
#1
Dec211, 07:38 AM

P: 192

[tex]\oint dz\frac{e^{ikz}}{z}[/tex]
How we know for [tex]k>0[/tex] is function analytic in upper or in lower half plane? 



#2
Dec211, 11:20 AM

P: 1,623

I am assuming you want to calculate:
[tex]\int_{\gamma}\frac{exp(ikz)}{z} \, \mathrm{d}z[/tex] where [itex]\gamma[/itex] is some closed loop such that [itex]0 \in \mathrm{Int}(\gamma)[/itex]. If this is the case, you can the integral using the Residue Theorem. That is, write [itex]exp(ikz)[/itex] as a power series. Divide each term of the power series by [itex]z[/itex] to obtain a meromorphic function. You can then perform the integration and the only term that contributes to the value of the integral is the residue. 



#3
Dec211, 11:36 AM

P: 192

I asked what I want to know. I don't understand why if I have
[tex]\oint \frac{e^{ikz}}{z}[/tex], [tex]k>0[/tex] function is analytic in upper half plane if I [tex]k<0[/tex] function is analytic in lower half plane? Why? 



#4
Dec211, 11:46 AM

P: 1,623

Analitic function
The contour integral is a number, not a function. So asking if the integral is analytic in the upper/lower halfplane doesn't seem to make much sense. If you want to consider the function all of whose values are equal to the contour integral, then this is just a constant function and is obviously analytic on the upper and lower halfplane.
So, unless your question is about trivialities, I think you need to be more precise. 



#5
Dec211, 11:52 AM

P: 192





#6
Dec211, 11:57 AM

P: 1,623

Well, exp(ikz) is entire and therefore analytic on the whole plane. If we divide the power series of exp(ikz) by z, we see that exp(ikz)/z has a simple pole at k = 0 but is welldefined everywhere else. Which means that our series expansion for exp(ikz)/z is valid on the upper halfplane.




#7
Dec211, 11:59 AM

P: 192

Why we close counture in upper half plane for k>0?




#8
Dec211, 12:01 PM

P: 1,623

What do you mean by 'close counture'?




#9
Dec211, 12:05 PM

P: 192

When we calculate integral which I wrote we use contour in upper half plane for k>0. Why?




#10
Dec211, 12:13 PM

P: 1,623

I do not know why you would choose to do this. Since the contour will be chosen over a closed curve and since exp(ikz)/z is analytic in the upper halfplane, this means if we integrate along any closed curve which lies entirely in the upper halfplane, the integral will necessarily be zero.




#11
Dec211, 12:53 PM

P: 192

I will also take a small conture around zero.




#12
Dec211, 01:24 PM

P: 192

Understand now?




#13
Dec211, 01:32 PM

P: 192

I want to calculate integral [tex]\int^{\infty}_{\infty}\frac{sinkx}{x}[/tex] use integration which I wrote. Why for k>0 in upper plane? Tnx.




#14
Dec211, 03:25 PM

P: 1,623

Ah! That is not a contour in the upper halfplane; the upper halfplane excludes the real axis. And you take the integral that way because you know exp(ikz)/z is analytic everywhere except 0 and there is a theorem that involves evaluating real integrals using complex integrals like the one you have.




#15
Dec211, 05:20 PM

P: 1,666

[tex]e^{ikz}[/tex] for [itex]z=Re^{it}[/itex] thats: [tex]e^{ikR(\cos(t)+i\sin(t))}[/tex] Now consider it's absolute value: [tex]e^{kR\sin(t)}[/tex] In the upper halfplane, sine is positive so that will tend to zero for k>0. And if k<0, they we'd have to divert the contour to the lower halfplane because then sin(t)<0. 



#16
Dec411, 06:10 AM

P: 192

Thanks mate. Sorry again. Is there some easy way to see that that integral will go to zero? When you see
[tex]\oint\frac{e^{ikz}}{z}[/tex]? 



#17
Dec411, 07:13 AM

P: 1,666

[tex]\mathop\oint\limits_{z=1} \frac{e^{ikz}}{z}dz[/tex] that's really clear or in the other case: [tex]\mathop\oint\limits_{D} \frac{e^{ikz}}{z}dz[/tex] then clearly specify in the text what D is. In regards to you question about the integral over the upper halfdisc around the semicircle, well, you just need to plug it all in and analyze it to see what happens as R goes to infinity. 



#18
Dec511, 08:02 AM

P: 192

Thanks. What about
[tex]\lim_{R\to\infty}e^{ikR\cos t}[/tex]? 


Register to reply 
Related Discussions  
Analitic Geometry  Differential Geometry  8  
Square of the Riemann zetafunction in terms of the divisor summatory function.  Linear & Abstract Algebra  6  
Give to an interpolation function an analitic function with mathematica  Math & Science Software  1  
help in finding a good forum for analitic geometry  Calculus & Beyond Homework  1  
Vector Algebra & Analitic Geometry in Space  General Math  1 