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Let $C_r$ be the positively oriented circle centered at origin with radius $2$. Let $f$ be a analytic function on $\{z : |z| > 1\}$ and $\lim_{z \to \infty} f(z)=0.$ I need to show that for $|z|>2,$ one has $$ f(z) = \frac{1}{2\pi i}\int_{C_2} \frac{f(s)ds}{z-s}.$$

I tried like this : Choose large $r$ such that $|z|<r$. We see that $\int_{C_r} \frac{ds}{s-z} = 2i\pi. $ So we can write $$f(z) + \frac{1}{2\pi i}\int_{C_r} \frac{f(s)ds}{z-s}= \frac{1}{2\pi i}\int_{C_r} \frac{f(s)-f(z)}{z-s}ds.$$ Here I am stuck. Mostly, I can use the condition $\lim_{z \to \infty} f(z)=0$ to make RHS go to zero as $r \to \infty$. But I am not able to complete the proof rigoursly.

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    $\begingroup$ I suggest two ways - one is use the Laurent series which is a series in $1/z$ only from the condition at infinity, or somewhat equivalently use $g(z)=f(1/z)$ and show it is analytic on the full unit disc by the condition at infinity and transform the integral and use regular Cauchy $\endgroup$
    – Conrad
    Commented Mar 3, 2023 at 13:38

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Since $\lim_{|z|\to\infty}f(z)$ exists, $f\left(\frac{1}{z}\right)$ has a removable singularity at $0$. Thus it can be extended to an analytic function on $|z|<1$. In particular, for $|s|<1$, we have $$f\left(\frac{1}{s}\right) = \frac{1}{2\pi i}\oint_{|\xi|=1/2}\frac{f(1/\xi)}{\xi-s}d\xi$$ by the Cauchy's integral formula. For $|z|>1$, let $s = 1/z$. Then we have \begin{align*} f(z) &= -\frac{1}{2\pi i}\oint_{|\zeta|=2}\frac{f(\zeta)}{1/\zeta-1/z}\left(-\frac{1}{\zeta^2}\right)d\zeta\\ &= -\frac{1}{2\pi i}\oint_{|\zeta|=2}\frac{zf(\zeta)}{\zeta(\zeta-z)}d\zeta \end{align*} by the change of variable $\zeta = 1/\xi$. Now we show that $$\oint_{|\zeta|=2}\frac{zf(\zeta)}{\zeta(\zeta-z)} - \frac{f(\zeta)}{\zeta-z}d\zeta = 0.$$ This holds because $$\frac{z}{\zeta(\zeta-z)}-\frac{1}{\zeta-z} = \frac{z-\zeta}{\zeta(\zeta-z)} = -\frac{1}{\zeta}$$ and $$-\frac{1}{2\pi i}\oint_{|\zeta|=2}\frac{f(\zeta)}{\zeta}d\zeta = \frac{1}{2\pi i}\oint_{|\xi|=1/2}-\frac{f(1/\xi)}{\xi}d\xi = -f\left(\infty\right) = 0.$$

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