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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

Ah ha, you did not see post #24...

**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: 15,392

You edited it after I posted.

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: 84,598

Ditto! Are you clear on the contour integration. Want to do another?

**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: 15,392

Wait, that's it? Are you sure? When I looked at some videos, I found that they split the integral into several integrals along the contour path...

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: 84,598

I am not following you? What 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.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,392

The contour integral. As far as I know tthe original integral and the contour integral are not the same.

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: 84,598

I am sorry, I am still not following. The original integral?

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: 15,392

bobbym wrote:

Please integrate that.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

Did we not get the right answer?

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: 15,392

But, where did we use the contour integration?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

We considered the poles that were inside the contour, see the graph of the imaginary points.

Then we used the residues and then the formula. That is how you do a contour integration.

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: 15,392

Ok. And why did we take those two poles and not the other two?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

Good question. See rule one of my signature.

It seems that the unit circle represents the interval -∞ to ∞. The top half represents 0 to ∞, we took the poles inside there. This is the best I can understand of it.

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: 15,392

Ok, but, why did I see some examples done with a rectangle path instead of a semicircle?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

That I do not know. But this I do know: One thing at a time!

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: 6,311

As I understand it if the integral is from -∞ to + ∞ you take all the poles.

In your example you only take the ones in the right side of the Argand diagram because those are the ones enclosed by your contour. As I'm still learning this, what I just said may be rubbish.

As for the shape of the contour, I don't think it matters. It has to enclose the poles, then there's a fiddly bit where the integral simplifies to just the residues at the poles because the rest is zero.

If we get desparate for the reason, I could ask David.

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: 84,598

Hi;

Yes, if you are going from -∞ to + ∞ you would use all the poles.

The only two poles that work are the ones on top of the real axis ( looks like the x axis ) see the drawing. See post #15

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: 15,392

Hi

It seems that rhe contour integral can be split up into several normal ones along the contour uaing the parametrization of that contour. One of those integrals will be the one we started with, but with limits from -R to R so then we should take the limit as R goes to infinity. Meanwhile, the other seems to go to zero for a reason unknown to me.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

Isn't that just the standard way of evaluating a limit with an infinite endpoint? It is more like a definition than a means to computation.

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: 15,392

Not exactly. I will try posting later what I mean, when I get on my laptop.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,598

Okay, but chances are it will be one of those methods mathematicians love but does not do anything.

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: 15,392

Here is what he did:

And, when we let R go to infinity, the second integral becomes zero and you get the value of

*Last edited by anonimnystefy (2013-03-29 00:28:35)*

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: 84,598

Those integrals look difficult the residue way does not require you to do the integral.

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: 15,392

But, are sure it always gives a correct solution?

And, how'd you get the idea that he is calculating the integral the hard way?

*Last edited by anonimnystefy (2013-03-29 13:42:08)*

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: 84,598

The residue method? I know it does for rational functions. And how come you are not sleeping at this late hour?

After you get your rest we will do more of 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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