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Consider the contour integral: $$I=\oint_\Gamma \frac{1}{\sqrt{z^2-1}}$$ Where $\Gamma$ is a circle at infinity and we have taken the branch cut to be between $z=\pm 1$. Now this function does not have any singularities (other then the branch points) so from Cauchy's residue theorem we would expect: $$I=0$$ However, $$I=2\pi i$$ What is the reason for this contradiction, i.e. why doesn't Cauchy's residue theorem hold in this case?

Edit - derivation of latter result

$$I=\oint \frac{1}{z} \frac{1}{\sqrt{1-z^{-2}}}dz$$ $$=\int^{2\pi}_0 \frac{1}{\sqrt{1-(Re^{i \theta})^{-2}}} i d \theta$$ as $R\rightarrow \infty$ this becomes: $$I=\int^{2 \pi}_0 i d\theta=2 \pi i$$

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  • $\begingroup$ @Dr.MV I meant to say circular so, it is closed (sorry). $\endgroup$ Commented Feb 12, 2016 at 5:33
  • $\begingroup$ how do you define your function for $Im(z) > 0$ ? if you say $\sqrt{r e^{i \theta} } = \sqrt{r} e^{i \theta / 2}$ for $r > 0$ and $\theta \in [0;2\pi[$ that should work $f(z)$ will be holomorphic in the half plane $arg(z) \in ]0;\pi[$ and $\oint_\Gamma f(z) dz = 0$ with $\Gamma$ being a closed contour contained in $arg(z) \in ]0;\pi[$ $\endgroup$
    – reuns
    Commented Feb 12, 2016 at 5:40
  • $\begingroup$ $\int_0^{2\pi}$ is not a semicircle closed contour at all, and .... $\endgroup$
    – reuns
    Commented Feb 12, 2016 at 5:43
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    $\begingroup$ You do realize there's a singularity at infinity? (... that your circle encloses once in the negative direction.) Also, the integral $\int \frac{\mathrm{d}x}{\sqrt{x^2 - a^2}}$ is elementary with antiderivative $\ln | x + \sqrt{x^2 - a^2}| + C$, so you can't avoid logarithmic involvement. What do you get when you enclose one logarithmic singularity? What do you get when you enclose two of them, but each containing a $\sqrt{x}$. (I.e., colorfully: when you enclose two halves of a logarithmic singularity.) $\endgroup$ Commented Feb 12, 2016 at 5:53
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    $\begingroup$ @robjohn While I guess you can say that, talking about "residues of sets" is not standard terminology. $\endgroup$
    – mrf
    Commented Feb 12, 2016 at 12:47

3 Answers 3

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In most complex analysis textbooks, residues are associated with functions, but in order to make sense of the residue at $\infty$, it's better to think of residues associated to $1$-forms. In your case (or in Yiorgos' example), the function is indeed holomorphic at $\infty$, but the form $$ \frac{dz}{\sqrt{1-z^2}} \qquad\big(\text{and } \frac{dz}{z} \big) $$ has a non-zero residue at $\infty$. Intuitively, you can think of this coming from the singularity of $z$ as in $dz$ at $\infty$ (the function $z$ has a pole at $\infty$).

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For the same reason $$ \oint\frac{dz}{z}=2\pi i. $$ The function $f(z)=z^{-1}$ is analytic in $\mathbb C_\infty\setminus\{0\}$.

Cauchy's Residue Theorem holds in simply connected regions in $\mathbb C$. Not in $\mathbb C_\infty$.

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  • $\begingroup$ The residue theorem holds for regions in the sphere, and being simply connected doesn't play a role (beyond making it unnecessary to state that the contour is nullhomologous). $\endgroup$ Commented Feb 12, 2016 at 11:14
  • $\begingroup$ So, what's wrong what my example? $\endgroup$ Commented Feb 12, 2016 at 11:19
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    $\begingroup$ I think you mean Cauchy's Integral Theorem rather than the Cauchy's Residue Theorem. $\endgroup$
    – robjohn
    Commented Feb 12, 2016 at 11:23
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For $|z|\gt1$, $$ \begin{align} \frac1{\sqrt{z^2-1}} &=\sum_{k=0}^\infty\frac1{4^k}\binom{2k}{k}z^{-2k-1}\\ &=\color{#C00000}{\frac1z}+\frac1{2z^3}+\frac3{8z^5}+\cdots \end{align} $$ So the residue of the branch cut from $-1$ to $1$ is $1$.

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