Transformation of contour integral $\int \frac{z^2}{e^{2\pi i z^3}-1} \operatorname dz$ over the circle $|z|=\sqrt[3]{n+\frac{1}{2}}$

Question

I would like to solve the following:

$$\int\limits_{|z|=\sqrt[3]{n+\frac{1}{2}}} \frac{z^2}{e^{2\pi i z^3}-1}\operatorname dz$$

I'm given an hint: "use a transformation $w=z^3$"

I would make use of the following theorem:

If $w$ holomorphic on $\Gamma$ and $f$ continous on $w(\Gamma)$ then $$\int\limits_{w(\Gamma)} f(w) \operatorname dw = \int\limits_\Gamma f(w(z)) w'(z) \operatorname dz$$

But I'm a bit confused as to the contour itself. Is the following right?

($\Gamma_1$ is the circle with radius $\sqrt[3]{n+\frac{1}{2}}$ and center $O$, while $\Gamma_2$ had a radius of $n+\frac{1}{2}$)

$$\begin{align} \int_{\Gamma_1} \frac{z^2}{e^{2\pi i z^3}-1}\operatorname dz &= \int_{\Gamma_2} \frac{w^{\frac{2}{3}}}{e^{2\pi i w}-1} \cdot \frac{1}{3}\cdot w^{-\frac{2}{3}}\operatorname dw\\ &= \frac{1}{3} \int_{\Gamma_2} \frac{\operatorname dw}{e^{2\pi i w}-1}\\ \end{align}$$

Using the residue theorem ($\operatorname*{res}_{w=k} f(w) = \frac{-i}{2\pi}$)

$$\int_{\Gamma_1} \frac{z^2}{e^{2\pi i z^3}-1}\operatorname dz = 1$$

Question

This looks to good to be true ;)

Could someone verify? Have I applied the theorem right?

Answer 1

Let $w=z^3$. Then since $\frac1{e^{2\pi iw}-1}$ has a residue of $\frac1{2\pi i}$ at each integer, we get $$ \begin{align} \int_{|z|=\left(n+\frac12\right)^{1/3}}\frac{z^2\,\mathrm{d}z}{e^{2\pi iz^3}-1} &=\frac13\cdot3\int_{|w|=n+\frac12}\frac{\mathrm{d}w}{e^{2\pi iw}-1}\\ &=2\pi i(2n+1)\frac1{2\pi i}\\ &=2n+1 \end{align} $$ since there are $2n+1$ integers inside $|w|=n+\frac12$. The factor of $3$ is because for each time $z$ traces the circle, $w=z^3$ circles it $3$ times.

Answer 2

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\oint_{\verts{z}\ =\ \pars{n + 1/2}^{1/3}} {z^{2} \over \expo{2\pi\ic z^{3}} - 1}\,\dd z:\ {\large ?}}$.

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I'll perform the evaluation without modifying the original contour. It'll use the well known expansion $\ds{{1 \over \expo{w} - 1}=\sum_{n\ =\ -\infty}^{\infty}{1 \over w - 2n\pi\ic}}$.

\begin{align}&\color{#66f}{\large% \oint_{\verts{z}\ =\ \pars{n + 1/2}^{1/3}}{z^{2} \over \expo{2\pi\ic z^{3}} - 1} \,\dd z} =\oint_{\verts{z}\ =\ \pars{n + 1/2}^{1/3}}z^{2}\sum_{m\ =\ -n}^{n} {1 \over 2\pi\ic z^{3} -2\pi m\ic}\,\dd z \\[5mm]&={1 \over 2\pi\ic}\oint_{\verts{z}\ =\ \pars{n + 1/2}^{1/3}} \bracks{{1 \over z} + 2\sum_{m\ =\ 1}^{n}{z^{5} \over z^{6} - m^{2}}}\,\dd z \\[5mm]&=1 + {1 \over \pi\ic}\sum_{m\ =\ 1}^{n} \oint_{\verts{z}\ =\ \pars{n + 1/2}^{1/3}}{z^{5} \over z^{6} - m^{2}}\,\dd z \\[5mm]&=1 + 2\sum_{m\ =\ 1}^{n} \sum_{k\ =\ 0}^{5}\lim_{\atop z\ \to\ m^{1/3}\expo{k\pi\ic/3}} \bracks{\pars{z - m^{1/3}\expo{k\pi\ic/3}}{z^{5} \over z^{6} - m^{2}}} =1 + 2\sum_{m\ =\ 1}^{n}\sum_{k\ =\ 0}^{5}{1 \over 6} \\[5mm]&=1 + 2\sum_{m\ =\ 1}^{n}1=1 + 2\times n=\color{#66f}{\LARGE 2n + 1} \end{align}