Strange (to me) integral...

[Char. Limit]

So, I was reading the book "Riemann's Zeta Function", by H.M. Edwards, and on page 10 I see an integral I don't quite understand. Here's the integral:

\int_{+\infty}^{+\infty} \frac{\left(-x\right)^s}{e^x - 1} \frac{dx}{x}

My first problem with this is that, I see an integral from some number to the same number. Shouldn't it be zero?However, the book says that those limits are "intended to create a path of integration which begins at positive infinity, moves to the left down the positive real axis, circles the origin once in the positive (counterclockwise) direction, and returns up the positive real axis to positive infinity." ...what? What does that mean and how can you get that from the integral?

Next, how is that evaluated? The book doesn't give an explanation, simply saying "this integral is equal to that integral, which is equal to this other integral" in essence. I don't understand... I can't even make it past page 10...

[Cyosis]

That is a very weird way to write down an integral. You should read it as a contour integral with a contour such that the origin is avoided.

Are you familiar with complex integration?

[Char. Limit]

Not really. I know some about path integrals, but not the complex version. This book was recommended (actually given) to me by a mathematician, so I thought I'd be able to understand it...

I was also wondering why they wrote dx/x seperately instead of putting the x^-1 with the main integral. Makes no sense...what was Riemann thinking? (The book says "next [Riemann] considers the contour integral", which I suppose should have cued me in, but that makes no sense as a contour integral either. Contour integrals are written differently, aren't they?)

[Cyosis]

Well I admit I have never seen a contour integral written like that before. Normally one would write it as :


\int_\gamma \frac{(-x)^s}{e^x-1}\frac{dx}{x}


With \gamma the contour described in the text. The circle around the origin would be given a certain radius \epsilon to avoid the singularities at \pm 2\pi i. You later take the limit \epsilon \rightarrow 0. You will have to read up on some complex analysis to understand this.

These links may be helpful to give you a general idea:

http://en.wikipedia.org/wiki/Cauchy_integral
http://en.wikipedia.org/wiki/Residue_%28complex_analysis%29

[Count Iblis]

I've seen this notation once before somewhere also in this same context (Riemann zeta function). So, I suspect that it may be traditional for this purpose...

[jackmell]

Not really. I know some about path integrals, but not the complex version. This book was recommended (actually given) to me by a mathematician, so I thought I'd be able to understand it...

I was also wondering why they wrote dx/x seperately instead of putting the x^-1 with the main integral. Makes no sense...what was Riemann thinking? (The book says "next [Riemann] considers the contour integral", which I suppose should have cued me in, but that makes no sense as a contour integral either. Contour integrals are written differently, aren't they?)

That's actually a beautiful integral and not too hard to express in a computable form if you parameterize the contour along it's Hankel path (or rather along it's inverse path to remove the -x). It's a "branch-cut" integral and the notation, although awkward, makes perfect sense if you understand what the Hankel path is. Look up Hankel Contour to see what I mean. Although I don't have it in front of me, I think he separates the x out in order to more easily represent a certain limit in the analysis later or that's just how Riemann originally wrote it. There is a copy of Riemann's paper in the back of the text. That section you're reading is just the first chapter and is difficult to understand even for graduate students (I think) so it's definitely not the book for you to read if you're starting.

First study Complex Variables and Complex Analysis (and really enjoy it), work many problem just to become familiar with Complex Analysis, then study some of the simple analytic properties of the zeta function and related functions, then begin to start using both Edwards and Titchmarsh and even after all that, they are still hard to read but if you persists, they gradually open up for you.