Curvilinear integral of a derivate 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? :(
 A: 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.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}:
$$\bbx{\ds{%
\oint_{\verts{z} = \pi}\totald{}{z}\pars{\expo{\ic z} \over z^{1/3}}\,\dd z =
\root{3}
\bracks{-\mrm{Ci}\pars{{2 \over 3},\pi} + \mrm{Si}\pars{{2 \over 3},\pi}\ic} +
{\root{3} \over \pi^{1/3}}\,\ic}}
$$
