How Can I Solve This Complex Integral with a Singularity?

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The forum discussion centers on solving the complex integral I(l,m;z) = ∫₀¹ (x^l)/(z - x^m) dx, where l and m are integers, and z = ω + i0⁺ is a complex number near the real axis. The primary challenge arises from the singularity in the integrand when ω approaches zero. Participants suggest using the formula 1/(ω + i0⁺ - x^m) = P(1/(ω - x^m)) - iπδ(ω - x^m) to separate the imaginary and real parts, but difficulties remain in handling the principal value integral. A change of variables from x^m to y is also proposed, yet the singularity issue persists.

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hiyok
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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
 
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hiyok said:
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?
 
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Simon Bridge said:
I'm more concerned with what ##x^\prime## represents.
It's not x', it's xl (x raised to the power of l).
 
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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 \nu = l/m&lt;1(assumption). However, the same problem exists. &lt;br /&gt; &lt;br /&gt; Then, how to do the principal value?
 
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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...
 
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