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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

That is correct.

For the second limit?

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**zetafunc.****Guest**

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

That is correct.

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

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,485

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

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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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: 90,485

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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.

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

Hi;

For the simple poles it is really easy.

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

What did you get for that one?

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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?

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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

**In mathematics, you don't understand things. You just get used to them.**

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**zetafunc.****Guest**

I thought that was the residue?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 90,485

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.

**In mathematics, you don't understand things. You just get used to them.**

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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: 90,485

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)*

**In mathematics, you don't understand things. You just get used to them.**

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