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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,489

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 15,657

You edited it after I posted.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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 agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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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...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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I am not following you? What integral?

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**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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The contour integral. As far as I know tthe original integral and the contour integral are not the same.

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**bobbym****Administrator**- From: Bumpkinland
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I am sorry, I am still not following. The original integral?

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
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bobbym wrote:

Please integrate that.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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Did we not get the right answer?

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**anonimnystefy****Real Member**- From: The Foundation
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But, where did we use the contour integration?

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**bobbym****Administrator**- From: Bumpkinland
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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.

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**anonimnystefy****Real Member**- From: The Foundation
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Ok. And why did we take those two poles and not the other two?

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**bobbym****Administrator**- From: Bumpkinland
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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.

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**anonimnystefy****Real Member**- From: The Foundation
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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
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That I do not know. But this I do know: One thing at a time!

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

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

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**bobbym****Administrator**- From: Bumpkinland
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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

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**anonimnystefy****Real Member**- From: The Foundation
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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
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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.

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**anonimnystefy****Real Member**- From: The Foundation
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Not exactly. I will try posting later what I mean, when I get on my laptop.

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**bobbym****Administrator**- From: Bumpkinland
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Okay, but chances are it will be one of those methods mathematicians love but does not do anything.

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**anonimnystefy****Real Member**- From: The Foundation
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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)*

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**bobbym****Administrator**- From: Bumpkinland
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Those integrals look difficult the residue way does not require you to do the integral.

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**anonimnystefy****Real Member**- From: The Foundation
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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)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
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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.

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