equality of complex integrals 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.
 A: 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
$$- 2 \pi i \int \limits _0^1 f(x)\text{ }dx + \int \limits _{|z| = 1}f(z) \text{ }dz = 0$$
$$2 \pi i \int \limits _0^1 f(x)\text{ }dx = \int \limits _{|z| = 1}f(z) \text{ }dz$$
