$f$ is continuous on $\gamma$ which is smooth & bounded. Let F be a function such that $F=\int_\gamma \frac{f(\beta)}{\beta-z} d\beta$ for all $z$ not in $\gamma$
Show F is holomorphic at $z$ not on the curve and that
F^'(z)=\int_\gamma \frac{f(\beta)}{(\beta-z)^2} d\beta
I used the basic definition:
F^'(z)=\lim_{h\to0}\frac{F(z+h)-F(z)}{h} $=\lim_{h\to0}\frac{\int_\gamma \frac{f(\beta)}{\beta-(z+h)} d\beta-\int_\gamma \frac{f(\beta)}{\beta-z} d\beta}{h}$ $=\lim_{h\to0}\int_\gamma \frac{(f(\beta)(\beta-z)-f(\beta)(\beta-z-h)}{h(\beta-z-h)(\beta-z)} d\beta$ $=\lim_{h\to0}\int_\gamma \frac{f(\beta)}{(\beta-z-h)(\beta-z)} d\beta$ $=\int_\gamma \frac{f(\beta)}{(\beta-z)^2} d\beta$
Just wondering is what I done correct (since it seems a bit trivial)? If so is it necessary that the curve is smooth & bounded, and that $f$ is continuous on it?