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I have $ g = e^{iz} / z^{1/3} $ and I have to calculate the integral of g' over a circumference of center (0,0) and radius $ \pi $ , oriented counterclockwise.

I really have no idea how to do this job. I tries using the residues theorem assuming

Integral = $ 2\pi i \sum_i^N Res(g',z_i)$ but the functions seems to have no poles.

I have the correct result which is $i \sqrt3/ \pi^{1/3}$

Can you please help me? :(

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    $g(z)$ has a singularity at the origin, hence $g'(z)$ has a singularity at the origin too. And I guess it should be $g(z)=e^{i\color{red}{z}} / z^{1/3}$. – 2017-01-24
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    Consider what happens by approximating $e^{iz}$ with $1+iz+\frac{1}{2}(iz)^2$ and computing the integral through a change of variable. – 2017-01-24
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    Yes it was z, I'm sorry for the mistake. Anyway the derivate has a z^(4/3) so I don't know how to calculate the residue. May I consider it as a order 4/3 pole? – 2017-01-24
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    Everything you need to do is to compute $$\oint_{\|z\|=\pi}z^{\alpha}\,dz $$ where $\alpha\in\mathbb{Q}$, and combine such result with a Taylor series, namely the Taylor series of the exponential function. – 2017-01-24
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    You may also consider that the substitution $z=u^3$ gives the integral of a meromorphic function over an arc of a circle. – 2017-01-24

2 Answers 2

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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{equation} \bbx{\ds{\oint_{\verts{z} = \pi}\totald{}{z}\pars{\expo{\ic z} \over z^{1/3}}\,\dd z = \ic\oint_{\verts{z} = \pi}z^{-1/3}\expo{\ic z}\,\dd z - {1 \over 3}\oint_{\verts{z} = \pi}z^{-4/3}\expo{\ic z}\,\dd z}} \label{1}\tag{1} \end{equation}

Lets choose the $\ds{z^{-\eta/3}}$-branch cut as follows: $$ z^{-\eta/3} = \verts{z}^{-\eta/3}\exp\pars{-\,{\eta \over 3}\,\mrm{arg}\pars{z}\ic}\,; \qquad -\pi < \,\mrm{arg}\pars{z} < \pi\,,\qquad z \not= 0\,;\quad\eta = 1,4. $$

Then, \begin{align} \oint_{\verts{z} = \pi}{\expo{\ic z} \over z^{1/3}}\,\dd z & = -\int_{-\pi}^{0}\expo{\ic x}\pars{-x}^{-1/3}\expo{-\pi\ic/3}\,\dd x - \int_{0}^{-\pi}\expo{\ic x}\pars{-x}^{-1/3}\expo{\pi\ic/3}\,\dd x \\[5mm] & = -\expo{-\pi\ic/3}\int_{0}^{\pi}{\expo{-\ic x} \over x^{1/3}}\,\dd x + \expo{\pi\ic/3}\int_{0}^{\pi}{\expo{-\ic x} \over x^{1/3}}\,\dd x = \root{3}\ic\int_{0}^{\pi}{\expo{-\ic x} \over x^{1/3}}\,\dd x \\[5mm] & = \root{3}\int_{0}^{\pi}{\sin\pars{x} \over x^{1/3}}\,\dd x + \root{3}\ic\int_{0}^{\pi}{\cos\pars{x} \over x^{1/3}}\,\dd x \\[5mm] & = \bbx{\ds{-\root{3}\ic \bracks{-\mrm{Ci}\pars{{2 \over 3},\pi} + \mrm{Si}\pars{{2 \over 3},\pi}\ic}}} \label{2}\tag{2} \end{align}

where $\ds{\mrm{Ci}}$ is a cosine integral function and $\ds{\mrm{Si}}$ is a sine integral function.

