Author 
Topic: double integral and essential singularity (Read 6158 times) 

immanuel78
Newbie
Gender:
Posts: 23


double integral and essential singularity
« on: Dec 27^{th}, 2006, 12:05am » 
Quote Modify

Let f be analytic in the region G={z:0<za]<R}. Show that if f has an essential singularity at z=a, then [double integral in G] f(x+iy)^2 dxdy = (infinity).


IP Logged 



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: double integral and essential singularity
« Reply #1 on: Jan 17^{th}, 2007, 10:45am » 
Quote Modify

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 lim_{za} (z  a)^{n}f(z) = 0. By the Riemann condition, that implies f has a pole of order n at a.


IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Ghost Sniper
Senior Riddler
Do not hide. It is useless.
Gender:
Posts: 599


Re: double integral and essential singularity
« Reply #2 on: Jan 18^{th}, 2007, 10:29am » 
Quote Modify

on Jan 17^{th}, 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?


IP Logged 
*sob* I miss my mommy... *blows nose* huh, I'm on? oh right.
(thinks to self) Time for my speech to these college kids.
"Reason is more important than all emotions..."



towr
wu::riddles Moderator Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730


Re: double integral and essential singularity
« Reply #3 on: Jan 18^{th}, 2007, 11:08am » 
Quote Modify

on Jan 18^{th}, 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 18^{th}, 2007, 11:09am by towr » 
IP Logged 
Wikipedia, Google, Mathworld, Integer sequence DB



Icarus
wu::riddles Moderator Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863


Re: double integral and essential singularity
« Reply #4 on: Jan 29^{th}, 2007, 11:07am » 
Quote Modify

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: _{0}^{}_{0}^{} f^{2}(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,  f^{2} dz  f^{2} dz, which is exactly the inside integral. Therefore f^{2} dz is finite. Define F(w) = zf^{2}(z)/(zw) 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) = zf^{2}(z) + a for some constant a. This shows that f^{2}(z) can have at most a pole of order 1 at 0. But it is easy to show that if f^{2} did have a pole at 0, the double integral would be infinite. Hence f^{2} 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 nonremovable singularity at a, the integral must be infinite.

« Last Edit: Feb 2^{nd}, 2007, 7:32pm by Icarus » 
IP Logged 
"Pi goes on and on and on ... And e is just as cursed. I wonder: Which is larger When their digits are reversed? "  Anonymous



Eigenray
wu::riddles Moderator Uberpuzzler
Gender:
Posts: 1948


Re: double integral and essential singularity
« Reply #5 on: Feb 2^{nd}, 2007, 4:54pm » 
Quote Modify

on Jan 29^{th}, 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) = e^{1/z}  1/z, then _{z=r} f^{2}(z) dz = 0 for all r>0. Quote:Define F(w) = zf^{2}(z)/(zw) 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) = zf^{2}(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 zf^{2}(z). The key to solving the problem is to consider the Laurent series of f: f(z) = _{n=} a_{n} z^{n}, converging absolutely and uniformly on compact subsets of 0<z<R. Then we have _{z=r} f(z)^{2} d = 2 a_{n}^{2} r^{2n}, by orthogonality of the functions {e : n }. So f(z)^{2} dxdy = _{0}^{R} r 2 a_{n}^{2}r^{2n} dr = 2_{=} a_{n}^{2} _{0}^{R} r^{2n+1}dr, which can only be finite if a_{n}=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 2^{nd}, 2007, 5:01pm by Eigenray » 
IP Logged 



