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#1 2013-03-28 01:54:04

anonimnystefy
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Contour integration

Can somebody explain contour integration to me a little bit? smile


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

#2 2013-03-28 01:57:40

bobbym
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Re: Contour integration

Do you want to see it work or you want a theoretical discussion?


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.

#3 2013-03-28 02:09:02

anonimnystefy
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Re: Contour integration

An example would be nice.


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

#4 2013-03-28 02:19:17

bobbym
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Re: Contour integration



Please integrate that.


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.

#5 2013-03-28 02:21:54

ShivamS
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Re: Contour integration

Stefy, check this out:
http://walet.phy.umist.ac.uk/MaMe/MMA/Contour.pdf


I have discovered a truly marvellous signature, which this margin is too narrow to contain. -Fermat
Give me a lever long enough and a fulcrum on which to place it, and I shall move the world. -Archimedes
Young man, in mathematics you don't understand things. You just get used to them. - Neumann

#6 2013-03-28 02:24:54

anonimnystefy
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Re: Contour integration

bobbym wrote:



Please integrate that.

How do I do 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

#7 2013-03-28 02:26:21

bobbym
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Re: Contour integration

Get the poles first.


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.

#8 2013-03-28 02:44:33

anonimnystefy
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Re: Contour integration

Okay, that is the easy part.

Last edited by anonimnystefy (2013-03-28 02:45:16)


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

#9 2013-03-28 02:46:47

bobbym
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Re: Contour integration

Take the positive 2, do you know why?


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.

#10 2013-03-28 02:56:22

anonimnystefy
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Re: Contour integration

No...


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

#11 2013-03-28 02:57:39

bobbym
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Re: Contour integration

The limits of integration are positive so we only take the positive poles.


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.

#12 2013-03-28 03:05:38

anonimnystefy
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Re: Contour integration

Since when can complex numbers be positive and negative? Do you want their real parts to be positive or...?


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

#13 2013-03-28 03:11:47

bobbym
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Re: Contour integration

I get the poles like this



You take the ones that do not have a minus sign in front. See the drawing provided later to tell which poles to use.


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.

#14 2013-03-28 12:19:37

anonimnystefy
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Re: Contour integration

I think those are the 1. and 3. in my list.


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

#15 2013-03-28 21:40:32

bobbym
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Re: Contour integration

Now get the residue of those two poles. This is done in a couple of ways. I prefer the formula. If you like your own way use it.

If you chose the right ones (the ones in red ) you will get:



Once you have the residues you are almost done. Tell me when you get mine.


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View Image: 2013-03-28_033058.gif      


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.

#16 2013-03-28 21:50:43

bob bundy
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Re: Contour integration

hi bobbym,

Would you mind explaining how to get the 'residues' ? Thanks.

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#17 2013-03-28 21:57:19

bobbym
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Re: Contour integration

Hi;

One way uses the Laurent series. I do not know it offhand. The other way zetafunc and I were using just a couple of days ago. It is just a formula. Hold on while I get it.

As usual I did not write it down but Wiki has it:



where c is a pole and n is its order.

Ex:



has 2 poles i and -i. To get the residue of i we say n =1 and c = i.

The whole formula simplifies to



which equals - i / 2


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.

#18 2013-03-28 22:16:04

bob bundy
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Re: Contour integration

Many thanks. 

Give me a week or so to get my brain around all of this and I may have more questions.  smile

Bob


You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

#19 2013-03-28 22:17:33

bobbym
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Re: Contour integration

Hi;

I added some stuff to post #17.

Give me a week or so to get my brain around all of this

If you can get it in a week then you will have far surpassed me. I never did get it, despite having it explained to me at least 5 times!

Rule 1 of my signature applies! To some, it only applies maybe 3 or 4 times. For me I stopped counting after 10000.


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.

#20 2013-03-28 22:33:21

anonimnystefy
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Re: Contour integration

Hi bobbym

Yes, I am getting those residues. What now?


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

#21 2013-03-28 22:35:16

zetafunc.
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Re: Contour integration

Sum the residues, and multiply the sum by 2iπ.

#22 2013-03-28 22:39:13

bobbym
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Re: Contour integration

Hi zeta;

I was going to ask you to come in here and help out. We were just working on this!

Hi anonimnystefy;

Normally 2 π i but here we are only taking half the contour so it is π i


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.

#23 2013-03-28 22:40:08

anonimnystefy
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Re: Contour integration

pi*sqrt(2)/4?

Last edited by anonimnystefy (2013-03-28 22:40:39)


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

#24 2013-03-28 22:41:26

bobbym
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Re: Contour integration

Correct! Wunderbar!


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.

#25 2013-03-28 22:43:40

anonimnystefy
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Re: Contour integration

Yes, I have. Did you see post #23?


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

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