Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Good day! Help me understan what's I'm wrong Consider a function $f$ that is holomorphic in the unit disk $|z|\le 1$. Prove that $$ \int\limits_{0}^{1} f(x)\,dx = \int\limits_{|z|=1} f(z) \log (z)\,dz,$$ where we choose the branch of the logarithm that take real values on the positive ray in the real line.

What I was doing: $$\int\limits_{|z|=1} f(z) \log (z)\,dz=\{ z=e^{2\pi i t} \}=2\pi i\int\limits_{0}^{1} f(e^{2\pi it}) \log(e^{2\pi it}) e^{2\pi it}\,dt=-4\pi^2 \int\limits_{0}^{1} f(e^{2\pi it})t e^{2\pi it}\,dt=$$ $$=-\left.\frac{2\pi}{i} f(e^{2\pi it})t e^{2\pi it}\right|_{0}^{1}+\frac{2\pi}{i}\int\limits_{0}^{1} e^{2\pi it} \bigl(f(e^{2\pi it})\bigr)'t\,dt+\frac{2\pi}{i}\int\limits_{0}^{1} e^{2\pi it}f(e^{2\pi it})\,dt=\frac{2\pi}{i} \int\limits_{0}^{1} e^{2\pi it} f(e^{2\pi it})\,dt=$$

Next, doing the inverse change: $z=e^{2\pi it}$ and $$=-\int\limits_{|z|=1} f(z)\,dz.$$

It's so strange. And what I wanted to prove to not work.

share|cite|improve this question
Since you've "deleted" the non-positive real axis and $\,\log(e^{2\pi it})\;$ gets there when $\,t=0.5\,$, I think your middle integral in the first line of your calculations has a problem...More so when, after the whole integration interval has been done, the argument of the logarithm has increased by $\,2\pi\,$ ... – DonAntonio Jul 2 '13 at 10:33
Is it possible the goal is to show that $2 \pi i \int \limits _0^1 f(x)\text{ }dx = \int \limits _{|z| = 1}f(z) \text{ }dz$ ? – bryanj Jul 2 '13 at 11:34
up vote 1 down vote accepted

Let's try to show that

$$2 \pi i \int \limits _0^1 f(x)\text{ }dx = \int \limits _{|z| = 1}f(z) \text{ }dz$$

Let $\gamma_{\epsilon}$ be a keyhole contour of radius 1, and with the horizontal segments on the non-negative real axis. Use the branch of log$(z)$ where $0 \lt \text{arg}(z) \lt 2 \pi$.

Because $f(z)$ is holomorphic on and inside the contour, we get $$ \int \limits _{ \gamma_{\epsilon} } f(z) \text{ }dz = 0 $$ for all $\epsilon$. Notice that the contributions from the segments near the positive real axis are close to: \begin{eqnarray} \int \limits _0^1 f(x) \text{log}(x) \text{ }dx \hspace{1cm}\text{ for the portion "above" the real axis where arg$(z) \approx 0$ } \\ -\int \limits _0^1 f(x) (\text{log}(x) + 2 \pi i)\text{ }dx \hspace{1cm}\text{ for the portion "below" the real axis where arg$(z) \approx 2 \pi i$ } \end{eqnarray}

In the limit as $\epsilon \to 0$ the sum of these goes to $- 2 \pi i \int \limits _0^1 f(x)\text{ }dx$, and so you obtain

$$- 2 \pi i \int \limits _0^1 f(x)\text{ }dx + \int \limits _{|z| = 1}f(z) \text{ }dz = 0$$

$$2 \pi i \int \limits _0^1 f(x)\text{ }dx = \int \limits _{|z| = 1}f(z) \text{ }dz$$

share|cite|improve this answer
Reiterative. Need to edit. But what we do with the "circle" integral? in normal situations makes Jordan's lemma.. – Vasili_Petrov Jul 2 '13 at 12:35
The radius of the almost circular part (there's a small gap near $z = 1$) circular does not change - the circular part is always taken over $|z| = 1$. There's no need to use Jordan's lemma. – bryanj Jul 2 '13 at 15:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.