$f(z)=\frac1{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{z-\zeta}d\zeta$ if $\lim_{z\to\infty}f(z)=0$ and $f$ is analytic in $|z|\ge1$ Let $f$ be analytic in $|z|\ge1$ and  $\lim_{z\to\infty}f(z)=0$. Then is
$f(z)=\frac1{2\pi i}\int_{\Gamma}\frac{f(\zeta)}{z-\zeta}d\zeta$,$\forall |z|\ge2$, where $\Gamma$ is the circle of radius $2$ with centre at the origin. 
I think yes, by the Cauchy integral formula. But, how do we ensure that the non analyticity of $f$ in $|z|<1$ dosent affect the formula? Thanks beforehand. 

Let $|z|\ge 2$. Let $\Gamma'$ be a circle with centre at $z$ and radius $R$ big enough that the circle lies within the region of analiticity of $f$.
Function $f$ is analitical in the region between the circles $\Gamma$ and $\Gamma'$,  and $z$ lies within this region, so, by the Cauchy formula:
$$ \frac{1}{2\pi i}\Big(\oint_{\Gamma'} - \oint_\Gamma\Big)\frac{f(\zeta)}{\zeta-z}d\zeta = f(z) $$
Let's parametrize circle $\Gamma'$ by $\zeta = z+ Re^{i\phi}$. We have
$$ \oint_{\Gamma'} \frac{f(\zeta)}{\zeta-z}d\zeta = \int_0^{2\pi} \frac{f(z+ Re^{i\phi})}{Re^{i\phi}} iRe^{i\phi}d\phi = i \int_0^{2\pi} f(z+ Re^{i\phi}) d\phi$$
since $R$ is arbitrary (as long as it is large enough) we can take the limit $R\rightarrow\infty$. But since $f$ vanishes at infinity, so does this integral. That means that
$$ \frac{1}{2\pi i}\oint_\Gamma\frac{f(\zeta)}{z-\zeta}d\zeta = f(z) $$
