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bowisP
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 Integral over contour   « on: Nov 27th, 2005, 7:37am » Quote Modify Remove

I need help with evaluating this integral

int: cos[z]/[z^2 - 4]  dz, along Coutour, where z is complex.

The contour is

|
/|-------\
/  |     \
\ |  \
-------|-------\--------|---\------->
-2  0 \   2    /
|\      /
| \ ____/
|

I don't think the integral exists away.
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Michael Dagg
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 Re: Integral over contour   « Reply #1 on: Nov 27th, 2005, 9:25am » Quote Modify

While that that contour is really not clear is it such that -2 is outside and 2 is inside the region or are both within the region?
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Michael Dagg
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 Re: Integral over contour   « Reply #2 on: Nov 27th, 2005, 11:21am » Quote Modify

The drawing is a bit better if you click on 'quote' (since you see the source)
The -2 lies outside and the 2 inside, as you suspected.
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Icarus
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 Re: Integral over contour   « Reply #3 on: Nov 27th, 2005, 11:21am » Quote Modify

Using my special moderator powers to look at the un-YaBB processed post, it appears that the contour includes 2 but not -2. (Actually, anyone can see it by quoting the post.) I think he intended a circle of radius 2 about z=2, but as Michael well knows, that doesn't matter. As long as the contour is simple and closed, and 2 is inside, while -2 is outside, the answer will be the same.

And yes, the integral does exist. To find it, use the Cauchy Integral Formula
 « Last Edit: Nov 27th, 2005, 11:27am by Icarus » IP Logged

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bowisP
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 Re: Integral over contour   « Reply #4 on: Nov 27th, 2005, 3:17pm » Quote Modify Remove

I tried to fix the graph but my daiilup has been so poor at times that I gave up. When I push post nothing happens.

Besides the point, the contour is not a ciricle but more like a a trapezoid only containing 2 and not -2. The integrand has two poles +/-2.

I know Cauchy's theorm but f(z) = cos[z] is analytic on the region like required by the theorem, you can't get it into the form need for integration. This is why I said I do not think it exists. Still don't.
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Michael Dagg
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 Re: Integral over contour   « Reply #5 on: Nov 27th, 2005, 4:08pm » Quote Modify

Icarus is correct that the integral exists.

The f(z) = cos(z) that you are attempting to relate to CIF is not the f(z) required and neither is the dominator thereof. This may be the reason you can't get it into the form needed for integration.

With g(z) = cos(z)/(z^2 - 4), you need h(z) = f(z)/(z - z0) to apply CIF which can be constructed from g. This should immediately convince you that f(z) is something other than cos(z). Note also that it is required that f(z) be analytic inside and on the boundary of the region - keep this in mind when you construct h(z), notably the numerator.
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bowisP
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 Re: Integral over contour   « Reply #6 on: Nov 27th, 2005, 5:38pm » Quote Modify Remove

Mucho gracias.

I couldn't be more embarrassed with what I was thinking and I wasn't thinking right about this problem. How does one think *right* about a problem anywho.

Making h(x) = f(z)(z - 2)^-1, with f(z) = cos[z](z + 2) ^-1 provides analytic f(z) * required *.

The solution is f(2) =  2pi i cos[2]/4

I am still confused about the issue with the poles for this problem or better yet problems like them that have no poles. Can you give me some guidelines?

Thx.
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Michael Dagg
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 Re: Integral over contour   « Reply #7 on: Nov 27th, 2005, 8:07pm » Quote Modify

I realize that it is likey some oversight on your part but if the function f on the left-hand side of your solution equality is the same f discussed previously, you are asserting that 2pi = 1/i.
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bowisP
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 Re: Integral over contour   « Reply #8 on: Nov 28th, 2005, 7:40am » Quote Modify Remove

Yeah, I thought to say I[2] = 2pi i cos[2]/4. I also made a typo with h(x): x.

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Michael Dagg
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 Re: Integral over contour   « Reply #9 on: Nov 28th, 2005, 2:01pm » Quote Modify

Your solution is correct for positive orientation.

I will let someone else take your questions about the poles and guidelines.

 « Last Edit: Nov 28th, 2005, 2:04pm by Michael Dagg » IP Logged

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Michael Dagg
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 Re: Integral over contour   « Reply #10 on: Nov 28th, 2005, 9:18pm » Quote Modify

I'm not sure quite what your issue with the poles is.

If a pole is inside the contour, then it will contribute to the integral. If the pole is outside the contour, then its contribution to the function is analytic everywhere inside the contour, and as a result, and as a result, its contribution to the integral is zero.

