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}} $

Question

I'm trying to find $f(t)$ such that

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

where $a \in (0, 1)$ and $f(t)$ has no $\theta$ terms (but it's fine to have $a, \phi$ terms).

Or, prove that no such $f(t)$ exists.

I've tried various contour integrals using $\ln(2 \sin \frac{t}{2}) = \frac{1}{2} \ln((1 - e^{it})(1 - e^{-it}))$, but nothing either leads me to the answer or convinces me that no such answer exists.

Answer 1

This is a straightforward Fourier series exercise using

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

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

Let $w=ae^{i\phi}, |w|<1$ so the RHS is $e^{2i\theta}\sum_{k \ge 0}e^{ik\theta}w^k$

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:

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

Note that $f$ has an absolutely convergent series with partial series bounded by $\frac{1}{(1-a)^2\pi a}+\frac{1}{\pi a}$ 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!