Tricky integral of a rational function

bobbym · 2011-07-18 01:47:20

Hi;

I'm not sure what the transformations do. It just looks like the coefficients are changed!

Then we have a major problem here because I have no idea what they are doing either.

gAr · 2011-07-18 02:04:02

The whole transformation thing is crazy!

But something is interesting there. With denominator the same, whatever be the numerator, the answer is a multiple of pi.

bobbym · 2011-07-18 13:22:06

Hi;

To show that the method could work, here it is.

We take the integral from - ∞ to ∞ because that is what the formula is set up for.

These are the poles ( the roots of the denominator), that lie in upper half of the contours plane?

These are the residues of the poles.

Sum them and times by 2 π i and you get:

Which is very close to just π. We divide it by two because our integral is symmetrical and goes from 0 to infinity.

Now we can see that unless we can get the poles analytically we will not get anything but a numerical answer. Since the denominator is a 12th degree polynomial there is no algebraic method to extract the roots ( poles ) so even a package is forced to go to numerics to get them.

So if anyone can get this integral analytically I would say he would have to first reduce the denominator to a combination of quartics.

Bob · 2011-07-18 20:19:42

hi bobbym,

Thanks for that. In my small way, it's what I was trying, inadequately, to do. I'll continue to work on the quartic solution, but, for me, that too, is full of approximate values.

If you can remember the journal (post #26) I'll try to get a look at it. Or even narrow it down to one from a list.

Bob
998

Last edited by Bob (2011-07-18 20:21:02)

bobbym · 2011-07-18 22:50:16

Hi Bob;

I am sorry but I do not have a clue as to where I saw it. Now I am wondering if I ever did see it.

I tried come up with a denominator consisting of 4 cubics but no luck there either.

gAr · 2011-07-18 22:58:05

Hi,

Since a CAS factors it to only 2 polynomials of degree 6, any further factoring will only be approximations, isn't it?

bobbym · 2011-07-18 23:05:18

Hi gAr;

The way I understand it that just means that those 6th degree polys are irreducible in the field of polynomials that the CAS is set to work in. In some other field they might be factorable.

gAr · 2011-07-18 23:21:51

Hi bobbym,

Okay.

bobbym · 2011-07-18 23:23:36

Hi gAr;

I made a small change to the above post. It is incorrect to think that the roots are transcendental. The roots are obviously algebraic and therefore not transcendental.

gAr · 2011-07-18 23:37:56

Okay.
Can we not just multiply 3 linear factors at a time which we found numerically?

Bob · 2011-07-19 03:20:55

hi bobbym and gAr

Ive computed the following using approximations, but Im pretty sure they can be established analytically.

If the first four poles are

then the next four are

and the final four are

Not quite what I was hoping for after 11 months of trying, but I had to post something.

Bob

gAr · 2011-07-19 04:03:07

Hi Bob,

Okay.

And congratulations!

bobbym · 2011-07-19 04:59:50

Hi Bob;

Congratulations on your 1000!

Hi gAr;

Can we not just multiply 3 linear factors at a time which we found numerically?

Yes, we could do that. But we would end up with cofficients that were approximations.

gAr · 2011-07-19 05:11:43

Hi bobbym,

Any cubic will be one among those, isn't it?
If there is any neat cubic factors, we can know from the approximate ones. That's what I thought.

bobbym · 2011-07-19 05:17:29

Hi gAr;

None that I can find so far. I have used PSLQ and the ISC. Neither can give a closed form for any number we have produced so far.

gAr · 2011-07-19 05:31:16

Okay.
I think no more luck with the factorization.

If there's anything simple, it can only be with properties of definite integrals or trigonometric substitution, I guess.

bobbym · 2011-07-19 20:41:49

Hi all;

Bumping this thread to let everyone know I found one way to do this. Here it is:

http://www.mat.uniroma2.it/~tauraso/AMM/AMM11148.pdf

Bob · 2011-07-19 21:16:05

hi bobbym,

Oh brilliant! I can move on now!

Bob

Au101 · 2011-07-21 07:43:41

bobbym wrote:

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

Hi guys, I'm sorry to revive this thread - which I'm sure you would like to put to bed - but I was following this thread out of interest. The link which bobbym provided is fascinating - if probably far beyond me, but something which I think is within my grasp - and yet which, at the moment, I can't follow - is this. I wonder if anybody could be so kind as to explain why this is the case, I can see that:

But I was unsure about the two numerators. I think I recall partial fractions being mentioned in this thread and this does remind me of a partial fraction - albeit an incredibly complicated one - but I've only ever had integers as numerators, and when I tried my method, I was at a loss as to how to reproduce this result. Of course, I wouldn't have been able to factorise:

Either - I merely verified that it was the case on a piece of paper. I think what I really want to do is to understand why this holds, even if I couldn't derive it myself, I think that that might be a bit beyond me. So what I'm asking is why are the two numerators what they are?

