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**zetafunc.****Guest**

That's great, how did you do it? What contour did you use?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hi;

First have you seen this page?

http://140.177.205.23/ContourIntegration.html

For these type they always use a half circle.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I tried that first but got the wrong answer (2π/3). The function is defined only for x ≥ 0 and f(x) ≥ 0 (i.e. the first quadrant)...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

That is because you picked the right pole but forgot that you are using only half a circle. You would get half that answer.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

I'm not sure I follow that, though... the reason you can normally halve your answer is because even functions have this property (if f(x) is even):

that's the reason I halved my answer, not because it is half a circle. In this example, wouldn't you use a quarter of a circle too? f(x) in this instance is neither odd nor even.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

It is not where the function is defined but the limits of integration. When you go from -∞ to ∞ you multiply the sum of the residues that apply by 2 π i. Here we are going from 0 to ∞, so it is π i.

I hope you understand that when I say something like that I am just using you for a sounding board. I am learning this topic too. I might be wrong with my explanations of things but not with the answers.

I am watching a nice pair of videos that cover this in a basic way. I do not know if they will answer all our questions but would you like to watch them too?

I watched the videos, they are at

http://www.youtube.com/watch?v=xIAyiP4LrDE

http://www.youtube.com/watch?v=LEzoKuwi0s4

See you in a bit, need some rest.

**In mathematics, you don't understand things. You just get used to them.Of course that result can be rigorously obtained, but who cares?Combinatorics is Algebra and Algebra is Combinatorics.**

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**zetafunc.****Guest**

Thanks for the videos -- I have watched them before although I do think it is missing something pretty important. The ML inequality to bound the integral over a path is essential I think, because they do not always tend to zero.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Using the drawing I chose the path (-1,0) to (1,0) back to (-1,0). That passes through the only pole, that is why I chose 1 / 2 the circle.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

If you mean the right half of the circle, I do not think that is correct. Why are you only integrating from 0 to 1 -- how does it relate to the original integral? z = 0 is also a branch point, yet your contour passes through it (if you used half a circle). This is not allowed.

I think I have a solution that uses a keyhole contour. I will post it tomorrow -- on my iPod at the moment.

**zetafunc.****Guest**

Okay, I am really tired but I just couldn't sleep without writing the solution in my notebook. I can't believe it worked (sort of). I ended up getting -pi/3 (even though it should be positive)... I'll sort that out tomorrow though. You need a keyhole contour in four parts, I'll explain tomorrow morning.

**zetafunc.****Guest**

Just re-did the residue calculations and now there is no negative in the final answer, which is great! I think I am getting the hang of it.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hi;

Okay, I will see you then.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

First, observe this diagram (remove spaces):

o i 4 2 . t i n y p i c . c o m / 1 4 j n m o g . j p g

Call this curve C, traversed along ABDEFGHJ. The smaller circle (the keyhole) is of radius r, the larger circle is of radius R. In other words:

We have a branch point at z = 0 (because of the square root) and the poles are at z = e[sup]iπ[/sup], z = e[sup]iπ/3[/sup] and z = e[sup]-iπ/3[/sup]. All of these lie in our contour. Summing the residues at these points and multiplying by 2iπ gives 2π/3.

So now we have

Note that:

Using the substitution z = Re[sup]iθ[/sup] in the second integral yields:

For the third integral, we replace z with ze[sup]2iπ[/sup], as we have gone through a rotation of 2π to get to this point:

For the fourth integral, we use the same substitution as we did for the second integral, except with z = re[sup]iθ[/sup]:

Putting it all together:

Now, letting r -> 0 and R -> ∞, the second and fourth integrals vanish, and we have:

which gives

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hi;

Okay, thanks. Any quick way on how you got that branch cut?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

When we plot y[sup]2[/sup] = x in the real plane, we plot two functions: y = √x and y = -√x, each single-valued.

If we consider w = z[sup]1/2[/sup], and let z = Re[sup]i(θ + 2kπ)[/sup], then w = Re[sup]i(θ + 2kπ)/2[/sup]. Letting k = 0 and k = 1 gives w[sub]1[/sub] = R[sup]1/2[/sup]e[sup]iθ/2[/sup] and w[sub]2[/sub] = -R[sup]1/2[/sup]e[sup]iθ/2[/sup], respectively. This corresponds to our branch cut.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Okay I need to work on this keyhole one.

I have a book problem that uses your method and another method. Would you like to see the other method?

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Sure -- I am guessing the other method uses the Cauchy integral formula?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Not exactly, I found this one.

Make the substitution x = x^2 and you get:

Taking advantage that the integrand is an even function:

The integral can be evaluated along the contour C along the real axis from -R to R , then circle back, counterclockwise, along a circle in the upper-half-plane back to -R.

The poles are simple:

You only need the ones that are positive on the imaginary axis:

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Thanks, that is a better way of doing it -- although it is nice to see both ways, as if there is an instance where you can't use a substitution it is helpful to practice on simpler problems. In fact, if you let u = x[sup]3[/sup] the problem reduces to a simple arctan integral (which can then be calculated via trig substitution or a simple semicircular contour).

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hi;

Yes, I thank you for making some of those concepts a little clearer to me.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

No problem -- have you worked through the problems in Spiegel's book? I am doing those now, the trig ones look interesting.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

No, I have not because I am working on two other problems that are not related to complex analysis but are a complete mystery to me. Also, I must look at your keyhole method and try to understand it. Lastly, I am reading a book on complex analysis. It is tough for me but I am hoping to pick up a bit here and there.

If you have some problem just post it and I will work on that with you.

But first I am going to get some sleep as soon as I answer this last guy's questions.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,219

Hope you are having no problems. See you later.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**

**zetafunc.****Guest**

I looked at my solution again and the sum of the residues is actually i/3, so my contour actually ends up yielding -π/3, not π/3... why is this?