Similarly $\ds{\pars{~\mbox{an indent around}\ z = 0\ \mbox{is included}~}}$, \begin{align} \oint_{\verts{z} = \pi}z^{-4/3}\expo{\ic z}\,\dd z & = -\int_{-\pi}^{-\epsilon}\expo{\ic x}\pars{-x}^{-4/3}\expo{-4\pi\ic/3}\,\dd x - \int_{\pi}^{-\pi}{1 \over \epsilon^{4/3}\expo{4\ic\theta/3}}\,\epsilon\expo{\ic\theta}\ic\,\dd\theta \\[5mm] & - \int_{\epsilon}^{-\pi}\expo{\ic x}\pars{-x}^{-4/3}\expo{4\pi\ic/3}\,\dd x \\[1cm] & = -\expo{-4\pi\ic/3}\int_{\epsilon}^{\pi}{\expo{-\ic x} \over x^{4/3}}\,\dd x + {3\root{3} \over \epsilon^{1/3}}\ic + \expo{4\pi\ic/3}\int_{\epsilon}^{\pi}{\expo{-\ic x} \over x^{4/3}}\,\dd x \\[5mm] & = -\root{3}\ic\int_{\epsilon}^{\pi}{\expo{-\ic x} \over x^{4/3}}\,\dd x + {3\root{3} \over \epsilon^{1/3}}\ic = 3\root{3}\ic\int_{x = \epsilon}^{x = \pi}\expo{-\ic x}\,\dd\pars{x^{-1/3}} + {3\root{3} \over \epsilon^{1/3}}\ic \\[5mm] & = 3\root{3}\ic\expo{-\ic\pi}\pi^{-1/3} - 3\root{3}\ic\expo{-\ic\epsilon}\epsilon^{-1/3} + {3\root{3} \over \epsilon^{1/3}}\ic \end{align} When $\ds{\epsilon \to 0^{+}}$, \begin{equation} \bbx{\ds{\oint_{\verts{z} = \pi}z^{-4/3}\expo{\ic z}\,\dd z = -\,{3\root{3} \over \pi^{1/3}}\,\ic}}\label{3}\tag{3} \end{equation}
With \eqref{1}, \eqref{2} and \eqref{3}: $$\bbx{\ds{% \oint_{\verts{z} = \pi}\totald{}{z}\pars{\expo{\ic z} \over z^{1/3}}\,\dd z = \root{3} \bracks{-\mrm{Ci}\pars{{2 \over 3},\pi} + \mrm{Si}\pars{{2 \over 3},\pi}\ic} + {\root{3} \over \pi^{1/3}}\,\ic}} $$

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    I had to integrate g'(z) – 2017-01-24
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    @PeppeDAlterio Sorry. I see the top of the question but I didn't realize it was $g'$ the involved function. I'll recheck a little bit later. – 2017-01-24
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    thank you. I'm waiting for you :) – 2017-01-24
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    @PeppeDAlterio Done. – 2017-01-26
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I will perform the computation of the harder integral $\oint_{\|z\|=\pi}g(z)\,dz$.

Assuming that $\frac{1}{z^{1/3}}$ stands for the principal branch of the cubic root, we have: $$ I=\oint_{\|z\|=\pi}\frac{e^{iz}}{z^{1/3}}\,dz = \sum_{n\geq 0}\frac{i^n}{n!}\oint_{\|z\|=\pi}z^{n-1/3}\,dz = \sum_{n\geq 0}\frac{i^n \pi^{n+2/3}}{n!}\int_{-\pi}^{\pi}i e^{i(n+2/3)\theta}\,d\theta$$ through a Taylor expansion and the substitution $z=\pi e^{i\theta}$. That leads to: $$ I = \sum_{n\geq 0}\frac{i^n \pi^{n+2/3}}{n!}\cdot\left.\frac{3}{3n+2}e^{(n+2/3)i\theta}\right|_{-\pi}^{\pi}$$ that simplifies to: $$ I = \sum_{n\geq 0}\frac{i^{n+1} \pi^{n+2/3}}{n!}\cdot\frac{3\cdot 2 \sin((n+2/3)\pi)}{3n+2}=\color{red}{3i \pi^{2/3}\sqrt{3}\sum_{n\geq 0}\frac{(-i\pi)^n}{n!(3n+2)}}.$$

I leave to you to perform the same steps for $g'(z)$, or just understand that $g(z)$ leads to a simplified series, that actually is a value of the complex exponential function.