Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -




Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno

Go back

Topic review (newest first)

2013-04-01 09:20:26

From a little pdf I downloaded.

2013-04-01 09:19:07

Where'd you get that idea?

2013-04-01 09:08:29

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

2013-04-01 02:38:42

Well, contour integration works on that one.

2013-03-31 19:29:53

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.

2013-03-31 17:38:37

It nowhere says that it is for those forms only.

2013-03-31 16:26:07

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

2013-03-31 14:42:11

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

2013-03-31 14:04:05

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.

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).

2013-03-31 14:01:18

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.

2013-03-31 13:53:03

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.

2013-03-31 13:49:55

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

2013-03-31 13:48:19


If you look here you will see the formula I am using or misusing and even that the contour is circular and encloses the poles.

2013-03-31 13:45:40

Because I did understand some parts of what I redlad earlier, and some of those parts were not included in your method.

By the way, now when I look at this stuff again, I think only the residue part confused me, so I do thank you for making that part clearer. It seems to be a common ground between your method and the contour integration method. I will have to read about it once more to make sure I understand it.

2013-03-31 13:39:10

Why do you think it is not contour integration?

Board footer

Powered by FluxBB