Find $f(t)$ such that $ int_0^{2 pi} f(t + theta) ln(2 sin frac{t}{2}) dt = frac{e^{i theta}}{e^{-i theta} - a e^{i phi}} $

This is a straightforward Fourier series exercise using

$g(t)=ln(2 sin frac{t}{2})=-sum_{n ge 1}frac{cos nt}{n}=-sum_{n ge 1}frac{e^{int}+e^{-int}}{2n}, 0<t< 2pi$,

while $g$ is absolutely integrable on $[0,2pi]$

Let $w=ae^{iphi}, |w|<1$ so the RHS is $e^{2itheta}sum_{k ge 0}e^{iktheta}w^k$

so we look for an $f(t)=sum _{k in mathbb Z}a_ke^{ikt}$ for which the integral equation is satisfied; we assume we can integrate term by term to get what $a_k$ would be good and then we show that the $f(t)$ we get works (so indeed the integration t6erm by term is valid)

(note that $f$ is not unique as any absolutely convergent sine series will integrate $g$ to zero as term by term integration is allowed by the absolute integrability of $g$, the dominated convergence theorem and integrability term by term of Fourier series - see below, while one could start by producing an $f$ out of thin air so to speak and show it works)

An easy computation (under our assumption about $f$ and integration term by term so only diagonal terms remain of course) gives:

$a_0=0, -pi (a_1+a_{-1})=0, -pi (frac{{a_k}+a_{-k}}{k})=w^{k-2}, k ge 2$

In particular one can choose $a_k =0, k le 1, a_k=-frac{kw^{k-2}}{pi}, k ge 2$ so a tentative $f$ would be:

$f(t)=-frac{1}{pi}sum_{k ge 2}kw^{k-2}e^{ikt}=-frac{1}{pi w^2}sum_{k ge 1}kw^{k}e^{ikt}+frac{e^{it}}{pi w}=-frac{e^{it}}{(1-we^{it})^2pi w}+frac{e^{it}}{pi w}$

Note that $f$ has an absolutely convergent series with partial series bounded by $frac{1}{(1-a)^2pi a}+frac{1}{pi a}$ so by the Lebesgue dominated convergence theorem, one can integrate term by term the series of $f$ against $g$ (which was absolutely integrable!); but now for a fixed $k ge 1$, $a_ke^{ikt}g(t)$ is also an integrable function and has Fourier series the obvious one obtained by shifting the terms of $g$ by $k$, so the integral of $a_ke^{ikt}g(t)$ is indeed the needed one done term by term as only the free term of $a_ke^{ikt}g(t)$ survives and we are done!