Show that the function tends to zero Let be $\gamma_{\epsilon}$ the circle $|z|=\epsilon$, positively oriented, $0<\alpha<1$ and $f$ a entire function. Show that
$$\displaystyle\lim_{\epsilon \to 0}{\displaystyle\oint_{\gamma_{\epsilon}}}{z^{-\alpha}f(z)dz}=0$$
My attempt: Let be $\gamma_{\epsilon}:|z|=\epsilon$, then $\gamma_{\epsilon}:z=\epsilon e^{i\theta}$, $0\leq\theta\leq2\pi$. So,
$\displaystyle\lim_{\epsilon \to 0}{\displaystyle\oint_{\gamma_{\epsilon}}}{z^{-\alpha}f(z)dz}=\int_{0}^{2\pi}{(\epsilon e^{i\theta})^{-\alpha}f(\epsilon e^{i\theta})\epsilon ie^{i\theta}d\theta}=i\epsilon^{1-\alpha}\int_{0}^{2\pi}{e^{i\theta(1-\alpha)}f(\epsilon e^{i\theta})d\theta}$...
But at this point I do not know how to follow.

Since $\epsilon^{1-a}\to 0$ as $\epsilon\to 0$, all you need is an upper bound for the integral, which shouldn't be difficult since the integrand is continuous and so it's modulus attains a maximum on a compact set.
