I have the following integral transform
$$ f(u) = \int_0^{2\pi} g(t)\cos(u-t)\,dt $$
where I know what $f(u)$ is (I have raw data rather than an analytical form) and I need to reconstruct $g(t)$.
While I can find online the inverse integral transforms for several kernels (e.g.) I can't find anything for this particular kernel.
Does anyone know the solution? Or how to derive it? Perhaps it can be rearranged into a well-known integral transform? Or a paper/textbook where the solution is given?
$$ f(u) = \int_0^{2\pi} g(t)\cos(u-t)\,dt \tag 1$$ $$ f(u) = \cos(u)\int_0^{2\pi} g(t)\cos(t)\,dt +\sin(u)\int_0^{2\pi} g(t)\sin(t)\,dt $$
$\int_0^{2\pi} g(t)\cos(t)\,dt=c_1 \tag 2$
$\int_0^{2\pi} g(t)\sin(t)\,dt=c_2 \tag 3$
$$f(u)=c_1\cos(u)+c_2\sin(u) \tag 4$$
If $f(u)$ is not sinusoidal the equation $(1)$ is not valid. The problem has no solution.
If $f(u)$ is sinusoidal the equation $(1)$ is valid. You can fit the data with the sinusoidal function $(4)$ which gives the values of $c_1$ and $c_2$ on the form of numerical approximates (Regression method).
Then one can find an infinity of functions $g(t)$ which satisfies Eqs.$(2)$ and $(3)$. The solution is not unique.