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:
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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 $\gamma_j$.
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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}}\;. $