# Analitic function

by matematikuvol
Tags: analitic, function
 P: 192 $$\oint dz\frac{e^{ikz}}{z}$$ How we know for $$k>0$$ is function analytic in upper or in lower half plane?
 P: 1,622 I am assuming you want to calculate: $$\int_{\gamma}\frac{exp(ikz)}{z} \, \mathrm{d}z$$ where $\gamma$ is some closed loop such that $0 \in \mathrm{Int}(\gamma)$. If this is the case, you can the integral using the Residue Theorem. That is, write $exp(ikz)$ as a power series. Divide each term of the power series by $z$ 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.
 P: 192 I asked what I want to know. I don't understand why if I have $$\oint \frac{e^{ikz}}{z}$$, $$k>0$$ function is analytic in upper half plane if I $$k<0$$ function is analytic in lower half plane? Why?
 P: 1,622 Analitic function 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.
P: 192
 Quote by 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.
My mistake. Why function $$\frac{e^{ikz}}{z}$$ is analytic in upper half plane for $$k>0$$?
 P: 1,622 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.
 P: 192 Why we close counture in upper half plane for k>0?
 P: 1,622 What do you mean by 'close counture'?
 P: 192 When we calculate integral which I wrote we use contour in upper half plane for k>0. Why?
 P: 1,622 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.
 P: 192 I will also take a small conture around zero.
 P: 192 Understand now? Attached Thumbnails
 P: 192 I want to calculate integral $$\int^{\infty}_{-\infty}\frac{sinkx}{x}$$ use integration which I wrote. Why for k>0 in upper plane? Tnx.
 P: 1,622 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.
P: 1,666
 Quote by matematikuvol I want to calculate integral $$\int^{\infty}_{-\infty}\frac{sinkx}{x}$$ use integration which I wrote. Why for k>0 in upper plane? Tnx.
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:

$$e^{ikz}$$

for $z=Re^{it}$

thats:

$$e^{ikR(\cos(t)+i\sin(t))}$$

Now consider it's absolute value:

$$e^{-kR\sin(t)}$$

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.
 P: 192 Thanks mate. Sorry again. Is there some easy way to see that that integral will go to zero? When you see $$\oint\frac{e^{ikz}}{z}$$?
P: 1,666
 Quote by matematikuvol Thanks mate. Sorry again. Is there some easy way to see that that integral will go to zero? When you see $$\oint\frac{e^{ikz}}{z}$$?
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:

$$\mathop\oint\limits_{|z|=1} \frac{e^{ikz}}{z}dz$$

that's really clear or in the other case:

$$\mathop\oint\limits_{D} \frac{e^{ikz}}{z}dz$$

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.
 P: 192 Thanks. What about $$\lim_{R\to\infty}e^{ikR\cos t}$$?

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