Let $\Gamma$ denote the positively oriented circle of radius $2$ with center at the origin. Let $f$ be an analytic function on $\{z\in\mathbb{C}\ |\ |z| \gt 1\},$ and let $$\lim_{z\to\infty} f(z)=0.$$ Prove that $$f(z)=\dfrac{1}{2\pi i}\int_{\Gamma}\dfrac{f(\zeta)}{z-\zeta}\ d\zeta$$ for all $z\in\mathbb{C}$ with $|z|>2.$
By the given condition it is clear that $\lim\limits_{z \to 0} f \left ( \dfrac 1 z \right ) = 0.$
Now define a function $g : B(0,1) \longrightarrow \Bbb C$ by
$$ g(z) = \left\{ \begin{array}{ll} f \left (\dfrac 1 z \right ) & \quad 0 \lt |z| \lt 1 \\ 0 & \quad z = 0 \end{array} \right. $$
Then $g$ is analytic on $B(0,1).$ Let $\Gamma' (t) = \frac {1} {\Gamma (t)},$ where $t$ varies over the parameter interval of $\Gamma.$ Then $\Gamma'$ is a circle of radius $\frac {1} {2}$ traversing in the clockwise direction. Since $g$ is analytic on and inside of $\Gamma'.$ So by Cauchy's integral theorem it follows that for all $z \in \Bbb C$ with $|z| \lt \frac {1} {2}$ we have $$g(z) = - \dfrac {1} {2 \pi i} \int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - z}\ d\zeta.$$
Hence for all $z \in \Bbb C$ with $|z| \gt 2$ we have $$f(z) = g \left ( \frac 1 z \right ) = - \dfrac {1} {2 \pi i} \int_{\Gamma'} \dfrac { g (\zeta) } {\zeta - \frac 1 z}\ d\zeta.$$
Can it be shown that $$-\displaystyle { \int_{\Gamma'} \dfrac {g(\zeta)} {\zeta - \frac 1 z}\ d\zeta = \int_{\Gamma} \dfrac {f(\zeta)} {z - \zeta}\ d\zeta}$$ for all $z \in \Bbb C$ with $|z| \gt 2\ $?
Any help in this regard will be highly appreciated. Thanks in advance.