Thanks

bobbym · 2011-07-21 08:07:01

Hi Au101;

A computer got that factorization because that is as far as you can go in the field of integers, 2 sixth order polynomials in the denominator.

If you read the first few posts I did warn the OP that his belief that because Mathematica did it easily that so should we. This is false, as a matter of fact it is the exact opposite.

I do not know how any part of this was done as no hand methods were used here

http://www.mathisfunforum.com/viewtopic.php?id=15996

The PDF I provided also looks suspiciously like large portions of his proof were done by some Computer Algebra System.

Au101 · 2011-07-21 08:14:17

Wow, I see, okay thanks bobbym, I was just interested mainly in a quick look at the proof and I agree that I don't think there's much that a human could do with a pen and paper, but since you seemed to have got that factorisation yourself I was wondering if I could verify it for myself and quickly failed .

Thanks

bobbym · 2011-07-21 08:15:58

Hi Au101;

Did you check it by doing the actual multiplications.

Au101 · 2011-07-21 08:44:55

Oh, well, what I did was - mainly for my own satisfaction - to expand

Which of course gives a massive expansion, if it's of interest:

Which, when you add it all together gives:

Which, of course, is:

So I had satisfied myself with the denominators and I tried using my standard partial fractions method and said

But then I got:

And gave up, because I knew that I must have been on the wrong track. And, well, that's when I asked you lol

Last edited by Au101 (2011-07-21 08:50:17)

bobbym · 2011-07-21 08:50:55

Hi;

The way I would do it would be by solving a set of simultaneous linear equations.

You would need two things:

1) To know the form for A and B beforehand.
2) The ability to solve a 6 x 6 set of linear eqautions.

Au101 · 2011-07-21 08:56:54

Hmmm, I can definitely do 2) in theory - although that's not to say that I might not have some trouble with doing a large, difficult, practical example - although I should think I would probably enjoy trying and, well, as for 1) I not only know the form, I know what they are, since I wish only to verify it, as much as I would like to be able to derive something like this - not being a computer - or at least an extremely experienced and intelligent professor of advanced mathematics - I think that to do so may well be somewhere beyond me.

Where I am stuck is with regards to where to go with these 2 pieces of knowledge.

Math Is Fun Forum

#51 2011-07-18 01:47:20

Re: Tricky integral of a rational function

#52 2011-07-18 02:04:02

Re: Tricky integral of a rational function

#53 2011-07-18 13:22:06

Re: Tricky integral of a rational function

#54 2011-07-18 20:19:42

Re: Tricky integral of a rational function

#55 2011-07-18 22:50:16

Re: Tricky integral of a rational function

#56 2011-07-18 22:58:05

Re: Tricky integral of a rational function

#57 2011-07-18 23:05:18

Re: Tricky integral of a rational function

#58 2011-07-18 23:21:51

Re: Tricky integral of a rational function

#59 2011-07-18 23:23:36

Re: Tricky integral of a rational function

#60 2011-07-18 23:37:56

Re: Tricky integral of a rational function

#61 2011-07-19 03:20:55

Re: Tricky integral of a rational function

#62 2011-07-19 04:03:07

Re: Tricky integral of a rational function

#63 2011-07-19 04:59:50

Re: Tricky integral of a rational function

#64 2011-07-19 05:11:43

Re: Tricky integral of a rational function

#65 2011-07-19 05:17:29

Re: Tricky integral of a rational function

#66 2011-07-19 05:31:16

Re: Tricky integral of a rational function

#67 2011-07-19 20:41:49

Re: Tricky integral of a rational function

#68 2011-07-19 21:16:05

Re: Tricky integral of a rational function

#69 2011-07-21 07:43:41

Re: Tricky integral of a rational function

#70 2011-07-21 08:07:01

Re: Tricky integral of a rational function

#71 2011-07-21 08:14:17

Re: Tricky integral of a rational function

#72 2011-07-21 08:15:58

Re: Tricky integral of a rational function

#73 2011-07-21 08:44:55

Re: Tricky integral of a rational function

#74 2011-07-21 08:50:55

Re: Tricky integral of a rational function

#75 2011-07-21 08:56:54

Re: Tricky integral of a rational function

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