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From: Joubert on 7 Jul 2010 21:19
On 07/08/2010 01:10 AM, Robert Israel wrote:

> Not quite...

f^(n)(a) :)

> This doesn't quite make sense. Paths in complex analysis have directions,
> and you haven't indicated any. I suspect you mean |z|>= r (otherwise
> you'll have convergence problems as z -> 0).

Yes, sorry.

> So I suppose you have a
> "keyhole" shaped contour containing two rays extending out to infinity. Now
> the trick is to replace this by a closed contour that goes out only a finite
> distance R, and use estimates on |f(z)| to show that the limit as R ->
> infinity is the
> integral over your contour.

Say I "close the keyhole" with the boundary of a circle of radius R
centered at the origin. Then since z|f(z)| goes to zero for R -> infty
(because Re(z) is negative) the integral over this portion of path is 0
by a Lemma which is called, where I live, big circle lemma. Am I raving
or does this make sense?
From: David C. Ullrich on 8 Jul 2010 09:40
On Wed, 07 Jul 2010 17:57:27 +0200, Joubert
<trappedinthecloset9985(a)yahoo.com> wrote:

>Suppose G=G(r,s) is a close, smooth enough path containing the origin.
>Consider the integral:
>
>1/i2Pi Int_G e^z/z^2 dz
>
>My book says this is e^0=1 because this is simply because of Cauchy
>formula. However it looks like he applies the cauchy formula to e^z/z
>whose value in zero is definitely not 1.
>
>Any hint?

If things were as you say (which you later say is not so) then
this integral would indeed be zero. You could get this from
the Residue Theorem. Or more or less equivalently, from
looking at the power series.

Or, trying to make sense of what you say the book says,
you might note that there's a thing commonly called
the "Cauchy Integral Formula", and there's also a
generalization sometimes called the "Cauchy Integral
Formula for Derivatives"...




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