The proof has two parts: analyticity off the contour follows from differentiating under the [integral](/page/Integral) sign (justified because the integrand is uniformly analytic in $z$ away from $\gamma$), and the decay estimate follows from expanding the Cauchy kernel for large $|z|$. **Step 1: Analyticity on $\mathbb{C} \setminus \gamma$.** [claim:[Differentiability](/page/Derivative) Of The [Cauchy Transform](/page/Cauchy%20Transform)] For every $z_0 \in \mathbb{C} \setminus \gamma$, $C$ is complex-differentiable at $z_0$ with \begin{align*} C'(z_0) = \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{(\xi - z_0)^2}\,d\xi. \end{align*} [/claim] [proof] Let $z_0 \in \mathbb{C} \setminus \gamma$ and let $d := \operatorname{dist}(z_0, \gamma) > 0$. For $|h| < d/2$, form the difference quotient: \begin{align*} \frac{C(z_0 + h) - C(z_0)}{h} = \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{h}\left(\frac{1}{\xi - z_0 - h} - \frac{1}{\xi - z_0}\right)d\xi = \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{(\xi - z_0 - h)(\xi - z_0)}\,d\xi. \end{align*} For $\xi \in \gamma$ and $|h| < d/2$, we have $|\xi - z_0| \ge d$ and $|\xi - z_0 - h| \ge d/2$, so \begin{align*} \left|\frac{f(\xi)}{(\xi - z_0 - h)(\xi - z_0)} - \frac{f(\xi)}{(\xi - z_0)^2}\right| = \frac{|f(\xi)| \cdot |h|}{|\xi - z_0 - h| \cdot |\xi - z_0|^2} \le \frac{2\|f\|_\infty |h|}{d^3}, \end{align*} where $\|f\|_\infty := \sup_{\xi \in \gamma}|f(\xi)|$ is finite by [continuity](/page/Continuity) of $f$ on the smooth (and hence compact) contour $\gamma$. Therefore, as $h \to 0$, the integrand [converges uniformly](/page/Uniform%20Convergence) on $\gamma$ to $f(\xi)/(\xi - z_0)^2$, and the [dominated convergence theorem](/theorems/4) gives \begin{align*} C'(z_0) = \lim_{h \to 0}\frac{C(z_0+h) - C(z_0)}{h} = \frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{(\xi - z_0)^2}\,d\xi. \end{align*} [/proof] Since $z_0 \in \mathbb{C} \setminus \gamma$ was arbitrary, $C$ is holomorphic on $\mathbb{C} \setminus \gamma$. The same argument applied inductively shows $C$ is infinitely differentiable with $C^{(n)}(z) = \frac{n!}{2\pi i}\int_\gamma \frac{f(\xi)}{(\xi - z)^{n+1}}\,d\xi$. **Step 2: Decay at infinity.** [claim:Decay Rate] $C(z) = O(1/z)$ as $z \to \infty$. [/claim] [proof] For $|z|$ sufficiently large that $|z| > \sup_{\xi \in \gamma}|\xi| =: M$, expand the Cauchy kernel: \begin{align*} \frac{1}{\xi - z} = \frac{-1}{z}\cdot\frac{1}{1 - \xi/z}. \end{align*} Since $|\xi/z| \le M/|z| < 1$ for all $\xi \in \gamma$, the geometric series converges and $|1/(1 - \xi/z)| \le 1/(1 - M/|z|)$. Therefore \begin{align*} |C(z)| = \left|\frac{1}{2\pi i}\int_\gamma \frac{f(\xi)}{\xi - z}\,d\xi\right| \le \frac{1}{2\pi}\cdot\frac{1}{|z|}\cdot\frac{1}{1 - M/|z|}\int_\gamma |f(\xi)|\,|d\xi| = \frac{\|f\|_{L^1(\gamma)}}{2\pi(|z| - M)}, \end{align*} where $\|f\|_{L^1(\gamma)} := \int_\gamma |f(\xi)|\,|d\xi|$ is finite since $f$ is continuous on the compact contour $\gamma$. For $|z| \ge 2M$ this gives $|C(z)| \le \|f\|_{L^1(\gamma)}/(\pi |z|)$, confirming $C(z) = O(1/z)$. More precisely, the geometric series expansion gives \begin{align*} C(z) = \frac{-1}{2\pi i z}\int_\gamma f(\xi)\,d\xi + O(1/z^2), \end{align*} so the leading-order coefficient is $(-1/(2\pi i))\int_\gamma f(\xi)\,d\xi$. In particular, if $\int_\gamma f\,d\xi = 0$ then $C$ decays as $O(1/z^2)$. [/proof]