Residue of function with $\frac{1}{\ln(z^{-2})}$ and $z^{-2}$ dependence

Question

In my current research, I have come across an integral that is refusing to submit to any method I can think of for evaluation. It is an integral that comes from an operator product expansion of gluon production in QCD, but I have gotten down to the following form

$$ \int_{-1}^1 e^{-i k x} \frac{1}{|x|^2 \ln^a\left(\frac{1}{|x|^2 \Lambda^2}\right) } dx $$ Where $\Lambda$ is a large number, and $a$ is a function of the coupling constant. At this point, I am trying to evaluate this integral (in a less than rigorous manner, I know) by taking the contour, but while I know that $|x|=\frac{1}{\Lambda}$ should be a pole, I can not figure out how to find this residue.

Any tips/help or is this not possible?

Answer 1

As long as you keep $|x|$ it's not a holomorphic function, so the concept of the residue doesn't apply to it at all. There's also a problem with the singularity at $x=0$ which makes the integral not covergent.

But let us consider $$ f(x) = \frac{e^{ikx}}{x^2 \ln^a(\frac{1}{x^2\Lambda^2})} $$ for $\Lambda > 0$ and analyze the singularity at $x= 1/\Lambda$. We have $$ \frac{1}{x\Lambda} = \frac{1}{1+ \Lambda(x-1/\Lambda)}= 1 - \Lambda (x-1/\Lambda) + \mathcal O\big((x-1/\Lambda)^2\big)$$ $$ \frac{1}{x^2\Lambda^2} = 1 - 2\Lambda (x-1/\Lambda) + \mathcal O\big((x-1/\Lambda)^2\big)$$ $$ \ln \frac{1}{x^2\Lambda^2} = - 2\Lambda (x-1/\Lambda) + \mathcal O\big((x-1/\Lambda)^2\big)$$ $$ \ln^a \frac{1}{x^2\Lambda^2} = 2^{a}\Lambda^{a} (1/\Lambda-x)^{a} + \mathcal O\big((1/\Lambda-x)^{a+1}\big)$$ $$ \frac{1}{\ln^a \frac{1}{x^2\Lambda^2}} = 2^{-a}\Lambda^{-a} (1/\Lambda-x)^{-a} + \mathcal O\big((1/\Lambda-x)^{-a+1}\big)$$ $$ \frac{e^{ikx}}{x^2} = \Lambda^2e^{ik/\Lambda} + \mathcal O\big((x-1/\Lambda)\big)$$ $$ \frac{e^{ikx}}{x^2\ln^a \frac{1}{x^2\Lambda^2}} = 2^{-a}\Lambda^{2-a}e^{ik/\Lambda} (1/\Lambda - x)^{-a} + \mathcal O\big((1/\Lambda-x)^{-a+1}\big)$$ In conclusion:

• if $a\notin\mathbb Z$ then point $x=1/\Lambda$ is not a pole, but a branch point - there's no such thing as a residuum in such point
• if $a\in\mathbb Z$, $a\le 0$ it's a regular point
• if $a\in\mathbb Z$, $a > 0$ then it's a pole of order $a$ and the residuum can be calculated, either by expanding the function $f(x)$ in Laurent series by any method, or from the formula $$ {\rm Res\,}_{z=1/\Lambda} f(z) = \frac{1}{(a-1)!}\frac{d^{a-1}}{dz^{a-1}}\big((z-1/\Lambda)^a f(z)\big)$$ For, example, for $a=1$ the residuum is $\frac12 \Lambda e^{ik/\Lambda}$.