If $f$ is holomorphic on the closed unit disc, prove $\int_Cf(z)\log(z)dz=2\pi i\int_0^1f(x)dx$ where $C$ is the unit circle.

Question

If $f$ is holomorphic on the closed unit disc, prove $\int_Cf(z)\log(z)dz=2\pi i\int_0^1f(x)dx$ where $C$ is the unit circle.

Hint is to use integration by part but I can't find a good reference for how to apply that to counter integrals. My first guess would be direct evaluation but I can't get the $\log$ to go away, and likewise with any approach involving Taylor expanding $f(z)$ at $0$. Please give hints not direct solutions.

Answer 1

One approach is to apply Cauchy's theorem to the loop $\gamma = \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$, where:

• $\gamma_1$ is an arc on the unit circle $C$, going anticlockwise from $\sqrt{1 - \epsilon^2}+i\epsilon$ to $\sqrt{1 - \epsilon^2}-i\epsilon$.
• $\gamma_2$ is the straight line segment, going from $\sqrt{1 - \epsilon^2}-i\epsilon$ to $-i\epsilon$
• $\gamma_3$ is an arc on the circle centred at $0$ with radius $\epsilon$, going clockwise from $-i\epsilon$ to $+i\epsilon$.
• $\gamma_4$ is the straight line segment, going from $+i\epsilon$ to $\sqrt{1 - \epsilon^2}+i\epsilon$.

Cauchy's theorem tells us that $\int_\gamma f(z) \log(z) dz = 0$.

(I'm assuming that the branch cut for $\log(z)$ runs along the positive real axis.)

Now consider the limit $\epsilon \to 0$.

• Argue that $\int_{\gamma_1} f(z) \log(z) dz \to \int_C f(z) \log(z)$ in this limit.
• Argue that $\int_{\gamma_3} f(z) \log(z) dz \to 0$ in this limit.
• Argue that $\int_{\gamma_2} f(z) \log(z) dz + \int_{\gamma_4} f(z) \log(z) dz \to - 2\pi i \int_0^1 f(x) dx$ in this limit.

In the third bullet point, the intuition is that fact that $\log(x + i\epsilon) \to \log(x)$ as $\epsilon$ tends to zero from above, but $\log(x - i\epsilon) \to \log(x) + 2\pi i$ as $\epsilon$ tends to zero from above.

It's quite possible that you'll have to amend some of the details as you work through this. However, I'm confident that the overall approach should work.

Edit: I thought a little how one might take these limits rigorously. It seems like the argument might be easier if we introduce two small parameters:

• $r$, the radius of the small circle that $\gamma_3$ wraps around
• $\epsilon$, the vertical separation between $\gamma_2$ and $\gamma_4$.

I'd be inclined to take the $\epsilon \to 0$ limit first, and then take the $r \to 0$ limit.

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Another approach

Here's another approach, based on FShrike's answer, which has been deleted for some reason. @FShrike - if you undelete your answer, I'll upvote.

Since $f$ is holomorphic on a domain containing the closed unit disk, $f$ has an antiderivative $F$ on the closed unit disk.

Let $\gamma_1$ be the same arc as above as in my first approach. Let $a_\epsilon = \sqrt{1 - \epsilon^2} + i\epsilon$ and $b_\epsilon = \sqrt{1 - \epsilon^2} - i\epsilon$ be the endpoints of this arc.

We have $$ \int_{\gamma_1} \frac d {dz}(F(z) \log(z)) dz = F(b_\epsilon) \log(b_\epsilon) - F(a_\epsilon) \log(a_\epsilon)$$

Applying the chain rule, we also have $$ \int_{\gamma_1} \frac d {dz}(F(z) \log(z)) dz = \int_{\gamma_1} f(z) \log(z) dz + \int_{\gamma_1} \frac{F(z)}{z} dz $$

The next step is to take the limit $\epsilon \to 0$ in both of the expressions above, to show that $$ \oint_C f(z) \log(z) dz = 2\pi i (F(1) - F(0)).$$

Finally, show that if $\gamma'$ is any path between $0$ and $1$, then $ F(1) - F(0) = \int_{\gamma'} f(z) dz$. You can derive the result you want by taking a particular choice for $\gamma'$.

I think this method is what is meant by the hint about "integrating by parts". Notice that we've effectively integrated by parts, without using an explicit parametrisation for $\gamma_1$. Which is nice.