The shape of the contour does not matter, as long the function is analytic on the curve itself (the curve cannot run through a singularity of the function), and the enclosed singularities are the same.

So suppose you have a function f(z) which is analytic in a region except for 2 points A and B that are simple poles. This means there is another function g(z) = (z-A)(z-B)f(z) that is analytic everywhere in the region.

Consider the three simple closed contours C1, C2, and C3:

` --------------------   |    C1  C2   |   |  -----    -----  |   |  | A |    | B |  |   |  | . |    | . |  |   |  |   |    |   |  |   |  -----    -----  |   |   |   --------------------       C3    `

//It seems that the old "modify and resave" technique for getting around YaBB's idiotic processing of preformatted text no longer works. The upgraded YaBB processes your text no matter what you do. Either that, or I've forgotten the proper technique.// C3 is intended to be a larger contour that completely encircles the other two.

Assume a positive orientation for all three.

By Cauchy's Theorem, the shape of the contour does not matter, as long as the same singular points are enclosed.
C1 encloses A, but not B. g(z)/(z-B) is analytic inside C1, so the integral of f(z) is given by

[int] f(z)dz = [int] (g(z)/(z-B)) / (z-A) dz = 2[pi]i (g(A)/(A-B))

Similarly, for contour C2,

[int] f(z)dz = [int] (g(z)/(z-A)) / (z-B) dz = 2[pi]i (g(B)/(B-A)).

For contour C3, both points are enclosed, so we pick up the contributions of both:

[int] f(x)dz = 2[pi]i ( g(A)/(A-B) + g(B)/(B-A) ).

It was because of this that Michael_Dagg was unsure what to say originally. From the fractured diagram, it was not clear if both poles were inside the contour, or just the one. If the pole at -2 had been within the contour, then you would have needed to add a contribution of 2[pi]i cos(-2)/-4 = -2[pi]i cos(2)/4, cancelling the pole at 2, and leaving a net result of 0 for the integral.

So he needed to know which poles were included before he could lead you to the correct answer.
 « Last Edit: Nov 28th, 2005, 9:25pm by Icarus » IP Logged

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bowisP
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 Re: Integral over contour   « Reply #11 on: Nov 30th, 2005, 6:31am » Quote Modify Remove

Thank you for the very good explanation. Funny thing is that my book Basic Complex Analysis, by J. Marsden, and in my class both, nowhere have I heard that if you transverse the contour more than once, say n times, the result would be n times the integral. Don't seem to be any theorems about this either.
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 Re: Integral over contour   « Reply #12 on: Nov 30th, 2005, 10:28pm » Quote Modify

Your book hasn't covered the more general forms of the Cauchy theorem yet, then. Marsden? That one has to be as old as Ahlfors, or close to it.

Its simple to prove: if g1 and g2 are two contours with the end point of g1 coincident with the beginning point of g2, the contour formed by first traversing g1 then g2 is commonly denoted by g1 + g2.

[ My apologies for those for whom the "symbol" font does not work. In the material below, the ò symbol would display as an integral sign if the symbol font worked with your system. ]

If f is any function (analytic or not) integrable along both g1 and g2, then f is also integrable along g1 + g2, and

òg1+g2 f = òg1 f + òg2 f.

This result follows immediately from the definition of integration along a contour, and the fact that for real integrals,
òac f(t)dt = òab f(t)dt + òbc f(t)dt.

Now consider a loop g. And suppose that instead of traversing it once, the integral contour traverses it n times (this contour is denoted by ng). But it should be clear that ng = g + g + g + ... + g. So

òng f = òg f + òg f + ... + òg f = nòg f.

This is true for any loop g and any f integrable along g.
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Michael Dagg
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 Re: Integral over contour   « Reply #13 on: Dec 1st, 2005, 7:17am » Quote Modify

I strongly doubt that the Marsden book does not contain that result. It's there but might be an exercise (wrote using summation notation - easier to miss expecially if it is an exercise).
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Michael Dagg
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 Re: Integral over contour   « Reply #14 on: Dec 1st, 2005, 7:57am » Quote Modify

Incidentally, over in the the putnam thread where I proposed using complex analysis to show that sqrt(pi) = int[exp(-x^2) dx, - oo, oo] utilizing contour integration can be accomplished by making use of the result that Icarus proved above in conjunction with some work, of course. One approach in doing so may require that you show that the Fresnel integrals exist. But, since we know they do, simply assume it.
 « Last Edit: Dec 1st, 2005, 8:46am by Michael Dagg » IP Logged

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 Re: Integral over contour   « Reply #15 on: Dec 1st, 2005, 7:31pm » Quote Modify

