In the proof of a lemma of the book I study, the derivative of the following integral is calculated :
$\frac{d}{dz}\frac{1}{2\pi} \int_{0}^{2\pi}\operatorname{Re}\left(\frac{re^{i\theta}+z}{re^{i\theta}-z}\right) \log \vert f(re^{i\theta})\vert \ d\theta$
and it is claimed that this derivative is equal to
$\left(\frac{1}{2\pi}\int _{0}^{2\pi}\frac{2re^{i\theta}}{(re^{i\theta}-z)^{2}}\log \vert f(re^{i\theta})\vert \ d\theta\right)$.
I don’t know how to derive this integral according to $z$ because of the real part inside. Can someone help me :)
Attempt at an answer, because there wasn't enough space in the comments, but I don't get the expected answer up to a $1/2$ factor so don't accept it straight away because I'm not sure, I just want to see if it leads somewhere.
For any complex number $\omega$, you can write $\text{Re}(\omega) = \dfrac{1}{2}(\omega + \bar{\omega})$. So I was thinking of using the "fact" that $\dfrac{\mathrm{d}}{\mathrm{d}z}\bar{z} = 0$ (which only makes sense with an appropriate definition of what this differential operator is; because the function $z \mapsto \bar{z}$ is anti-holomorphic, hence non-differentiable). But if you set $\dfrac{\partial}{\partial z} = \dfrac{1}{2}\big(\dfrac{\partial}{\partial x} - i \dfrac{\partial}{\partial y})$, then any anti-holomorphic function gets taken to $0$ by $\dfrac{\partial}{\partial z}$.
So assuming that this is the "convention" taken for $\dfrac{\mathrm{d}}{\mathrm{d} z}$ (and dropping the $\dfrac{1}{2\pi}$ factor for convenience):
\begin{align*} &\dfrac{\mathrm{d}}{\mathrm{d}z} \displaystyle\int_0^{2\pi} \text{Re}\big(\dfrac{re^{i\theta} + z}{re^{i\theta}-z}\big) \log|f(re^{i\theta})|\mathrm{d}\theta \\ =&\dfrac{1}{2}\displaystyle\int_0^{2\pi}\dfrac{\mathrm{d}}{\mathrm{d}z}\big(\dfrac{re^{i\theta} + z}{re^{i\theta}-z} + \dfrac{re^{-i\theta} + \bar{z}}{re^{-i\theta}-\bar{z}}\big)\log|f(re^{i\theta})|\mathrm{d}\theta \end{align*}
(we differentiate under the integral sign because the integrand is smooth). So, still assuming that $\dfrac{\mathrm{d}}{\mathrm{d}z}\bar{z} = 0$, the second term in the parentheses differentiates to $0$. Differentiating the first term gives exactly what we want, except there's a factor of $2$ coming in and cancelling the half, whereas in the expression you're looking for, the $2$ stays.
So either I got something wrong, or there is a typo. Waiting for someone to confirm or deny what I just wrote, so please do not validate yet! :)