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Analitic function

  1. Dec 2, 2011 #1
    [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?
    Last edited: Dec 2, 2011
  2. jcsd
  3. Dec 2, 2011 #2


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    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.
  4. Dec 2, 2011 #3
    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?
  5. Dec 2, 2011 #4


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    The contour integral is a number, not a function. So asking if the integral is analytic in the upper/lower half-plane 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 half-plane.

    So, unless your question is about trivialities, I think you need to be more precise.
  6. Dec 2, 2011 #5
    My mistake. Why function [tex]\frac{e^{ikz}}{z}[/tex] is analytic in upper half plane for [tex]k>0[/tex]?
  7. Dec 2, 2011 #6


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    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 well-defined everywhere else. Which means that our series expansion for exp(ikz)/z is valid on the upper half-plane.
  8. Dec 2, 2011 #7
    Why we close counture in upper half plane for k>0?
  9. Dec 2, 2011 #8


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    What do you mean by 'close counture'?
  10. Dec 2, 2011 #9
    When we calculate integral which I wrote we use contour in upper half plane for k>0. Why?
  11. Dec 2, 2011 #10


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    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 half-plane, this means if we integrate along any closed curve which lies entirely in the upper half-plane, the integral will necessarily be zero.
  12. Dec 2, 2011 #11
    I will also take a small conture around zero.
  13. Dec 2, 2011 #12
    Understand now?

    Attached Files:

  14. Dec 2, 2011 #13
    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.
  15. Dec 2, 2011 #14


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    Ah! That is not a contour in the upper half-plane; the upper half-plane 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.
  16. Dec 2, 2011 #15
    Need to be more clear mate. We choose the upper half-contour for k>0 in that integral because we wish the integral over the large semi-circle to tend to zero as R goes to infinity. Consider the expression:


    for [itex]z=Re^{it}[/itex]



    Now consider it's absolute value:


    In the upper half-plane, 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 half-plane because then sin(t)<0.
  17. Dec 4, 2011 #16
    Thanks mate. Sorry again. Is there some easy way to see that that integral will go to zero? When you see

  18. Dec 4, 2011 #17
    Ok, that one "looks" like you're just going around the origin but really you mean the half-disc contour in either half-plane. Around the origin, the integral is 2pi i. Otherwise if it's around the discs. You could be more specific like:

    [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 half-disc around the semi-circle, well, you just need to plug it all in and analyze it to see what happens as R goes to infinity.
  19. Dec 5, 2011 #18
    Thanks. What about

    [tex]\lim_{R\to\infty}e^{ikR\cos t}[/tex]?
  20. Dec 5, 2011 #19
    How about you answer that assuming k, R and t are real numbers. Use the Euler identity:


    What is the maximum in absolute value that expression attains? Does it ever settle down to a limit no matter how large x gets?
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