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May 2nd, 2012, 03:50 PM   #1
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About Zeta integral expression

So the formula for the Zeta function is . Now I understand the contour comes from infinity just above the real axis, turn around the origin and comes back to infinite just below the real axis. I have multiple questions about this!
1) So it does not matter how much above/below the real axis the contour is, or what is the size of the circle around the origin is, the value of the integral is the same, is that true? As long as we have the singularity inside the contour and we do not intersect the real axis(because teh function has to be singled valued), the integral has the same value? Then is the integral equal to some residue at zero? Here, I guess the pole has no order defined if s is not an integer right?
2) If for instance s=2, then the function to integrate is analytic no? (because 1/(e^x-1) has a simple pole ). But if it is true then why Cauchy theorem does not apply here? In other words, why is this integral not zero...because the contour is not closed? But then I am not sure to understand why the shape of the contour does not matter.
3) Finally, I don't understand why this integral converge for any s. I understand it for s>1, because even if the circle tends to zero , the integral on the circle goes to zero. If s<1, it seems that if the contour around the origin was going to zero it would blow up at that point....So is this integral simply converging because the contour does not pass right to the pole? But how to explain then that it would blow up if we go close to me pole...

As you see I am very lost! Complex analysis is one of my weak point in maths. Even answer to one of this question would be greatly appreciated!
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May 6th, 2012, 10:42 AM   #2
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Re: About Zeta integral expression

Oups, Inknew it was not a trivial question, but this is simply the definition of the Zeta function on the complex plane....I guess most people don't really understand it finally, me included..!! I will work on it. cheers.
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