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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.

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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
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1 Answer 1

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

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

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

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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
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