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If $f$ is analytic everywhere on an annulus, then is the corresponding Cauchy formula for an annulus analytic everywhere on $\mathbb{C}$?$\:\:\:$

$$\frac{1}{2 \pi i} \oint_{C(z;r_2)} \frac{f(w)}{w-z} dw - \frac{1}{2 \pi i} \oint_{C(z;r_1)} \frac{f(w)}{w-z} dw$$

And does it follow that every function analytic on an annulus may be extended to a function analytic on $\mathbb{C}$?

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  • $\begingroup$ If the annulus is $0 < r_1 < |z| < r_2$, then $f=f_2-f_1$, where $f_2$ is analytic in $|z| < r_2$ and $f_1$ is analytic in $r_1 < |z|$. So $f_2 = \sum_{n=0}^{\infty}a_{n}z^{n}$ and $f_1=\sum_{n=1}^{\infty}a_{n}z^{-n}$. $\endgroup$
    – Disintegrating By Parts
    Mar 8, 2014 at 23:43
  • $\begingroup$ The book I am reading seems to imply that if $f$ is a continuous function on some $C(a;r)$, then the following function is analytic on its domain $\mathbb{C} - C(a;r)$. $$ \oint_{C(a;r)} \frac{f(w)}{w-z} dw$$ $\endgroup$
    – Darren
    Mar 8, 2014 at 23:57
  • $\begingroup$ Take $f=1$ defined on $\mathbb{C}$, then subtraction of the two integrals is equal to zero which is obviously equal to the value of function.Am I missing something? $\endgroup$
    – BigM
    Mar 9, 2014 at 0:22
  • $\begingroup$ Actually it isn't. The two integrals will be different for a point inside the annulus. $\endgroup$
    – Darren
    Mar 9, 2014 at 0:28
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    $\begingroup$ @Darren Naylor: The integral you stated in your comment is analytic everywhere in $\mathbb{C}\setminus C(a;r)$. However, the radial limits from inside the contour (if they exist) will not generally equal those from outside the contour. If this integral function extended continuously to $\mathbb{C}$, then you'd have a problem, especially considering that the integral expression vanishes at $\infty$. The jump function across the contour is amazingly general. Look at the contour integral on the unit circle and you'll see a Fourier series expressing the difference under reasonable conditions. $\endgroup$
    – Disintegrating By Parts
    Mar 9, 2014 at 1:04

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ok, to summarize then, I think that the Cauchy integral formula is analytic on each of the three open regions separated by the two concentric circles, but it doesn't follow that it has an extension analytic on $\mathbb{C}$.

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  • $\begingroup$ Correct. A bit more can be said if you consider two integrals separately. One is analytic in $|z|<r_2$, the other in $|z|>r_1$. The intersection of two domains is the original annulus. This question is related. $\endgroup$
    – user127096
    Mar 12, 2014 at 22:11

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