I went through that one when I was in college, though I admit I don't remember the details now. As I recall, you set up a half-circle with diameter along the real axis, then let the radius increase without bound. The integral on the half-circle itself goes to zero as the radius increases, leaving the entire contour integral equal to its value along the real axis. Of course, I must be forgetting something here, because e-x^2 is unbounded along the imaginary axis.
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 Re: Integral over contour   « Reply #16 on: Dec 2nd, 2005, 4:25am » Quote Modify

It's fairly straightforward to show
int(exp(-x2) dx, x=0..oo) = epi i/4 int(exp(-ix2) dx, x=0..oo), or
int(cos(x2) - i sin(x2) dx) = (1-i)/sqrt(2) int(exp(-x2) dx),
demonstrating the equivalence of computing the Fresnel integrals.

One only needs to show that the integral of exp(-z2) along the arc from R to Repi i/4,
int (exp(-(Reit)2)Rieit, t=0..pi/4 ),
goes to 0 as R -> oo.  The integrand has norm
Rexp(-R2cos(2t)).
For convenience, change variables u=pi/2 - 2t; we are to bound
int( R exp(-R2sin t), t=0..pi/2 ).
Picking 0<t'<pi/2 such that sin(t')=R-3/2, and breaking up the integral into (0,t') and (t',pi/2) seems to do the trick.

As for computing int(exp(-x2)dx,  x=-oo..oo) itself, I don't know if there's anything easier than squaring and using polar coordinates.  But it's equivalent to calculating Gamma(1/2), which you can do by performing a sequence of change of variables, starting with the double integral for Gamma(s)Gamma(1-s), and finally doing a contour integral to get  Pi/sin(Pi s).
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bowisP
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 Re: Integral over contour   « Reply #17 on: Dec 2nd, 2005, 5:45am » Quote Modify Remove

on Dec 1st, 2005, 7:17am, Michael_Dagg wrote:
 It's there but might be an exercise.

Sorry for my oversight once again, it is an exercise not using sums though. There is another one like it about integrating around k parts of a contour which makes complete sense.
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bowisP
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 Re: Integral over contour   « Reply #18 on: Dec 2nd, 2005, 6:18am » Quote Modify Remove

I just noticed and it is pleasing that Euler's identity
exp(ix) = cos(x) + i sin(x) although not like the Fresnel forms I wonder if it could be put in those forms.

Makes you think that exp((ix)^2) = exp(-x^2) = cos(x^2) + (i sin(x^2))^2
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bowisP
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 Re: Integral over contour   « Reply #19 on: Dec 2nd, 2005, 6:22am » Quote Modify Remove

Correct:

Makes you think that exp((ix)^2) = exp(-x^2) = cos(x^2) + i^2 sin(x^2)  = cos(x^2) - sin(x^2)

No?
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 Re: Integral over contour   « Reply #20 on: Dec 2nd, 2005, 3:45pm » Quote Modify

No!

e-x^2 = ei(ix^2) = cos(ix2) + i sin(ix2).
 « Last Edit: Dec 2nd, 2005, 3:49pm by Icarus » IP Logged

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 Re: Integral over contour   « Reply #21 on: Dec 2nd, 2005, 7:01pm » Quote Modify

There is something that looks similar:

exp(-x^2) = cosh(x^2) - sinh(x^2).

Unfortunately, it won't do you much good in trying to find int[exp(-x^2),-00,00)] since you will end up with exp(-x^2) once again.
 « Last Edit: Dec 2nd, 2005, 11:02pm by Michael Dagg » IP Logged

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Michael Dagg
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 Re: Integral over contour   « Reply #22 on: Dec 4th, 2005, 7:07am » Quote Modify

on Dec 2nd, 2005, 5:45am, bowisP wrote:
 ...is an exercise not using sums though.

It would be unthinkable for that result not to be in Marsden in some form. And as you can see from what Icarus did, the proof is not difficult using what you already know. Both the Ahlfors and Marsden books are considered harder than most, particularly the former.

About books, let me mention a delightful book: Visual Complex Analysis,  Needham, Oxford Press. It has been very well accepted and if you reference it in conjunction with a stardard book on the subject, you will spend less time fishing around for geometrical interpretations.

Here is a free online CA book:

http://www.math.gatech.edu/~cain/winter99/complex.html

Perhaps this thread should be moved to the CA one.
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 Re: Integral over contour   « Reply #23 on: Dec 4th, 2005, 12:07pm » Quote Modify

True. I tend to forget that William finally made us moderators of all the forums. Orginally, we were just moderators of the riddle forums, and not the general ones.
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