Contour integration

anonimnystefy · 2013-03-30 14:49:55

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

bobbym · 2013-03-30 14:53:03

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.

anonimnystefy · 2013-03-30 15:01:18

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.

bobbym · 2013-03-30 15:04:05

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).

anonimnystefy · 2013-03-30 15:42:11

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

bobbym · 2013-03-30 17:26:07

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

anonimnystefy · 2013-03-30 18:38:37

It nowhere says that it is for those forms only.

bobbym · 2013-03-30 20:29:53

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.

anonimnystefy · 2013-03-31 03:38:42

Well, contour integration works on that one.

bobbym · 2013-03-31 10:08:29

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

anonimnystefy · 2013-03-31 10:19:07

Where'd you get that idea?

bobbym · 2013-03-31 10:20:26

From a little pdf I downloaded.

Math Is Fun Forum

#101 2013-03-30 14:49:55

Re: Contour integration

#102 2013-03-30 14:53:03

Re: Contour integration

#103 2013-03-30 15:01:18

Re: Contour integration

#104 2013-03-30 15:04:05

Re: Contour integration

#105 2013-03-30 15:42:11

Re: Contour integration

#106 2013-03-30 17:26:07

Re: Contour integration

#107 2013-03-30 18:38:37

Re: Contour integration

#108 2013-03-30 20:29:53

Re: Contour integration

#109 2013-03-31 03:38:42

Re: Contour integration

#110 2013-03-31 10:08:29

Re: Contour integration

#111 2013-03-31 10:19:07

Re: Contour integration

#112 2013-03-31 10:20:26

Re: Contour integration

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