Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,648

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.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**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: 81,648

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.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**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: 81,648

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.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**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: 81,648

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 81,648

Hi;

Okay, I will see you then.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 81,648

Hi;

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 81,648

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?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

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

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,648

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:

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 81,648

Hi;

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

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**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: 81,648

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.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**Online**

**zetafunc.****Guest**

Okay, see you later.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 81,648

Hope you are having no problems. See you later.

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

**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?