# Cauchy integral formula for a unbounded domain

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 $$|z|. We see that $$\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.

• 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 Commented Mar 3, 2023 at 13:38

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 $$-\frac{1}{2\pi i}\oint_{|\zeta|=2}\frac{f(\zeta)}{\zeta}d\zeta = \frac{1}{2\pi i}\oint_{|\xi|=1/2}-\frac{f(1/\xi)}{\xi}d\xi = -f\left(\infty\right) = 0.$$