Find $f(t,w)$ such that $\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = \ln|w - \delta(\theta)|$ Problem
As part of a 2d Stokes flow fluid simulation I'm working on I'm trying to find a $f(t, w)$ such that
$$\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = \ln|w - \delta(\theta)|, \forall \theta \in [0, 2\pi)$$
where $w \in \mathbb{C}, |w| < 1$ and $\delta(t)$ is just a complex finite fourier series: $$\delta(t) = \sum_{n=-N}^N c_n e^{int}, c_n \in \mathbb{C}$$
Partial solutution
If $\delta(t) = e^{it}$ I have that
$$f(t, w) = \frac{1}{\pi} Re\Big(\frac{1}{1 - w e^{-it}}\Big)$$
to give some flavor of what $f(t,w)$ might look like (this value for $f(t,w)$ is not unique).  I'm trying to generalize this result to a wider class of $\delta(t)$, ideally all finite Fourier series but maybe it's only possible for a certain class, I'm not sure.
Possible approach?
I thought maybe I could factor $\delta(t) - \delta(\theta)$ as $a e^{-iNt} \prod_{n=-N}^N (e^{it} - \alpha_n)$.  Then the log term is linearly seperable:
$$ \ln|a e^{iNt} \prod_{n=-N}^N (e^{it} - \alpha_n) | = \ln|a| + \sum_{n=-N}^N \ln|e^{it} - \alpha_n| $$
I can then split the integral into smaller, solvable pieces:
$$\int_0^{2 \pi} f(t, w) \ln|\delta(t) - \delta(\theta)| dt = $$
$$\ln|a| \int_0^{2 \pi} f(t, w) dt + \sum_{n=-N}^N \int_0^{2 \pi} f(t, w) \ln|e^{it} - \alpha_n| dt$$
Assuming $|\alpha_n| = 1$, (which is a big assumption!), I think I can use my solution for $\delta(t) = e^{it}$ and a little bit of linear algebra to build up a more complex solution.  But the $\theta$ term disappears in the factorization (I have to pick some arbitrary value of $\theta$ to even do the factorization).  Maybe that's okay?  My solution when $\delta(t) = e^{it}$ should hold for all $\theta$ so maybe picking a value for $\theta$ arbitrarily is okay?  It's unclear to me and I'm not sure how to justify or refute the idea.
Any help would be appreciated.  Mostly I'm out of tools in my toolbox and I'm not sure if I'm on the right track or if this is even possible in principle.
 A: It's a bit unfortunate that you have named your function $\delta$ since I'm going to talk about the Dirac delta distribution, normally also denoted by $\delta$. But I will use $u$ for the Dirac delta distribution so that you can keep your notation.
The relation looks quite similar to the definition of a translated Dirac delta distribution $u$:
$$
\int_{-\infty}^{\infty} f(x) \, u(x-a) \, dx = f(a).
$$
So if we allow $f$ to be a distribution we would expect
$$
f(t,w) = u(\delta(t)-w)
$$
to work.
A: I've managed to solve this for a slightly larger class of curves.  The result is interesting so I thought I'd share:
First, generalize $\delta(t) - \delta(\theta)$ to the interior of the unit disk and call it $\gamma(z)$:
$$\gamma(z) = \left( \sum_{n=-N}^N c_n z^n \right) - \delta(\theta), c_n \in \mathbb{C}.$$
Basically, $\gamma(e^{it}) = \delta(t) - \delta(\theta)$.  Crucially, assume that $\gamma(z)$ has no roots in the unit disk.  Note that $\gamma(z)$ is analytic and $\ln|\gamma(z)|$ is harmonic.
Find a value $y$ such that $\gamma(y) = w - \delta(\theta)$ and $|y| < 1$, assuming there is one (it helps here to relax the requirements on $w$ from the original question and instead place those requirements on $y$).
Then we can construct a Poisson Integral:
$$ \log|w - \delta(\theta)| = \log|\gamma(y)| = \frac{1}{2 \pi} \int_0^{2\pi} P_r(\phi - t) \log|\gamma(e^{it})|dt, \ \ \ \ y = r e^{i \phi}$$
where $P_r(\theta)$ is the Poisson Kernel and is given by
$$ P_r(\phi) = \Re\left( \frac{1 + r e^{i \phi}}{1 - r e^{i \phi}} \right), \ \ \ \ 0 \leq r < 1.$$
Therefore
$$ \boxed{f(t, w) = \frac{1}{2 \pi} \Re\left( \frac{1 + y e^{-it}}{1 - y e^{-it}} \right)}$$
Obviously there are a lot of limitations here.  $\gamma(z)$ can't have any roots in the unit disk for any value of $\theta$ (if there are roots the Poisson Integral gets some extra terms that I don't know how to account for).  Also you need to be able to find a $y$ in the unit circle, because the Poisson Kernel is only available there.  Otherwise you're free to select $w$ and $\delta(t)$, though.
