contour integration with dogbone, branch cut: $\frac1{2\pi\mathrm{i}}\int_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z$

Question

Compute the following integral $$\frac1{2\pi\mathrm{i}}\int_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z.$$

Taking a branch of $\sqrt{z^2-1}$, satisfying $\sqrt{z^2-1}>0$ for $z>0$, I tried this problem with a 'dogbone' contour and I get, $$\int_C\frac{\sqrt{z^2-1}}{z-3}\,\mathrm{d}z=-2\int_{-1}^1 \frac{\sqrt{x^2-1}}{x-3}\,\mathrm{d}x,$$ considering the integrations at the branch points are tends to zero as $\epsilon$ goes to zero.

After that, I got stuck because I cannot use the Cauchy integral theorem because the singularity is outside the domain. Please give an idea for this kind of problem. I feel I am wrong, and I want to know the right figure for the contour.

Answer 1

For $R>3$, Cauchy's Integral Theorem guarantees that

$$\begin{align} \oint_{\text{Dogbone}}\frac{\sqrt{z^2-1}}{z-3}\,dz&=\oint_{|z|=2}\frac{\sqrt{z^2-1}}{z-3}\,dz\\\\ &=\oint_{|z|=R}\frac{\sqrt{z^2-1}}{z-3}\,dz-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=3\right)\\\\ &=-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=\infty\right)-2\pi i \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=3\right) \end{align}$$

where the integral around the dog bone contour is taken counter clockwise.

The Residue at Infinity of $f(z)=\frac{\sqrt{z^2-1}}{z-3}$ is equal to the residue at $z=0$ of $-\frac1{z^2}f\left(\frac1z\right)=\frac{\sqrt{1-z^2}}{z^2(3z-1)}$. Therefore, we have

$$\begin{align} \text{Res}\left(\frac{\sqrt{z^2-1}}{z-3},z=\infty\right)&=\text{Res}\left(-\frac1{z^2}\frac{\sqrt{1/z^2-1}}{1/z-3},z=0\right)\\\\ &=\lim_{z\to 0}\frac{d}{dz}\left(\frac{\sqrt{1-z^2}}{3z-1} \right)\\\\ &=-3 \end{align}$$

and the reside at $3$ is $2\sqrt 2$.

Hence, we find that

$$\oint_{\text{Dogbone}}\frac{\sqrt{z^2-1}}{z-3}\,dz=2\pi i (3-2\sqrt 2)$$

where we have tacitly selected the branch of the square root on which $\sqrt{z^2-1}$ is of positive sign when $z\in \mathbb{R}$, $z>1$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ Compute the following integral: $\ds{\bbox[5px,#ffd]{% \int_{\verts{z} = 2}{\root{z^{2} - 1} \over z - 3}\, {\dd z \over 2\pi\ic}}}$. I'll choose de principal branchs of $\ds{\root{z \pm 1}}$. Namely, $$ \left\{\begin{array}{rcl} \ds{\root{z \pm 1}} & \ds{=} & \ds{\root{\verts{z \pm 1}}\expo{\ic\arg\pars{z \pm 1}}} \\[1mm] \ds{\arg\pars{z \pm 1}} & \ds{\in} & \ds{\pars{-\pi,\pi},\qquad z \not= \mp 1} \end{array}\right. $$ The above path $\ds{\braces{z\ \mid\ \verts{z} = 2}}$ doesn't enclose any pole.

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• Once the branch cuts are set in place, we have to add the contributions from paths slightly above and below the cuts.

• Once it's done, the integral is evaluated along a closed contour: It vanishes out.

• Of course, we have to subtract the previous adding $\ds{\pars{~\mbox{see the first}\ \bullet\ \mbox{above}~}}$. \begin{align} &\bbox[5px,#ffd]{% \int_{\verts{z} = 2}{\root{z^{2} - 1} \over z - 3}\, {\dd z \over 2\pi\ic}} = \int_{\verts{z} = 2} {\root{\pars{z + 1}\pars{z - 1}} \over z - 3}\, {\dd z \over 2\pi\ic} \\[5mm] = & \require{cancel} \cancel{-\int_{-2}^{-1}{\pars{\root{-x - 1}\expo{\ic\pi/2}} \pars{\root{1 - x}\expo{\ic\pi/2}} \over x - 3}\, {\dd x \over 2\pi\ic}}\label{1}\tag{1} \\[2mm] & -\int_{-1}^{1}{\root{x + 1} \pars{\root{1 - x}\expo{\ic\pi/2}} \over x - 3}\, {\dd x \over 2\pi\ic} \\[2mm] & -\int_{1}^{-1}{\root{x + 1} \pars{\root{1 - x}\expo{-\ic\pi/2}} \over x - 3}\, {\dd x \over 2\pi\ic} \\[2mm] & \cancel{-\int_{-1}^{-2}{\pars{\root{-x - 1}\expo{-\ic\pi/2}} \pars{\root{1 - x}\expo{-\ic\pi/2}} \over x - 3}\, {\dd x \over 2\pi\ic}}\label{2}\tag{2} \\[5mm] = & -\,{1 \over \pi}\int_{-1}^{1} {\root{1 - x^{2}} \over x - 3}\,\dd x = \bbx{3 - 2\root{2}} \approx 0.1716 \\ & \end{align} Integrals in lines (\ref{1}) and (\ref{2}) cancel each other. This happens because the combined branch cuts leaves a branch cut in $\ds{\bracks{-1,1}}$.