Analitic function

  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. jgens

    jgens 1,621
    Gold Member

    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. 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. jgens

    jgens 1,621
    Gold Member

    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. My mistake. Why function [tex]\frac{e^{ikz}}{z}[/tex] is analytic in upper half plane for [tex]k>0[/tex]?
  7. jgens

    jgens 1,621
    Gold Member

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

    jgens 1,621
    Gold Member

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

    jgens 1,621
    Gold Member

    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. I will also take a small conture around zero.
  13. Understand now?

    Attached Files:

  14. 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. jgens

    jgens 1,621
    Gold Member

    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. 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. Thanks mate. Sorry again. Is there some easy way to see that that integral will go to zero? When you see

  18. 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. Thanks. What about

    [tex]\lim_{R\to\infty}e^{ikR\cos t}[/tex]?
  20. 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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