Contour integrals using residues

Question

The question I'm working on is the following: Let $C_R$ be a contour in the shape of a wedge starting at the origin, running along the real axis to $x=R$, then along the arc $0 \leq \theta \leq 2\pi/3$, then back down to the origin along the ray $\theta=2\pi/3$.

Evaluate the limit as $R$ approaches infinity of $\int_{C_R} \frac{dz}{(1+z^3)^3} dz$.

So far I have the integrals set up using parameterization along $C_1,C_2,C_3$ going anticlockwise from the origin respectively. I know the middle integral approaches $0$ as $R$ approaches infinity. And also the first and third integral can be factorised into $(1-\exp(2i\pi/3)) \int_0^R \frac{dx}{(1+x^3)^3}$. I have found the pole to be $\exp(i\pi/3)$ but I am struggling to compute the residue. Any help would be appreciated. Please show full workings if you can as I am finding it all a bit too confusing at the moment.

Answer 1

As you noted, the only pole within the contour is at $z=e^{i \pi/3}$. The residue at this pole may be computed as follows:

$$\lim_{z\to e^{i \pi/3}} \frac1{2!} \frac{d^2}{dz^2}\frac{(z-e^{i \pi/3})^3}{(z^3+1)^3}$$

Now,

$$z^3+1 = (z-e^{i \pi/3})(z+1)(z-e^{-i \pi/3}) = (z-e^{i \pi/3}) [z^2+(1-e^{-i \pi/3}) z - e^{-i \pi/3} ]$$

Thus, the derivative is

$$\begin{align} \frac1{2!}\frac{d^2}{dz^2} \frac1{[z^2+(1-e^{-i \pi/3}) z - e^{-i \pi/3} ]^3} &= -\frac{3}{2} \frac{d}{dz} \frac{2 z+1-e^{-i \pi/3}}{[z^2+(1-e^{-i \pi/3}) z - e^{-i \pi/3} ]^4} \\ &= -\frac{3}{[z^2+(1-e^{-i \pi/3}) z - e^{-i \pi/3} ]^4} \\ &+ 6 \frac{[2 z+1-e^{-i \pi/3}]^2}{[z^2+(1-e^{-i \pi/3}) z - e^{-i \pi/3} ]^5} \end{align}$$

Now plug in $z=e^{i \pi/3}$. The result is a mess that ultimately simplifies to $(5/27) e^{i 4 \pi/3}$.

The integral over the real line is then

$$-i 2 \pi \frac{5}{27} \frac{e^{i \pi/3}}{1-e^{i 2 \pi/3}} = \frac{5 \pi}{27 \sqrt{3}}$$

Answer 2

It's a third order pole, so you need to use the higher order formula if you aren't doing a series expansion.

$$Res(f,c)=\frac{1}{(n-1)!}\lim_{z\rightarrow c}\frac{d^{n-1}}{dz^{n-1}}((z-c)^{n}f(z))$$

So in this case, $$Res(f,e^{\frac{i\pi}{3}})=\frac{1}{2!}\lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{d^{2}}{dz^{2}}\left(\frac{(z-e^{\frac{i\pi}{3}})^{3}}{(1+z^{3})^{3}}\right)\\=\frac{1}{2}\lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{d^{2}}{dz^{2}}\left(\frac{1}{(1+z)^{3}(z-1+e^{\frac{i\pi}{3}})^{3}}\right)\\=\frac{1}{2}\lim_{z\rightarrow e^{\frac{i\pi}{3}}}\frac{-3+3i\sqrt{3}+42ze^{\frac{i\pi}{3}}+42z^{3}}{(1+z)^{5}(z-1+e^{\frac{i\pi}{3}})^{5}}\\=-\frac{5}{27}e^{\frac{i\pi}{3}}$$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ There is another way to evaluate the integral with $\large\tt\mbox{just one pole}$:

\begin{align}&\color{#c00000}{% \int_{0}^{\infty}{\dd x \over \pars{1 + x^{3}}^{3}}}= {1 \over 3}\int_{0}^{\infty}{x^{-2/3} \over \pars{x + 1}^{3}}\,\dd x \\[3mm]&={1 \over 3}\,2\pi\ic\,{1 \over 2!}\,\lim_{x \to \expo{\ic\pi}} \totald[2]{x^{-2/3}}{x} -{1 \over 3}\int_{\infty}^{0} {x^{-2/3}\expo{-4\pi\ic/3} \over \pars{x + 1}^{3}}\,\dd x \\[3mm]&={10\pi\ic \over 27}\,\expo{-8\pi\ic/3} +\expo{-4\pi\ic/3}\color{#c00000}{% \int_{0}^{\infty}{\dd x \over \pars{1 + x^{3}}^{3}}} \end{align}

\begin{align}& \color{#c00000}{\int_{0}^{\infty}{\dd x \over \pars{1 + x^{3}}^{3}}} ={10\pi\ic\expo{-8\pi\ic/3}/27 \over 1 - \expo{-4\pi\ic/3}} ={10\pi\ic \over 27}\,\expo{-2\pi\ic}\, {1 \over \expo{2\pi\ic/3} - \expo{-2\pi\ic/3}} ={10\pi\ic \over 27}\,{1 \over 2\ic\sin\pars{2\pi/3}} \\[3mm]&={5\pi \over 27}\,{1 \over \root{3}/2} \end{align}

$$ \color{#66f}{\large\int_{0}^{\infty}{\dd x \over \pars{1 + x^{3}}^{3}} ={10\root{3} \over 81}\,\pi} $$