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   Author  Topic: double integral and essential singularity  (Read 6158 times)
immanuel78
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double integral and essential singularity  
« on: Dec 27th, 2006, 12:05am »
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Let f be analytic in the region G={z:0<|z-a]<R}.
Show that if f has an essential singularity at z=a, then [double integral in G] |f(x+iy)|^2 dxdy = (infinity).
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Re: double integral and essential singularity  
« Reply #1 on: Jan 17th, 2007, 10:45am »
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This hasn't be responded to for awhile, and I want to try out towr's new script for inserting math symbols on a real problem, so I thought I'd give my current thoughts on it.
 
I would suggest assuming that G |f(x+iy)|2 dxdy < . From this, you ought to be able to show that limza (z - a)nf(z) = 0. By the Riemann condition, that implies f has a pole of order n at a.
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Re: double integral and essential singularity  
« Reply #2 on: Jan 18th, 2007, 10:29am »
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on Jan 17th, 2007, 10:45am, Icarus wrote:
This hasn't be responded to for awhile, and I want to try out towr's new script for inserting math symbols on a real problem, so I thought I'd give my current thoughts on it.
 

 
So how do you insert the math symbols?
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Re: double integral and essential singularity  
« Reply #3 on: Jan 18th, 2007, 11:08am »
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on Jan 18th, 2007, 10:29am, hiyathere wrote:
So how do you insert the math symbols?
To save another thread from going offtopic, I've summarized it here
« Last Edit: Jan 18th, 2007, 11:09am by towr » IP Logged

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Re: double integral and essential singularity  
« Reply #4 on: Jan 29th, 2007, 11:07am »
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Okay - here is an approach that seems to work - though I doubt it was the intended solution:
 
To ease the notation, we can assume wlog that a = 0. Assume the integral is finite. Express it in polar coordinates:  00 |f2(re)| rddr.  For the outer integral to exist, the inner integral must be finite for almost all values of r. Again wlog, we can assume that it is finite for r = R. Now taking the circle of Radius R about 0 as the path, | f2 dz | |f2| |dz|, which is exactly the inside integral. Therefore f2 dz is finite. Define F(w) = zf2(z)/(z-w) dz. We see that F(w) is finite at all points inside the circle, including 0. F(w) is also holomorphic (a common lemma used in proving Cauchy's formula), and in fact, F(z) = zf2(z) + a for some constant a. This shows that f2(z) can have at most a pole of order 1 at 0. But it is easy to show that if f2 did have a pole at 0, the double integral would be infinite. Hence f2 is analytic at 0. So f is the square root of an analytic function, and hence f has a removable singularity at 0, or has 0 as a branch point. But the latter condition is impossible, since f is analytic everywhere in G. Therefore f must have a removable singularity.
 
Taking the contrapositive gives that if f has a non-removable singularity at a, the integral must be infinite.
« Last Edit: Feb 2nd, 2007, 7:32pm by Icarus » IP Logged

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Re: double integral and essential singularity  
« Reply #5 on: Feb 2nd, 2007, 4:54pm »
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on Jan 29th, 2007, 11:07am, Icarus wrote:
For the outer integral to exist, the inner integral must be finite for almost all values of r.

This can't be enough to show that f doesn't have an essential singularity.  Indeed, if f is any function continuous on {0<|z|<R}, then the inner integrals will all be finite, as the integral of a continuous function over a compact interval.
 
In fact, if f(z) = e1/z - 1/z, then |z|=r f2(z) dz = 0 for all r>0.
 
Quote:
Define F(w) = zf2(z)/(z-w) dz. We see that F(w) is finite at all points inside the circle, including 0. F(w) is also holomorphic (a common lemma used in proving Cauchy's formula), and in fact, F(z) = zf2(z) + a for some constant a.

F(z) is holomorphic, but only because it is the regular part (nonnegative powers in z) of the Laurent series for zf2(z).
 
The key to solving the problem is to consider the Laurent series of f:
 
f(z) = n=- an zn,
 
converging absolutely and uniformly on compact subsets of 0<|z|<R.  Then we have
 
|z|=r |f(z)|2 d = 2 |an|2 r2n,
 
by orthogonality of the functions {e : n }.  So
 
|f(z)|2 dxdy
 = 0R r 2 |an|2r2n  dr
 = 2=- |an|2 0R r2n+1dr,
 
which can only be finite if an=0 whenever n<0, that is, if f actually has a removable singularity at 0.
 
Of course one needs to check that things converge appropriately to justify interchanging the order of integration and summation, but this boring.
« Last Edit: Feb 2nd, 2007, 5:01pm by Eigenray » IP Logged
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