How Can I Solve This Complex Integral with a Singularity?

[hiyok]

hi,

I have difficulty in figuring out the following integral:

$I(l,m;z) = \int^1_0 dx~\frac{x^l}{z - x^m}$,

where $l$ and $m$ are integers, while $z = \omega + i0_+$ is a complex number that is infinitely close to the real axis. What is interesting to me is when $\omega$ is close to zero, so that the integrand bears a singularity in the domain.

Could somebody help me out ?

Thanks a lot !

hiyok

 

[Simon Bridge]

hi,

I have difficulty in figuring out the following integral:

$I(l,m;z) = \int^1_0 dx~\frac{x^l}{z - x^m}$,

where $l$ and $m$ are integers, while $z = \omega + i0_+$ is a complex number that is infinitely close to the real axis.

You mean:$$z=\lim_{k\rightarrow 0^+}\omega + ik$$... but since z is not a function of x, what is the problem?

I'm more concerned with what $x^\prime$ represents.

What is interesting to me is when $\omega$ is close to zero, so that the integrand bears a singularity in the domain.

The point x=0 is not part of the integral if that's what you were worried about. The integration is from 0<x<1. z does not take part in the integral anyway. If you are interested in what happens to the result for ω=0 put it in and see.

Could somebody help me out ?

What have you tried so far?
How does this integral come up in the first place?

 

[D H]

I'm more concerned with what $x^\prime$ represents.

It's not x', it's x^l (x raised to the power of l).

 

[hiyok]

Thanks for response.

1. Yes, I mean $\lim_{k\rightarrow 0}(\omega+ik)$.

2. As pointed out by D H, it is $x^l$, x raised to the power of l.

3. Initially, I tried to do it by this formula, $\frac{1}{\omega+i0_+-x^m} = \mathcal{P}\left(\frac{1}{\omega-x^m}\right)-i\pi \delta(\omega - x^m)$, with $\delta(x)$ denoting the Dirac function and $\mathcal{P}$ indicating the principal value. The imaginary part can thus be easily worked out. But I do not know how to handle the real part (i.e., the principal value part), which is supposed to contain a singularity at $x^m = \omega$.

4. I have also tried to make a change of variable, $x^m \rightarrow y$. In terms of $y$, the integral becomes something like $\int dy ~ \frac{y^{\nu}}{z-y}$, with [itex]\nu = l/m<1$(assumption). However, the same problem exists. <br /> <br /> Then, how to do the principal value?$[/itex]

 

[Simon Bridge]

Have you tried rationalizing the denominator or converting it to a complex exponential?
have you tried u=z-x^m ... of course this makes u a complex number...