Suppose $f$ is holomorphic inside and on a positively oriented contour $\gamma$ . Let $a$ lie inside $\gamma$

Question

Suppose $f$ is holomorphic inside and on a positively oriented contour $\gamma$ . Let $a$ lie inside $\gamma$ . Show that $f'(a)$ exists and $$f'(a)=\frac{1}{2\pi i} \int _{\gamma} \frac{f(w)}{(w-a)^2} \;dw$$

My Attempt:

i know it enough to show that

$\left| \frac{f(a+h)-f(a)}{h}-\frac{1}{2\pi i} \int _{\gamma} \frac{f(w)}{(w-a)^2} \;dw\right|\rightarrow 0$ as $h\rightarrow 0$

How to prove this ? i think we can also use cauchy's integral formula

Answer 1

You are allowed to differentiate Cauchy's formula under the integral sign: The Cauchy formula is about an integral of the form $$\Phi(a):=\int_\gamma {f(w)\over w-a}\>dw\ ,$$ whereby the integrand is a nice function along $\gamma$, as long as the parameter value $a$ does not come near $\gamma$. Now if a point $a_0$ off $\gamma$ is given the integrand (or its pullback to the $t$-axis) is for each $w\in\gamma$ (resp., each $t\in[a,b]$) a differentiable function of $a$ in the neighborhood of $a_0$. Therefore you can differentiate under the integral sign with respect to $a$ and obtain $$\Phi'(a_0)=\int_\gamma {f(w)\over (w-a_0)^2}\>dw\ .$$

Answer 2

If you're allowed to use the fact that holomorphic functions are analytic, then we have

$$ f(w) = f(a) + f'(a)(w-a) + \frac{1}{2}f''(a)(w-a)^2+\dotsb+\frac{f^{(n)}(a)}{n!}(w-a)^n+\dotsb $$ in a neighborhood of $a$.

Then

$$ \frac{1}{2\pi i}\int \frac{f(w)}{(w-a)^2}\,dw=\\ \frac{f(a)}{2\pi i}\int \frac{1}{(w-a)^2}\,dw + \frac{f'(a)}{2\pi i}\int \frac{1}{w-a}\,dw+\dotsb +\frac{f^{(n)}(a)}{2\pi i n!}\int (w-a)^{n-2}\,dw+\dotsb $$

By the Cauchy integral formula, we know the integral in the second term is 1. The integrands in the subsequent terms are all holomorphic so the integrals vanish. And we can check directly that the first integral vanishes.