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## #101 2013-03-31 13:49:55

anonimnystefy
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### Re: Contour integration

I think the problem is that you are treating a regular integral as a contour integral.

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

## #102 2013-03-31 13:53:03

bobbym

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### Re: Contour integration

In order to evaluate real integrals, the residue theorem is used in the following manner: the integrand is extended to the complex plane and its residues are computed (which is usually easy), and a part of the real axis is extended to a closed curve by attaching a half-circle in the upper or lower half-plane, forming a semicircle. The integral over this curve can then be computed using the residue theorem. Often, the half-circle part of the integral will tend towards zero as the radius of the half-circle grows, leaving only the real-axis part of the integral, the one we were originally interested in.

It is not a problem, looks like you convert it into a contour integral or something like that. Anyway the method can be used for real integrals.

Do you think we have done 5 of these and got the right answer by accident? Sure we are lacking in rigor in the approach but it does work for real integrals of that type.

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #103 2013-03-31 14:01:18

anonimnystefy
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### Re: Contour integration

Do you see the part after "Often". That's where your problem is. It will not always tend to zero. Also, there are some things called branch points, which I am trying to figure out, which cannot be handled regularly, So a different path must be chosen.

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

## #104 2013-03-31 14:04:05

bobbym

Online

### Re: Contour integration

You are missing the point. This will obviously not do every integral. No method does. But often is good enough. The whole integral is reduced to a line on the complex plane. That page I sent you uses the same method  we are using to do an integral. We are lacking the knowledge of when this can applied.

I think you will find that it depends on the principal part of the Laurent series of the integrand.

http://mathworld.wolfram.com/ContourIntegration.html

Look at equation 12. I remember saying for rational functions only. You will also see it is a definite integral. The method can be extended to some other forms, see (14),(15) and (16).

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #105 2013-03-31 14:42:11

anonimnystefy
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### Re: Contour integration

It seems to me the only condition is that the function is holomorphic.

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

## #106 2013-03-31 16:26:07

bobbym

Online

### Re: Contour integration

(14)(15) and (16) show that it is only for those forms.

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #107 2013-03-31 17:38:37

anonimnystefy
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### Re: Contour integration

It nowhere says that it is for those forms only.

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

## #108 2013-03-31 19:29:53

bobbym

Online

### Re: Contour integration

it only mentions 3 forms, (14)(15) and (16) have the exact method I am using. Your integral is not of that form, so other methods have to be used.

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #109 2013-04-01 02:38:42

anonimnystefy
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### Re: Contour integration

Well, contour integration 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

## #110 2013-04-01 09:08:29

bobbym

Online

### Re: Contour integration

Yes, it does but not using residues. There are a couple of ways to do a contour integration.

In mathematics, you don't understand things. You just get used to them.
Some cause happiness wherever they go; others, whenever they go.
If you can not overcome with talent...overcome with effort.

## #111 2013-04-01 09:19:07

anonimnystefy
Real Member

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### Re: Contour integration

Where'd you get that idea?

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

bobbym

Online