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I have to calculate the integral of $g'$ extended to the circumference with center at the origin and radius $\pi$, oriented counterclockwise, having $$ g = \frac{e^{iz}}{z^{1/3}} \ \ \ \ , \ \ arg(z) \in (-\pi,\pi) $$ I thought I could use the Residue teorem, but $ Res(g';0) = 0 $ and the result to this question should be $$ \int_{+\gamma} g' = i \frac{\sqrt3}{\pi^{1/3}} $$

Can somebody please help me?

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    What is $+\gamma$? – 2017-02-22
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    circumference with center at the origin and radius π, oriented counterclockwise – 2017-02-22
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    Is $g'$ the derivative of $g$? – 2017-02-22
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    Yes, it is. Where g is the function given. Anyway, according to the teacher, the solution should be very quick – 2017-02-22
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    Lol, do excuse me, but my complex analysis is not very good. – 2017-02-22
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    I was thinking about Cauchy's integral formula, but I don't think it works... – 2017-02-22
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    Nope, it's the derivate – 2017-02-22
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    Of course you can use the residue calculation. Apply it to calculate the integral. – 2017-02-22
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    @RafaBudría I get 0 as Residual but the result is not 0 :( – 2017-02-22
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    The function can be analytical or not. If not, the residue can be zero or not. But in any case, the contour integral it's proportional to the residue. – 2017-02-22

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I think I got something, but not sure about the correctness of some steps: Let's parametrize $\gamma$ as $z=\pi e^{i\theta}$ with $-\pi\le\theta<\pi$

$$\oint_\gamma g'(z)\mathrm dz=\int_{-\pi}^\pi g'(\pi e^{i\theta})i\pi e^{i\theta}\mathrm d\theta=$$

$$=\left.g\pi e^{i\theta}\right|_{-\pi}^\pi=\left.\frac{\exp({i\pi e^{i\theta}})}{\pi^{1/3}e^{i\theta/3}}\right|_{-\pi}^\pi=$$

$$=\frac{\exp(i\pi e^{i\pi})}{\pi^{1/3}e^{i\pi/3}}-\frac{\exp(i\pi e^{-i\pi})}{\pi^{1/3}e^{-i\pi/3}}=\frac{\exp(-i\pi)e^{-i\pi/3}}{\pi^{1/3}}-\frac{\exp(-i\pi)e^{i\pi/3}}{\pi^{1/3}}=$$

$$=\frac{1}{\pi^{1/3}}\left(-e^{-i\pi/3}+e^{i\pi/3}\right)=$$

$$=\frac{1}{\pi^{1/3}}\left(-\cos\left(\frac{-\pi}{3}\right)-i\sin\left(\frac{-\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)=$$

$$=\frac{1}{\pi^{1/3}}\left(-\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)+\cos\left(\frac{\pi}{3}\right)+i\sin\left(\frac{\pi}{3}\right)\right)=$$

$$=\frac{i\sqrt 3}{\pi^{1/3}}$$