Proving Limit of Integral of Continuous f w/ Jordans Lemma

[gtfitzpatrick]

Homework Statement

Suppose that f is continuous and that there exist constants A,B ≥ 0 and k>1 such that |f(z)|≤A|z|−k for all z such that |z|>B. let CR denote the semicircle given by |z| = R, Re(z) ≥ 0. Prove that limR→∞∫f(z)dz=0

Homework Equations

The Attempt at a Solution

I don't understand what I'm prooving here. I think it has something to do with jordans lemma but I haven't a clue! any ideas...please...

 

[gabbagabbahey]

It is against forum rules to create multiple threads for the same problem.

Have you learned the estimation lemma yet?

 

[gtfitzpatrick]

Hi!
Sorry about the multiple thread but I called it jordans lemma then thought, it might have nothing to do with it but I couldn't delete the thread after creating it!

I haven't done the estimation lemma but I looked it up
Let f : U\rightarrowC be continuous (where U is some subset of C), let \gamma be a path in U, and suppose |f(z)| < M for all z \in \lambda. Let length( \lambda)= L. Then

|\intf(z)dz| \leq ML.

 

[vela]

I deleted the other thread. Next time, just hit the report button and ask a mentor to edit or delete the thread for you.

 

[gabbagabbahey]

I haven't done the estimation lemma but I looked it up
Let f : U\rightarrowC be continuous (where U is some subset of C), let \gamma be a path in U, and suppose |f(z)| < M for all z \in \lambda. Let length( \lambda)= L. Then

|\intf(z)dz| \leq ML.

It is probably in your textbook, but just not named. Many texts (c.f. Mathematical Methods for Physicists A Concise Introduction Tai L. Chow p. 255) simply list it as a property of complex integrals and give a short proof of it.

What is the arclength L of your semicircular contour CR? If |f(z)| \leq A|z|^{-k} (which is what I assume you meant when you wrote |f(z)|≤A|z|−k) and |z|&gt;B&gt;0, what is an upper bound for |f(z)| on your semi-circular contour?

What does that tell you about \left| \int_{CR} f(z) dz \right|?

 

[gtfitzpatrick]

The arc length is\pi r

|f(z)| \leq A|z|^{-k} the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of \frac{A}{|z|^{k}} gets smaller so as R\rightarrow\infty then |f(z)| \rightarrow 0
I guess that's what ther're saying but i guess i better try and word it better

 

[gabbagabbahey]

The arc length is\pi r

|f(z)| \leq A|z|^{-k} the semicircle is given by |z| = R so for k>1, if the value of R gets bigger, then the value of \frac{A}{|z|^{k}} gets smaller so as R\rightarrow\infty then |f(z)| \rightarrow 0

Right or, using the estimation lemma with the fact that |f(z)| \leq A|R|^{-k}, you have

\left| \lim_{R \to \infty} \int_{CR} f(z)dz \right| = \lim_{R \to \infty} \left| \int_{CR} f(z)dz \right| \leq \lim_{R \to \infty} \left( A|R|^{-k} \right) = 0