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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

That is correct.

For the second limit?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

That is correct.

Now you have to add them, what did you get?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Adding and then multiplying by 2*pi*i, I get

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

You only multiply by π i, see post #9420.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Why? I thought it was 2*pi*i*(sum of the residues), not pi*i...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Yes, but that is from -∞ to ∞. This integral is symmetrical and goes from 0 to ∞.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Oh, okay...

Also, I get the same answer as you but only if I take the real part of my answer and ignore the imaginary part. Am I supposed to do that or is that just an error in my arithmetic?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Something is wrong with the arithmetic. What are you using to compute the sum of the two limits?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

Hmm, WolframAlpha is getting the right answer, so there is a problem with me, I'll check my arithmetic again.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Okay, I am going to have to get some sleep. See you later. You should be able to do some others using this same 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**

Okay, see you later.

I got it, I was being silly, it is not the arithmetic that is an issue -- I just wasn't using De Moivre's to get the complex number form of those powers of i. This is a great technique, I will try it out on some other interesting integrals.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Hi;

For the simple poles it is really easy.

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 it for something like

and I cannot get it to work...

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

What did you get for that one?

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 am getting the residue to be zero but I am guessing I need to do something else... there is only one pole and it is a removable singularity (no C[sub]-1[/sub] co-efficient in the Laurent series). The answer should be \pi and DUIS can be used to show this, but I wanted to approach it with contour integration.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Yes, no Laurent series. I am getting zero also. It is obviously wrong so this method might not apply here

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 remember reading some PDF where a guy managed to do it but I can't remember what he did...

I think there is something missing in our methodology -- aren't there a number of steps that have to be done first, e.g. showing that the function is holomorphic?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Yes, I believe so. Are you sure you could do it using Residues? This one does not have a Laurent series. Doesn't that mean no poles?

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

But that function is clearly not defined at x = 0... surely it has a pole there? What else would be there? I thought that if it doesn't appear in the series expansion about that point, then it just means that it is a removable singularity.

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

A pole is the coefficient of the a(-1) term. There is no such term here.

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 thought that was the residue?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Oh excuse me, I meant residue. If the residue is 0 then the integral is 0 according to that formula.

0 is a removable singularity but it is not a pole.

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

Combinatorics is Algebra and Algebra is Combinatorics.

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**zetafunc.****Guest**

But this integral is π, not zero... what do we do?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 87,192

Have to find another way to evaluate it. What I am saying is that for some reason this method does not apply here. Possibly because it has no Laurent series.

http://mathworld.wolfram.com/Pole.html

Second paragraph helps.

*Last edited by bobbym (2013-03-22 12:58:33)*

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

Combinatorics is Algebra and Algebra is Combinatorics.

**Online**