I am studying Cauchy integrals and their value in points on the integration curve. The principal Cauchy value is defined on these points, however in part of the proof it is used that
$\int_{γ} \frac{1}{|z-z_{0}|^\alpha} dz$
exists, where $γ$ is a smooth open or closed curve, $z_{0}$ it is a point of $γ$ (which is not an extreme if $γ$ is open) and $0<\alpha<1$. I don't understand why this statement is true, does anyone have an idea?
The main point is that the only part of the integral that can give us trouble is in a neighbourhood of $z_0$. So, without loss of generality, we'll consider a parameterisation of $\gamma$ such that $\gamma(0) = z_0$ and $\gamma'(0) \neq 0$.
Now, we can Taylor expand $\gamma$ around $0$ to find that, for some $c$ between $0$ and $t$, \begin{align*} \lvert \gamma(t) - z_0 \rvert &= \biggl\lvert \gamma'(0)t + \frac{\gamma''(c)}{2}t^2 \biggr\rvert \\ &\geq \lvert \gamma'(0)\rvert\lvert t\rvert \biggl(1 - \frac{\lvert \gamma''(c)\rvert}{\lvert \gamma'(0)\rvert} \frac{\lvert t\rvert}{2}\biggr). \end{align*}
Since $\gamma$ is smooth, $\gamma''$ is bounded on $[-1, 1]$, so when $\lvert t \rvert \leq 1$, we have that $\lvert \gamma''(c)\rvert \leq C$ for some $C$ independent of $t$. Hence, let $\epsilon = \min\{\lvert \gamma'(0)\rvert/C, 1\}$ so that, when $\lvert t\rvert < \epsilon$, $$ \lvert \gamma(t) - z_0 \rvert \geq \frac{\lvert\gamma'(0)\rvert}{2} \lvert t\rvert. $$
So finally, we can decompose the integral as \begin{align*} \int_\gamma \frac{1}{\lvert z - z_0 \rvert^\alpha} \mathrm{d}z = \int_I \frac{\gamma'(t)}{\lvert \gamma(t) - z_0 \rvert^\alpha} \mathrm{d}t = \int_{-\epsilon}^\epsilon \frac{\gamma'(t)}{\lvert \gamma(t) - z_0 \rvert^\alpha} \mathrm{d}t + A, \end{align*} where $A$ is the integral taken over the parts of the curve away from $z_0$, and so finite.
But now, taking $D = \sup_{\lvert t \rvert < \epsilon}\gamma'(t)$, we have from our previous bound on $\lvert \gamma(t) - z_0\rvert$ that \begin{align*} \biggl\lvert\int_{-\epsilon}^\epsilon \frac{\gamma'(t)}{\lvert \gamma(t) - z_0 \rvert^\alpha} \mathrm{d}t\biggr\rvert &\leq \frac{2D}{\lvert \gamma'(0) \rvert} \int_{-\epsilon}^\epsilon \frac{1}{\lvert t\rvert^\alpha}\mathrm{d}t, \end{align*} which is finite since $0 < \alpha < 1$.