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From: Joubert on 7 Jul 2010 11:57
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?
From: Ray Vickson on 7 Jul 2010 14:50
On Jul 7, 8:57 am, Joubert <trappedinthecloset9...(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?

What is [n!/(2*Pi*i)]*int f(z)/(z-a)^(n+1) dz, integrated over a
'nice' closed path having 'a' in its interior?

R.G. Vickson
From: Axel Vogt on 7 Jul 2010 14:59
Joubert 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?

Another hint: what does your book say about residue and Laurent
series and what do you get by dividing the Taylor series of exp
by z^2 (doing it for each term)?
From: Joubert on 7 Jul 2010 16:13
On 07/07/2010 08:50 PM, Ray Vickson wrote:

> What is [n!/(2*Pi*i)]*int f(z)/(z-a)^(n+1) dz, integrated over a
> 'nice' closed path having 'a' in its interior?
>
> R.G. Vickson

f'(a) I guess.
Now the problem is (sorry I cheated) G is NOT closed. G is instead
something like:

G(r,s)={z in C: |arg(z)|=s, |z| >= 0} + {z in C:|z|=r, |arg(z)| <= s}

with s is in the interval ]Pi/2,Pi[


Is it conceivable that the integral may still be 1 even over such path?
From: Robert Israel on 7 Jul 2010 19:10

> On 07/07/2010 08:50 PM, Ray Vickson wrote:
>
> > What is [n!/(2*Pi*i)]*int f(z)/(z-a)^(n+1) dz, integrated over a
> > 'nice' closed path having 'a' in its interior?
> >
> > R.G. Vickson
>
> f'(a) I guess.

Not quite...

> Now the problem is (sorry I cheated) G is NOT closed. G is instead
> something like:
>
> G(r,s)={z in C: |arg(z)|=s, |z| >= 0} + {z in C:|z|=r, |arg(z)| <= s}
>
> with s is in the interval ]Pi/2,Pi[

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). 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.

> Is it conceivable that the integral may still be 1 even over such path?
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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