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I have been trying to evaluate the residue at infinity of the following function:

$\exp(-\frac{1}{3}Log(1-z))\exp(-\frac{2}{3}Log(1+z))$

The first Log (multiplied by $-\frac{1}{3}$) has branch cut $[0,\infty)$ and the second has branch cut $(-\infty,0]$, i.e the principal branch. Therefore the left term has branch cut at $(-\infty ,1]$ and the right term has branch cut at $(-\infty,-1]$.

Therefore the product of those two terms has branch cut $[-1,1]$ and one can show this by determining the continuity of that product over the interval where both branch cuts of the individual terms coalesce.

I successfully evaluated the residue at infinity of that function by evaluating the integral of that product over an arbitrarily large circle centre $0$, and got $-exp(-\pi/3)$ which I believe is correct, integrating clockwise over the circle.

My question is, that in order to evaluate this integral over this arbitrarily large circle, I made the substitution $Log(1-R\exp(i \theta)) \approx Log(-R\exp( i \theta)$ and $Log(1 + R\exp(i \theta)) \approx Log(R\exp(i \theta))$ which I thought to be true (with respect to the individual branch cuts of each Log) to a degree of error that would produce an overall error going to $0$ of the integral I wanted to evaluate.

Now I would like to formalise these replacements, and I think perhaps one way to do so would be to find Laurent series with arbitrarily large exterior radii for the product, but I'm not sure how to do so at all, without just calculating the coefficients by hand term by term but that seems like a painful process... and I'm not even sure if this is a possible approach or if there is an easier justification for these replacements

Any help would be greatly appreciated!

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  • $\begingroup$ I think you meant to write that the branch cuts are at $[1,\infty)$ and $(-\infty,-1]$. $\endgroup$ – John Doe Jun 1 at 13:26
  • $\begingroup$ Sorry, yes those are the branch cuts of the actual left and right product terms, I was only disclosing which branch of each Logarithm I'm using in each term, but I will add an edit to my post with your addendum. $\endgroup$ – Aneesh Jun 1 at 13:29
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As a multi-valued function, $f(z) = (1 - z)^{-1/3} (1 + z)^{-2/3}$.

$z = \infty$ is not a branch point, so there are three branches which are regular at infinity. With the principal value of the power function, they can be written analytically as $$\frac 1 {e^{i (\pi + 2 \pi k)/3} z \left( 1 - \frac 1 z \right)^{1/3} \left( 1 + \frac 1 z \right)^{2/3}}.$$ The residue at infinity is $-e^{-i (\pi + 2 \pi k)/3}$. Your choice of the logarithms corresponds to $k = 0$.

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  • $\begingroup$ So is this final expression $f(z)$? or is it $\frac{1}{z^2}f(1/z)$? $\endgroup$ – Aneesh Jun 5 at 14:42
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    $\begingroup$ It's $f(z)$. $ $ $\endgroup$ – Maxim Jun 5 at 15:04
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You can use some log rules to change the form of the logarithm. (I replaced $Re^{i\theta}$ with $r$ here, since the complex bit really doesn't affect anything). (There are some extra $2k\pi i$'s missing, they are irrelevant after taking the limit. $$\log(1-r)=-\log\left(\frac1{1-r}\right)=-\log \left(\frac1r\frac1{(1/r)-1}\right)=\log(- r)+\log\left(1-\frac1r\right)$$Then use the Taylor series $$\log(1-x)=x+\frac {x^2}2+\frac{x^3}3+\cdots\qquad|x|<1$$to expand the last term. You'll see that as $r\to\infty$, this term vanishes, and you get just $\log(-r)$, just as you got. (Alternatively you can just let $r\to\infty$ in that second logarithm term, and it clearly goes to $\log 1=0$. I showed the series approach first since you mentioned Laurent series in your question, but I believe taking the limit in the logarithm is much easier).

The same method works for $\log(1+r)$.


Edit in response to comments:

With regards to your comment on whether replacement in the integral is OK, I thought you're evaluating a residue? In that case, you aren't directly replacing it into an integral, rather you're just doing a simplification step when evaluating a limit. But to answer your question anyway, sometimes you can make a replacement (provided with a limit).

For instance$$\begin{align}\lim_{\lambda\to\infty} \int_0^\infty e^{-\lambda f(x)}dx&\sim \lim_{\lambda\to\infty}\lim_{\epsilon\to0}\int_{x_0-\epsilon}^{x_0+\epsilon}e^{-\lambda f(x)} dx\\&=\lim_{\lambda\to\infty}\lim_{\epsilon\to0}\int_{x_0-\epsilon}^{x_0+\epsilon}e^{-\lambda f(x_0)}e^{-\lambda x^2f''(x_0)/2+\cdots}dx\\&\sim\lim_{\lambda\to\infty}\int_{-\infty}^{\infty}e^{-\lambda f(x_0)}e^{-\lambda x^2f''(x_0)/2} dx\end{align}$$ where $f(x)$ is minimised at $x_0\in(0,\infty)$. You can then evaluate this (its a Gaussian integral). You can google asymptotic analysis of integrals to find more.

I don't think this applies in your case, because as I said above, you aren't actually evaluating the integral in this way, you're doing it via complex analysis and residues.

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  • $\begingroup$ Just to clarify, these laws hold for the single valued logarithm with some choice of branch cut up to a term $2k\pi i$ and since each Log is inside an $\exp$ term anyway, this term isnt an issue for us? $\endgroup$ – Aneesh Jun 1 at 16:57
  • $\begingroup$ Actually dont we need to be careful here since although the additional term is inside $\exp$ there is $-\frac{1}{3}$ $\endgroup$ – Aneesh Jun 1 at 17:07
  • $\begingroup$ @Aneesh Well once a branch has been chosen, the extra $2k\pi i$ isn't there. But yes, all these calculations would be done with some specific choice of branch. After this point, the answer follows the same path as however you did it the first time - this answer just shows why the approximation is fine. $\endgroup$ – John Doe Jun 1 at 18:20
  • $\begingroup$ But in the step $Log(1-z) = -Log(\frac{1}{1-z})$ if we choose $Log$ with the positive real axis as our branch, shouldn't this actually read $Log(1-z) = -Log(\frac{1}{1-z}) + i2\pi$? Forgive me if I am confusing things $\endgroup$ – Aneesh Jun 1 at 18:29
  • $\begingroup$ Another question I have is $Log(a) + Log(b) = Log(ab)$ for the $Log$ with branch cut the positive real axis, isn't an identity is it? Surely if $a = \pi + 1$ and $b = \pi$ say, we need $Log(a) + Log(b) - i2\pi = Log(ab)$? Secondly the taylor expansion you use at the end should only work for the principal branch of $Log$ where you can take the maclaurin series at $z=0$ in a disc of radius $1$ at the origin since it is holomorphic in that disc ($Log(1-z)$ has branch cut $[1,\infty)$), but this series expansion doesn't hold for $Log$ with branch cut the positive real axis does it? $\endgroup$ – Aneesh Jun 2 at 16:35

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