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The following theorem is stated in a book:

If $f$ is continuous on the arcs $\gamma_r=\{a+re^{i\theta}\,:\,\theta_1\leq\theta\leq\theta_2\}$ where $a,\theta_1,\theta_2$ don't depend on $r$, and if $\displaystyle\lim_{z\to a}(z-a)f(z)=b$, then $\displaystyle\lim_{r\to0}\int_{\gamma_r}f(z)\ dz=ib(\theta_2-\theta_1).$

It seems very useful so I would like to understand it correctly.

The proof given is just one line of symbols without any words:

$\displaystyle\int_{\gamma_r}f(z)\ dz=i\int_{\theta_1}^{\theta_2}[b+\varepsilon(a+re^{i\theta})]\ d\theta=ib(\theta_1-\theta_2)+o(1).$

I don't understand this proof. Could you explain it to me?

I think I understand the theorem more or less though.

Close to $a$, I have $\displaystyle f(z)=\sum_{k=-\infty}^{-2}c_k(z-a)^k+\frac b{z-a}+\sum_{k=0}^\infty c_k(z-a)^k$. So when I integrate $f$ around $a$, only the $\frac b{z-a}$ term doesn't vanish so I get

$$\int_{\gamma_r}f(z)\ dz=\int_{\gamma_r}\frac b{z-a}\ dz=\int_{\theta_1}^{\theta_2}\frac b{a+re^{i\theta}-a}ire^{i\theta}\ d\theta=ib(\theta_1-\theta_2).$$

But this is more that I was supposed to prove. This is for all $r$ that are small enough, not a limit. So I think this must be wrong, and the proof from the book is the right way to do it. But I don't understand the proof.

I'm also not sure I understand the hypothesis. $f$ clearly must be holomorphic near $a$ because otherwise nothing will work. So why is it additionally assumed that it is continuous on the arcs?

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up vote 1 down vote accepted

Note that for a parameterized curve $C$ given by $z(t)$, we have

$$ \int_C f dz = \int_a^b f(z(t)) \dot{z}(t) dt$$

Your curve is given by $z(t) = a+re^{it}$, so the proof is just carrying out the integration directly (I.e. Without CIF). The $o(1)$ term comes from the epsilon part of the integral. Basically, as $r$ gets small, that term vanishes, but the preceding term does not.

This is a different approach than the proof in your book.

Let $z(t) = a+re^{it}$. Then, $\dot{z}(t) = ire^{it}$. Therefore, $$\int_{\gamma_r} f\ dz = \int_{t_1}^{t_2} f(a+re^{it}) ire^{it}\ dt.$$

Likewise, note that $z = a+re^{it}$ implies $re^{it}=z-a$, so as $z\to a$, $r \to 0$.

Hence, $\lim_{z\to a} (b-a)f(z) = \lim_{r\to 0} re^{it}f(a+re^{it}) = b$. So then we take the limit, and can re-write the integral as

$$\lim_{r\to 0} \int_{\gamma_r} f\ dz = \lim_{r\to 0} i\int_{t_1}^{t_2} f(a+re^{it}) re^{it}\ dt = i\int_{t_1}^{t_2} b\ dt = ib(t_2-t_1).$$

As far as the approach in your book, I want to say that they get the epsilon from the definition of continuity, but I might be wrong. It's not explicit that $f$ is analytic, just that it is continuous on the arc, so the assumption of existence of a power series representation is not guaranteed.

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But I don't understand what the epsilon is! Somehow $f$ disappeared from the integral and I don't understand it... – Bartek Feb 4 '13 at 16:22
I'll edit my answer to make it more explicit :) – Emily Feb 4 '13 at 16:37
I'll continue thinking on exactly how they get their result. – Emily Feb 4 '13 at 17:04

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