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## #76 2013-03-31 11:38:27

bobbym

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### Re: Contour integration

Nope, that is how he solved them in the book. Do you think I invented this? Also, I check the correct answer and correct choice of poles by trying the others in combination. They do not work.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #77 2013-03-31 11:41:07

anonimnystefy
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### Re: Contour integration

Can you post how he exactly solved it?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #78 2013-03-31 11:44:04

bobbym

Online

### Re: Contour integration

I have! The only thing I am lacking is a coherent idea on which poles to take. That I did not write down because at the time I understood it better and it was self evident to me.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #79 2013-03-31 11:50:57

anonimnystefy
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### Re: Contour integration

Word to word?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #80 2013-03-31 11:53:26

bobbym

Online

### Re: Contour integration

Yes, as I have written them down. Me and zetafunc did a few also. He came up with essentially the same method. The only problem is the ambiguity in explaining the choice of pole. For the contour I believe they always use the unit circle in these examples.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #81 2013-03-31 12:17:39

anonimnystefy
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### Re: Contour integration

No, actually, the contour is chosen and the poles inside the contour are the ones you calculate the residue of.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #82 2013-03-31 12:21:19

bobbym

Online

### Re: Contour integration

Okay, choose some arbitrary contour and try to get the answer. This method gets the answer to every one of these problems.

For the pole I choose the red point. Answer will be π / 3.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #83 2013-03-31 12:24:39

anonimnystefy
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### Re: Contour integration

I did not say that the choice of contour is arbitrary. I said that the contours are chosen, not residues. In fact, sometimes, you may need to calculate the residue of more than one pole.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #84 2013-03-31 12:33:39

bobbym

Online

### Re: Contour integration

Yes, we just did a problem where 2 poles were inside the unit circle. Sometimes you do all the poles. But as you can see from the drawing a circle is the simplest shape to go through the 3 points. It happens to be a unit circle.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #85 2013-03-31 12:38:33

anonimnystefy
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### Re: Contour integration

You seem to be arbitrarily choosing which poles you are taking and which not.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #86 2013-03-31 12:46:08

bobbym

Online

### Re: Contour integration

Nope, I have a rule. I just do not have any solid justification for it.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #87 2013-03-31 12:54:09

anonimnystefy
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### Re: Contour integration

What rule is that?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #88 2013-03-31 12:56:14

bobbym

Online

### Re: Contour integration

Because we have intervals of 0 to infinity the bottom pole is out. Then I take the poles that have a positive imaginary part. That leaves the one pole I chose.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #89 2013-03-31 13:05:08

anonimnystefy
Real Member

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### Re: Contour integration

Hm, okay. Then, how would you integrate log(x)/((1+x^2)^2) from 0 to infinity?

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #90 2013-03-31 13:09:29

bobbym

Online

### Re: Contour integration

We have not finished the other one. If we jump around we will get confused. We should calculate the residues for the other problem before moving on to the next one.

But you have not been listening to me, as far as I know this method only works on rational functions. At least that is what some site I was looking at said. He defines a rational function as p(x) / q(x) where both p(x) and q(x) are polynomials and of course q(x) ≠ 0.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #91 2013-03-31 13:11:46

anonimnystefy
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### Re: Contour integration

That site is not correct. All sources I have found have it working on other types if functions as well.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #92 2013-03-31 13:16:40

bobbym

Online

### Re: Contour integration

I was going to say that but zetafunc shows a rather simple one where this method falls flat on its face. There must be some conditions when can use it and when you can not.

I will look at the integral.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #93 2013-03-31 13:19:45

anonimnystefy
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### Re: Contour integration

#### bobbym wrote:

I was going to say that but zetafunc shows a rather simple one where this method falls flat on its face. There must be some conditions when can use it and when you can not.

I will look at the integral.

Actually, that was the point of the integral I gave you. As you will see, the method you will try on it will probably fail, but the Wiki article has no problem doing it.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #94 2013-03-31 13:29:43

bobbym

Online

### Re: Contour integration

The residue method does not work on all cases. There are conditions.

The method I am using there will not work but what method works all the time.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #95 2013-03-31 13:31:03

anonimnystefy
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### Re: Contour integration

Yes, but the contour integration methid works on that one.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #96 2013-03-31 13:36:20

bobbym

Online

### Re: Contour integration

It will fail on others. Either because of violations or impossibility to do.

This method is a not a general method for all integrals.It works on the examples provided.

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #97 2013-03-31 13:38:26

anonimnystefy
Real Member

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### Re: Contour integration

This was just to show that the method you have is not contour integration. But I do not know if it works.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #98 2013-03-31 13:39:10

bobbym

Online

### Re: Contour integration

Why do you think it is not contour integration?

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

## #99 2013-03-31 13:45:40

anonimnystefy
Real Member

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### Re: Contour integration

Because I did understand some parts of what I redlad earlier, and some of those parts were not included in your method.

By the way, now when I look at this stuff again, I think only the residue part confused me, so I do thank you for making that part clearer. It seems to be a common ground between your method and the contour integration method. I will have to read about it once more to make sure I understand it.

The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment

## #100 2013-03-31 13:48:19

bobbym

Online

### Re: Contour integration

Hi;

If you look here you will see the formula I am using or misusing and even that the contour is circular and encloses the poles.

http://en.wikipedia.org/wiki/Residue_theorem

In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.