Example of integration over path on Riemann surface Let $X$ be a Riemann surface
$$
   X = \left\{ (z,w) \in \mathbb{C}^2 \mid z^3 + w^3 = 1 \right\}.
$$
Then we have $z^2 dz + w^2 dw = 0$ and we can define a holomorphic form $\omega$ on $X$ by
$$
   \omega = \left\{ \begin{array}{rl} \frac{dz}{w}, & w \neq 0 \\ -\frac{wdw}{z^2}, & z \neq 0 \end{array} \right.
$$
Let $j = e^{\frac{2 i\pi}{3}}$ and $\gamma$ be a path on $\mathbb{C}_{z}$ given by the following image:
$\hskip4cm $ 
Let $\Gamma$ denote a closed path on $X$ obtained by lifting $\gamma$ and such that $\Gamma$ passes through $(0,1)$, and let $\Gamma_j$ denote a lifting of its part $\gamma_j$. 
$\hskip3cm $ 
We will integrate $\omega$ over $\Gamma$. Let's start from $z = 0$ and move right. We have
$$
   \int\limits_{\Gamma_1} \omega = \int\limits_{0}^{1-\varepsilon} \frac{dt}{(1-t^3)^{1/3}} =: I_{\varepsilon}\;.
$$
Next we compute $\int_{\Gamma_2} \omega$. We can use the parametrisation 
$z = 1 + \varepsilon e^{i \varphi}$, $\phi \in [-\pi,\pi]$ and
$$w = (1-(1+\varepsilon e^{i \varphi})^3)^{1/3} = (-\varepsilon e^{i \varphi}(3 + 3 \varepsilon e^{i \varphi} + \varepsilon^2 e^{2i \varphi}))^{1/3}\;.$$ 
For sufficiently small $\varepsilon$ we have $|w| \geqslant |\varepsilon|^{1/3} C_{1}$ on $\Gamma_2 = \Gamma_2(\varepsilon)$. Then 
$$
   \left| \int\limits_{\Gamma_2} \omega \right| \leqslant \int\limits_{-\pi}^{\pi} \frac{| \varepsilon i e^{i \varphi} | dt}{|w|} \leqslant C_{2} \varepsilon^{2/3} \to 0\text{ when }\varepsilon \to 0.
$$
Using same techniques we obtain that $\int_{\Gamma_5} \omega$ and $\int_{\Gamma_8}\omega$ tend to $0$ when $\varepsilon \to 0$. Now we compute $\int_{\Gamma_3} \omega$. Let's remark that $w = \sqrt[3]{1-z^3} = \sqrt[3]{(1-z)(j-z)(j^2-z)} = \sqrt[3]{1-z}\sqrt[3]{(j-z)(j^2-z)}$. When $z$ goes around $1$ on the curve $\gamma_2$, the first factor's argument increases by $\frac{2 \pi}{3}$ because the argument of $(1-z)$ increases by $2 \pi$. The argument of the second factor in unchanged. Then the integral over $\Gamma_3$ is just
$$
  \int\limits_{\Gamma_3} \omega = \int\limits_{1-\varepsilon}^{0} \frac{dt}{j(1-t^3)^{1/3}} = -j^2 I_{\varepsilon}
$$
since $j^2 = \frac{1}{j}$. To take the integral over $\Gamma_4$ let's remark that $t^3 = (jt)^3$ and then $1-t^3 = 1-(jt)^3$. Then 
$$
   \int\limits_{\Gamma_4} \omega = \int\limits_{0}^{1-\varepsilon} \frac{ j dt}{ j (1-(jt)^3)^{1/3}} = I_{\varepsilon}\;.
$$
Analogously to $\Gamma_3$ we have that $\int_{\Gamma_6} \omega = \int_{\Gamma_9} \omega = -j^2 I_{\varepsilon}$ and $\int_{\Gamma_7} \omega = I_{\varepsilon}$. Since the form $\omega$ is holomorphic on $X$ we must have
$$
   \sum\limits_{k=1}^{9} \int\limits_{\Gamma_k} \omega = I
$$
where $I$ is a so-called period, that doesn't depend on $\varepsilon$. Passing to the limit when $\varepsilon \to 0$ we obtain
$$
   3(1-j^2) \int\limits_{0}^{1} \frac{dt}{(1-t^3)^{1/3}} = I\;.
$$
My question is how to find such period $I$? 
P.S. By the way, using other techniques (not related to Riemann surfaces) we can show that
$$
   \int\limits_{0}^{1} \frac{dt}{(1-t^3)^{1/3}} = \frac{2 \pi}{3 \sqrt{3}}\;.
$$
 A: The path is homotopic in the complement of the branch points $1,j,j^2$ to a large circle $|z|=R$, parametrized in the mathematically positive sense. Since the homotopy does not pass through branch points, it can be lifted to the surface $X$. Since the form is holomorphic, it is closed, so the integral only depends on the homotopy class, and it is equal to $$ \int_{|z|=R} \frac{dz}{\sqrt[3]{1-z^3}} = \int_{|z|=R} \frac{dz}{z\sqrt[3]{z^{-3}-1}} = \frac{1}{e^{\frac{i \pi}{3}}} \int_{|z|=R} \frac{dz}{z \sqrt[3]{1-z^{-3}}}$$
where I hope I got the branch of the third root right, and in the last expression it is the branch that maps $1$ to $1$. Expanding this out in a power series and using the fact that $\int\limits_{|z|=R} z^n \, dz = 0$ for $n \ne -1$ and $\int\limits_{|z|=R} z^{-1} \, dz =2\pi i$ (basically using the residue theorem at $\infty$) gives
$$ \frac{1}{e^{\frac{i \pi}{3}}} \int_{|z|=R} \frac{1}z\left(1-\frac13z^{-3}\pm\ldots\right) dz  = 2\pi ie^{-\frac{i \pi}{3}}$$ and I think this is the same result you got.
