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## Topic review (newest first)

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.

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

, where z is complex and n is a constant.

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?

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