Let \(\gamma(t)\) be a piecewise smooth curve in \(\mathbb{C}, t \in[0,1]\). Let \(F(w)\) be a continuous function on \(\gamma\). Show that \(f(z)\) defined by \[ f(z):=\int_{\gamma} \frac{F(w)}{w-z} d w \] is analytic on the complement of the curve \(\gamma\).
Solution: Using Morera
By Morera’s theorem, it suffices to show \(\int_\Delta f(z) \dz = 0\) for all triangles \(\Delta \subseteq \gamma^c\). Claim: \[ \int_\Delta f(z) \dz &= \int_\Delta \int_\gamma {F(w) \over w-z} \dw \dz \\ &= \int_\gamma \int_\Delta {F(w) \over w-z} \dz \dw \\ &= \int_\gamma F(w) \qty{ \int_\Delta {1 \over w-z} \dz} \dw \\ &= \int_\gamma F(w) \cdot 0 \dw \\ &= 0 .\]
That the inner integral is zero follows from the fact that the function \(z\mapsto {1\over w-z}\) is holomorphic on \(\gamma^c\), since it has only a simple pole at \(w\) where \(w\in \gamma\) is fixed.
That the interchange of integrals is justified follows from Fubini’s theorem: these are continuous functions on compact sets, which are uniformly bounded and thus Lebesgue measurable and integrable.
Solution: Using limit definition
The claim is that \(f\) is complex differentiable, thus smooth, thus holomorphic and equal to its Taylor series expansion. The quick justification: \[ \dd{}{z} f(z) &= \dd{}{z} \int_\gamma {F(w) \over w-z}\dw \\ &= \int_\gamma \dd{}{z} {F(w) \over w-z} \dw \\ &= \int_\gamma {F(w) \over (w-z)^2} \dw ,\] where differentiating through the integral is justified since the integrand is a continuous function of \(z\) on \(\gamma\) since \(w\neq z\) on \(\gamma\), and \(\gamma\) is a compact set.
Slightly more rigorously, one can equivalently pass a limit through the integral to show that the defining limit exists: \[ f(z+h) - f(z) &= \int_\gamma {F(w) \over w+h-z} \dw - \int_\gamma {F(w) \over w-z}\dw \\ &= \int_\gamma {(w-z)F(w) - (w+h-z)F(w) \over (w+h-z)(w-z) } \dw \\ &= \int_\gamma F(w) {h \over (w+h-z)(w-z)} \dw \\ &\convergesto{h\to 0} \int_\gamma {F(w) \over (w-z)^2}\dw ,\] since the term involving \(h\) goes to 1.