Analitic function

matematikuvol

[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?

jgens

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.

matematikuvol

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?

jgens

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.

matematikuvol

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

jgens

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.

matematikuvol

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

jgens

What do you mean by 'close counture'?

matematikuvol

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

jgens

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.

matematikuvol

I will also take a small conture around zero.

matematikuvol

Understand now?

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matematikuvol

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.

jgens

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.

jackmell

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:

[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 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.

matematikuvol

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]?

jackmell

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.