16 01:28:22

Hi;

This is the denominator expanded out using your coefficients.

Here are my equations.

P=-.776376329716313367678396729764340830599615711886

Q=.38281829417943271655606084094051556034456756849056

R=-.6084053916280706468434233044383803346185796004872

S=.1083620659490676250704100110401129566047127759067

T=-8.615218278655615985478179965797278834781804687627

U=24.106271297265356993763009485782695345494455926176

Bob · 2011-07-16 01:44:22

hi bobbym,

Oh wow! The advantages of having computing power. It took me ages to get my (grossly inadequate) values.

But what are these numbers? Apart from just long decimals; does mathematica 'recognise' them as roots or trig functions etc of something exact? Do you see what I'm asking? I had originally hoped to get P, Q etc as whole numbers, and when that proved impossible, as rationals, then maybe sequential trig values like sine 15, sine 30, sine 45 ..... but nothing. Ggggrrr!

So, back to the OP. How does mathematica come up with an analytical answer?

I shall re-try the partial fractions idea now I know the factorisation is, in principle, a 'right step'.

Bob
890

bobbym · 2011-07-16 02:16:38

Hi Bob;

Both mathematica and maple do not recognize them in terms of radicals, simple trig values or constants such as pi, e, ln(2) etc.

gAr · 2011-07-16 02:27:23

Even this didn't find any.

bobbym · 2011-07-16 02:33:06

Hi gAr;

Yes, that too! Most likely those values might be transcendental.

gAr · 2011-07-16 02:50:21

Hi bobbym,

Yes, and I wonder who made up those polynomials, which evaluates to pi?!

Even with ostrogradsky method, I couldn't proceed.
There must be a trick, something like replacing x with some other polynomial. Otherwise what can be done, where even complex analysis fails?!

gAr · 2011-07-16 03:08:15

This page tells about few methods: http://mpec.sc.mahidol.ac.th/radok/phys … /sec41.htm

Math Is Fun Forum

#1 2010-08-08 04:29:36

Tricky integral of a rational function

#2 2010-08-08 09:37:50

Re: Tricky integral of a rational function

#3 2010-08-08 19:11:15

Re: Tricky integral of a rational function

#4 2010-08-08 20:07:11

Re: Tricky integral of a rational function

#5 2010-08-08 21:49:05

Re: Tricky integral of a rational function

#6 2010-08-09 06:54:06

Re: Tricky integral of a rational function

#7 2010-08-14 00:55:53

Re: Tricky integral of a rational function

#8 2010-08-14 02:16:38

Re: Tricky integral of a rational function

#9 2010-08-15 08:37:54

Re: Tricky integral of a rational function

#10 2010-08-17 23:45:25

Re: Tricky integral of a rational function

#11 2010-08-18 01:35:26

Re: Tricky integral of a rational function

#12 2010-08-22 22:21:18

Re: Tricky integral of a rational function

#13 2011-07-15 23:42:19

Re: Tricky integral of a rational function

#14 2011-07-16 00:04:48

Re: Tricky integral of a rational function

#15 2011-07-16 00:14:57

Re: Tricky integral of a rational function

#16 2011-07-16 00:20:57

Re: Tricky integral of a rational function

#17 2011-07-16 00:57:47

Re: Tricky integral of a rational function

#18 2011-07-16 01:06:57

Re: Tricky integral of a rational function

#19 2011-07-16 01:28:22

Re: Tricky integral of a rational function

#20 2011-07-16 01:44:22

Re: Tricky integral of a rational function

#21 2011-07-16 02:16:38

Re: Tricky integral of a rational function

#22 2011-07-16 02:27:23

Re: Tricky integral of a rational function

#23 2011-07-16 02:33:06

Re: Tricky integral of a rational function

#24 2011-07-16 02:50:21

Re: Tricky integral of a rational function

#25 2011-07-16 03:08:15

Re: Tricky integral of a rational function

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