Let \(\gamma\) be a smooth curve joining two distinct points \(a, b \in \mathbb{C}\), and let \(g\) be continuous on the range of \(\gamma\). Prove that the function defined by the formula \[ f(z) = \int_\gamma \frac{g(w)\,dw}{w - z} \] is analytic off the range of \(\gamma\). Justify every step.
Erratum: the source does not say what \(g\) is. The corrected statement adds the hypothesis that \(g\) is continuous on the range of \(\gamma\); under this hypothesis the Cauchy-type integral is defined for every \(z\) off that range.
Solution
Choose a smooth parametrization \[ \gamma:[0,1]\longrightarrow\CC \] and write \[ \Gamma=\gamma([0,1]). \]
1. The set \(\Gamma\) is compact, and for every \(z\in\CC\sm\Gamma\) theintegral\[ f(z)=\int_\gamma\frac{g(w)}{w-z}\,dw \]is well defined.
The interval \([0,1]\) is compact and \(\gamma\) is continuous, so \(\Gamma\) is compact. Fix \(z\notin\Gamma\). Then \(w-z\) never vanishes on \(\Gamma\), so \[ w\longmapsto\frac{g(w)}{w-z} \] is continuous on \(\Gamma\). Hence its line integral along the smooth curve \(\gamma\) exists.
2. Fix \(z_0\in\CC\sm\Gamma\) and set\[ \delta=\operatorname{dist}(z_0,\Gamma)>0. \]If \(\abs{\eta}<\delta/2\), then \(z_0+\eta\notin\Gamma\) and\[ \abs{w-z_0-\eta}\geq\delta/2 \]for every \(w\in\Gamma\).
Since \(\Gamma\) is compact and \(z_0\notin\Gamma\), the continuous function \(w\mapsto\abs{w-z_0}\) attains a strictly positive minimum \(\delta\) on \(\Gamma\). For \(w\in\Gamma\) and \(\abs{\eta}<\delta/2\), the reverse triangle inequality gives \[ \abs{w-z_0-\eta} \geq \abs{w-z_0}-\abs{\eta} > \delta/2. \] In particular, \(z_0+\eta\) does not lie in \(\Gamma\).
3. For \(0<\abs{\eta}<\delta/2\),\[ \frac{f(z_0+\eta)-f(z_0)}{\eta} = \int_\gamma \frac{g(w)} {(w-z_0-\eta)(w-z_0)} \,dw. \]
By step 2, both integrals are defined. Linearity of the line integral and the identity \[ \frac1\eta \left( \frac1{w-z_0-\eta} - \frac1{w-z_0} \right) = \frac1{(w-z_0-\eta)(w-z_0)} \] give the formula.
4. As \(\eta\to0\), the integrands in step 3 converge uniformly along\(\Gamma\) to\[ \frac{g(w)}{(w-z_0)^2}. \]
Because \(g\) is continuous on the compact set \(\Gamma\), there is \[ M=\max_{w\in\Gamma}\abs{g(w)}<\infty. \] For \(0<\abs{\eta}<\delta/2\), step 2 gives \[ \begin{aligned} &\abs{ \frac{g(w)} {(w-z_0-\eta)(w-z_0)} - \frac{g(w)}{(w-z_0)^2} }\\ &\qquad= \frac{\abs{g(w)}\abs{\eta}} {\abs{w-z_0-\eta}\abs{w-z_0}^2}\\ &\qquad\leq \frac{2M\abs{\eta}}{\delta^3} \end{aligned} \] for every \(w\in\Gamma\). The right-hand side is independent of \(w\) and tends to zero with \(\eta\), proving uniform convergence.
5. The complex derivative of \(f\) exists at \(z_0\) and satisfies\[ f'(z_0) = \int_\gamma\frac{g(w)}{(w-z_0)^2}\,dw. \]
Let \[ L=\int_0^1\abs{\gamma'(t)}\,dt, \] the finite length of the smooth curve. By step 4, the difference between the line integral in step 3 and the proposed derivative is bounded by \[ L\, \sup_{w\in\Gamma} \abs{ \frac{g(w)} {(w-z_0-\eta)(w-z_0)} - \frac{g(w)}{(w-z_0)^2} }, \] which tends to zero as \(\eta\to0\). Therefore the difference quotient in step 3 converges to the displayed line integral.
6. The function \(f\) is analytic on \(\CC\sm\Gamma\).
The point \(z_0\in\CC\sm\Gamma\) was arbitrary. Step 5 shows that \(f\) has a complex derivative at every point of \(\CC\sm\Gamma\). Since \(\Gamma\) is compact, its complement is open. Hence \(f\) is analytic off the range of \(\gamma\).
7. Q.E.D.
Step 6 is the required conclusion.