Residue of exp[i.kx] / [ 1 - k^2 ]: Find Answer & Fix Mistake

[Master J]

Ive run into some residue problems, I can't seem to find a clear answer anywhere on this...

I need to find the residue of exp[i.kx] / [ 1 - k^2 ], where k is my complex variable, and x is positive.

I have poles at 1 and -1 in my integral. Now everywhere I look, a pole of order n is when one has say, in my case, ( 1 - k^2)^n...the n being outside the bracket. In what I have above, 1 - k^2, is this still of order 2?


Ive tried computing the residue but I can't get the correct answer, sin(x). My method is as follows:

multiply the above by (k - 1)^2, and then evaluate at k = 1, -1...what am I doing wrong here?

 

[Amok]

I have poles at 1 and -1 in my integral. Now everywhere I look, a pole of order n is when one has say, in my case, ( 1 - k^2)^n...the n being outside the bracket. In what I have above, 1 - k^2, is this still of order 2?

I think it's a first oder pole. To be honest I don't really remember everything that goes on with Laurent series, but I do remember a simple set of rules that allows you to find residues. Do you know these rules?Well, 1 is a 0th order zero of:

p(k)= e^{ikx}

and a 1st order zero of

q(k)=1-k^2

Which means you have a 1st order pole and the residue at 1 is:

Res_1 = \frac{p(1)}{q'(1)} = \frac{-e^{ix}}{2}

The same reasoning should give you

Res_{-1} = \frac{p(-1)}{q'(-1)} = \frac{e^{-ix}}{2}

I hope I have used all the correct English terminology.

Ive tried computing the residue but I can't get the correct answer, sin(x).

Sin(x) is not "the residue", but rather the integral of your function divided by 2pi over any closed path in C containing both your singularities ( 1 and -1). It also is the sum of both your residues multiplies by i.

My method is as follows:

multiply the above by (k - 1)^2, and then evaluate at k = 1, -1...what am I doing wrong here?

I dunno, you tell me. Multiply the above what by (k - 1)^2?

 

[mathwonk]

If f is a function analytic and non zero at p, then the residue of f/(z-p) at p is sort of obviously the value of f at p. (divide a power series expanded at p by (z-p) and ask yourself what the coefficient is of (z-p)^-1.) so in your case, you have f/(z^2-1), where f has no zeroes at all. since (1-z^2) has simple zeroes at 1 and -1, the quotient has simple poles at those points. so if you have, at 1 say, the analytic function e^(ixz)/(-1-z), divided by (z-1), ...is that enough?

 

[HallsofIvy]

Ive run into some residue problems, I can't seem to find a clear answer anywhere on this...

I need to find the residue of exp[i.kx] / [ 1 - k^2 ], where k is my complex variable, and x is positive.

I have poles at 1 and -1 in my integral. Now everywhere I look, a pole of order n is when one has say, in my case, ( 1 - k^2)^n

You didn't mean k^2 here did you? A pole of order n at k= a involves (a- k)^n in the denominator. At a= 1, that is (1- k) and at a= -1, (-1- k)^n= -(1+ k)^n if n is odd.

...the n being outside the bracket. In what I have above, 1 - k^2, is this still of order 2?

The important point is that 1- k^2= (1- k)(1+ k) so that at k= 1, it is 1- k that is what you want to look at and at k= 1, 1+ k. Each has power 1.

Ive tried computing the residue but I can't get the correct answer, sin(x). My method is as follows:

multiply the above by (k - 1)^2, and then evaluate at k = 1, -1...what am I doing wrong here?

Multiply by k- 1 and evaluate at k= 1, multiply by k+ 1 and evaluate at k= -1.

 

[Marin]

Hi all!

Just because,Master J, you mentioned you're using the residue thm in this case to evaluate an integral, here's an amusing question for you about an integral of a similar type:

First, x is still a non-negative number. We are interesting in the integral over all R of e^{ikx}k/(1+k^2), k being the integration variable. [NOTE: it's k/(1+k^2) instead of 1/(1-k^2)]. Now, obviously there are two 'loops' you can take to apply the residue thm, both surrounding a singularity at +/- i, respectively, with a semicircle (one in the upper, the other one in the lower half plane) and both close down by following the real axis. However, a simple computation of the residue shows they yield a different result for the integral.

So, what is the true value of the integral and why?

Remark: If you try viewing the integral as a Fourier transform of an L^1 function, the Riemann-Lebesgue lemma tells you that the transformed function has to be continuous and decaying at infinity, showing which 'loop' is the correct one! What's wrong with the other one?

PS: I myself had some headache two weeks ago while figuring out where the problem lies, but all in all, I found it helpful towards understanding

 

[Redbelly98]

I think the problem, and solution, can be pretty well summarized as follows:

... multiply the above by (k - 1)^2, and then evaluate at k = 1, -1...what am I doing wrong here?

Multiply by k- 1 and evaluate at k= 1, multiply by k+ 1 and evaluate at k= -1.

That will get you the residue at k=1, as well as the residue at k=-1.