I wish to find a finite form for
$$\int_0^{2\pi} f(e^{i(t+\theta)}) \ln(a - e^{-it}) dt $$
where $f(z)$ is holomorphic in the unit disk and $|a| > 1$.
My first thought was to do a contour integral. For a slightly different integral this works: $$\begin{matrix} \int_0^{2\pi} f(e^{i(t+\theta)}) \ln(a - e^{it}) dt \\ = \oint f(z e^{i\theta}) \ln(a - z) \frac{dz}{iz} \\ = 2 \pi \ln(a) \cdot f(0) \end{matrix}$$
But the same idea fails for the original integral when I get a $\frac{1}{z}$ pole that I can't evaluate or cancel out.
Note that $\log (a-z)-\log a=-\int_0^z\frac{dw}{a-w}$ for a fixed value of $\log a$ and $|z| \le 1 <|a|$ defines uniquely and holomorphically the function $\log (a-z)$ on the closed unit disc (so in a small neighborhood of it also) - here since we do not know $a$ we cannot specify better since for example, $a$ could be a negative real where the principal branch is not defined (in all the other cases we can use the principal branch of course).
Hence $\log (a-z)=\log a-(1/a)\int_0^z\sum_{k \ge 0}\frac{w^k}{a^k}dw=\log a-\sum_{k \ge 1}\frac{z^k}{ka^k}$ or:
$\log (a-e^{-it})=\log a-\sum_{k \ge 1}\frac{e^{-itk}}{ka^k}$
Using the Taylor expansion $f(e^{i(t+\theta)})=\sum_{m \ge 0}c_me^{itm}e^{im\theta}$, we can multiply and integrate term by term by absolute convergence (here the OP implies $f$ has good boundary properties on the unit circle), so we get:
$\frac{1}{2\pi}\int_0^{2\pi} f(e^{i(t+\theta)}) \ln(a - e^{-it}) dt=c_0\log a-\sum_{k \ge 1}\frac{c_ke^{ik\theta}}{ka^k}=f(0)\log a-h(\frac{e^{i\theta}}{a})$ where $h$ is the integral convolution of $f$ with $z/(1-z)$ or if you want $\int_0^z\frac{f(w)-f(0)}{w}dw$