Given that $f:\mathbb{R} \rightarrow \mathbb{R}$, find $F(\omega)$ such that, $$ f(t) = \int_{-\infty}^{\infty} F(\omega)cos(\omega \hspace{1mm}log(t)) d\omega $$
Also please find $f_k$ such that,
$$ f(t) = \sum_{k=0}^{\infty} f_k \hspace{1mm}cos(k \hspace{1mm} log(t)) $$
The first thing I noticed is that this is a very similar problem to finding the Taylor Series coefficients of $f(t)$ with complex powers of $t$.
$$ f(t) = \sum_{k=0}^{\infty} f_k \hspace{1mm} Re\{e^{k \hspace{1mm} log(t) i}\} = \sum_{k=0}^{\infty} f_k \hspace{1mm} Re\{t^{ki}\} = Re\{ \sum_{k=0}^{\infty} f_k \hspace{1mm} t^{ki}\} $$
This led me to try the same trick of taking the derivative of this expression. However, since this is a polynomial with complex powers, taking the regular derivative is not helpful. Since,
$$ \frac{d}{dt} t^{ki} = t^{ki-1} $$
Maybe try taking an imaginary order derivative https://en.wikipedia.org/wiki/Differintegral#Definitions_via_transforms ?
New Approach:
Define a new function $g(t) = f(t^{-i})$.
$$ g(t) = f(t^{-i}) = Re\{ \sum_{k=0}^{\infty} f_k \hspace{1mm} (t^{-i})^{ki}\} = Re\{ \sum_{k=0}^{\infty} f_k \hspace{1mm} t^{-i^2 k}\} = Re\{ \sum_{k=0}^{\infty} f_k \hspace{1mm} t^k\} = \sum_{k=0}^{\infty} Re \{ f_k \} \hspace{1mm} t^k $$
Now the coefficients we are searching for are the coefficients of $g(t)$.
$$ Re \{ f_k \} = \frac{1}{k!} \big[\frac{d^k}{dt^k} g(t)\big]_{t=0} $$
$$ \therefore Re \{ f_k \} = \frac{1}{k!} \big[\frac{d^k}{dt^k} f(t^{-i})\big]_{t=0} $$