Is the Analytic Function e^{ikz}/z Upper or Lower Half Plane for k>0?

[matematikuvol]

\oint dz\frac{e^{ikz}}{z}

How we know for k>0 is function analytic in upper or in lower half plane?

 

[jgens]

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.

 

[matematikuvol]

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?

 

[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]

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?

 

[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?

 

[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.

 

[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]

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.

 

[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}?

 

[jackmell]

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.

 

[matematikuvol]

Thanks. What about

\lim_{R\to\infty}e^{ikR\cos t}?

 

[jackmell]

Thanks. What about

\lim_{R\to\infty}e^{ikR\cos t}?

How about you answer that assuming k, R and t are real numbers. Use the Euler identity:

e^{ix}=\cos(x)+i\sin(x)

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