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- zetafunc.
- 2012-10-26 03:10:09
Oh wait, I just noticed...
and since
for n ≠ 1, z ∈ Z, then every term reduces to zero, so the contour integral is zero... right?
- zetafunc.
- 2012-10-26 03:03:37
Thanks for that, I did not know of this method. Is that the only way to show a function like this is holomorphic on C? Would I have to do this before I evaluate any contour integral?
- scientia
- 2012-10-26 02:42:19
Let so where and .
Since the partial derivatives are continuous for all and satisfy the Cauchy–Riemann equations and , is holomorphic on the complex plane.
Since the partial derivatives are continuous for all and satisfy the Cauchy–Riemann equations and , is holomorphic on the complex plane.
- zetafunc.
- 2012-10-26 00:08:39
Why is it holomorphic on C?
- scientia
- 2012-10-25 23:04:59
Then shouldn't it be 0 by Cauchy's integral theorem since is holomorphic on ?
- zetafunc.
- 2012-10-25 22:58:05
C is some simple closed curve about 0.
- scientia
- 2012-10-25 22:50:28
What is C?
- zetafunc.
- 2012-10-25 20:52:56
I was trying to find
Let z = ln(t), then dz = (1/t)dt. This transforms the integral to
Is this correct? And what would the graph of this look like?