Analysis - limits and integrals

[saus]

This is stuff I should know but am too intimidated to ask the professor.

In Bak/Newman's Complex Analysis, pg 148, it says \lim_{\epsilon\rightarrow 0, M\rightarrow\infty}\int_{C} R(z)\log z dz=\int_0^\infty R(x)\log x dx, where C is the horizontal line segment from i\epsilon to \sqrt{M^2-\epsilon^2}+i\epsilon. R(x) is a rational function with the degree of the numerator greater or equal to the degree of the denominator + 2, but I don't think this matters here.

No further details are given. I tried to work it all out completely, but had some issues. 1. I assume this limit means take one limit, then the next limit. Doesn't this sometimes depend on which limit you take first? Why does it not matter in this case? 2. It seems like we'll need two applications of either uniform convergence or the dominated convergence theorem to take care of the two limits. But it wasn't immediately obvious to me what I should do. I'm also unsure of whether one of these theorems (uniform convergence thing and dominated convergence thm) is stronger than the other or if they're completely different things.

I'll try to work it out more once I get time. Thanks for any help.

[jestingrabbit]

What do you mean by R(z) here? The real part of z?

[++$_]

It says: "R(x) is a rational function with the degree of the numerator greater or equal to the degree of the denominator + 2."

[jestingrabbit]

It says: "R(x) is a rational function with the degree of the numerator greater or equal to the degree of the denominator + 2."

Yes, yes it does... must have been blind when I first read it.

[Yakk]

I'd guess, at a far remove, to look at this as a closed curve problem. Then again, maybe you are trying to prove a closed curve property. In which case, this is putting the cart before the horse -- however, if there is any way to solve it via simpler closed curves, I'd be tempted. (the "gap" between the end points of the left and right sides of that equation end up collapsing... it shouldn't be hard to argue that you can talk about the limit of a closed curve, and it is 0 iff your theorem is true).

You can do double limits. Do you have much real analysis background?

If f:RxR->R, we can talk about lim a->b of f(a) = X, where b is an element of the extension of RxR that includes the various points at infinity. This ends up being defined as "for every neighborhood N1 of X, there is a neighborhood N2 of b such that the image of N2\{b} under f is contained in N1. Neighborhoods of points-at-infinity are like neighborhoods of infinity in R -- a collection of neighborhoods of infinity in R are the sets (M, infinity). So (0, epsilon) x (M, infinity) produces a neighborhood of (0, infinity).

That make any sense?

[saus]

I was thinking about issues like \lim_{n\rightarrow\infty m\rightarrow\infty} \dfrac{m}{n}.

It looks like I should have given more context for the problem. It's part of calculating a real integral using the residue theorem. We know a certain integral will equal 2pi*i times the residues, then we show the integral along certain parts of the curve go to 0. This part of it is showing that the integral on the complex plane over particular curves converging to the positive real axis is equal to the integral from 0 to infinity.

Yakk, your idea about the simpler closed curves seems like a very easy way to show this.

[jestingrabbit]

The big issue that I kept worrying about was roots along the positive real axis. If you know those aren't there, I would replace that integral with one that finds the real axis quickly ie, along a circle around 0 with radius \epsilon, then along the reals out to M.

[saus]

>_< Forgot to mention, there are no poles on the positive real axis.

[jestingrabbit]

Okay, that means that there is some epsilon beyond which there are no poles in the region R^+ \times (-\epsilon, \epsilon). So, the integral along your path equals the integral along (\epsilon, M) plus the integral along the other bits. So, do some analysis on the bits.

But tell me, is there possibly a root of R at 0?