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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

Can somebody explain contour integration to me a little bit?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

An example would be nice.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.**

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,416

Stefy, check this out:

http://walet.phy.umist.ac.uk/MaMe/MMA/Contour.pdf

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

bobbym wrote:

Please integrate that.

How do I do that?

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

Okay, that is the easy part.

*Last edited by anonimnystefy (2013-03-27 03:45:16)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Take the positive 2, do you know why?

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

No...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.

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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,126

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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

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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,126

Many thanks.

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

Bob

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

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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.

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

Hi bobbym

Yes, I am getting those residues. What now?

Here lies the reader who will never open this book. He is forever dead.

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**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

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

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

pi*sqrt(2)/4?

*Last edited by anonimnystefy (2013-03-27 23:40:39)*

Here lies the reader who will never open this book. He is forever dead.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,771

Correct! Wunderbar!

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.

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,849

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

Here lies the reader who will never open this book. He is forever dead.

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