Eigenvalues of an integral operator $e^{\cos(x-y)+\beta \cos(x)}$ I would like to find the eigenvalues of the operator
$$T[f]= \int_{0}^{2\pi} e^{2k\cos(x-y)+\beta \cos(x)}f(y)dy.$$ I don't know how to approach it.
For an easier version
$$T[f]= \int_{0}^{2\pi} e^{2k\cos(x-y)}f(y)dy$$ I do an educated guess of $e^{imy}$ and with some algebraic manipulations I arrive at the final result.
But for this one I am stuck.
Uptade:
Following the indications of @Gary I realised that the fourier transform defines a Homomorphisme for convolution and multiplication. Thus I identify that my problem can be recasted as
\begin{equation}
e^{\beta cos(x)}[e^{2kcosx}*f(y)] = \lambda f(x)
\end{equation}
Taking the Fourier transform on both sides, we obtain that
\begin{equation}
\lambda F(\omega)=conv(H(\omega),F(\omega)H_2(\omega)) = \int_{0}^{2\pi}H(\omega-k)F(\omega)H_2(\omega)dk
\end{equation}
Now we have that
\begin{equation}
g(x)= \int_{0}^{2\pi}e^{im\omega}d\omega\int_{0}^{2\pi} H_1(\omega-k)H_2(k)F(k)dk
\end{equation}
after a change of variables $t=\omega-k $ we obtain that
\begin{equation}
g(x)= \int_{0}^{2\pi}e^{im(t+k)}H_1(t)dt\int_{0}^{2\pi} H_2(k)F(k)dk
\end{equation}
so we now have some progress and we see that:
\begin{equation}
g(x)= h_1(x)\int_{0}^{2\pi} H_2(k)F(k)e^{imk}dk
\end{equation}
and so :
\begin{equation}
g(x)= h_1(x)(h_2(x)*g(x))
\end{equation}
which is the stating point. So still stuck
 A: $$\int e^{2k \times \cos(x-y)+\beta \times \cos x} f(y) dy=e^{\beta \times \cos x} \int e^{2k \times \cos(x-y)} f(y) dy  = e^{\beta \times \cos x}  \times  (e^{2k \times \cos(x-y)} \star f(y)).$$ So coming to ur simplified problem:
So taking fourier transform $$\Rightarrow
  g(x) = \int e^{2k \times \cos(x-y)} f(y) dy $$ will give $G(\omega) = H(\omega) F(\omega)$ where $H(\omega) = $ Fourier transform of $e^{2k \times \cos(x-y)}$ and similarly $F(\omega) = $ Fourier transform of $f(x)$. So for eigenfunction: $$G(\omega) = H(\omega) F(\omega) = H(\omega) \delta(\omega-\omega_0) = H(\omega_0)\delta(\omega-\omega_0)\Rightarrow F(\omega) = \delta(\omega-\omega_0)$$ is an eigen function $\Rightarrow f(y) = e^{i \omega y}$ is an eigen function. This is the solution for ur simplified problem. Since its just a convolution, we could have directly told that this is an impulse response of an LTI system and wrote the answer.
Now ur original problem in Fourier transform is of the form: $G(\omega) = \operatorname{conv}(H_1(\omega),(H(\omega) F(\omega))$ where $H_1(\omega)$ is the Fourier transform of $e^{\beta \times \cos x}$. Now assuming $\beta=2k$, we get: $G(\omega) = \operatorname{conv}(H(\omega),(H(\omega) F(\omega))$. Let me know if this is useful. Did you solve the simplified problem? It is not clear from what you wrote.
If $f(x)$ is an eigenfunction with eigenvalue $\lambda$ then it can be shown that $f(-x)$ is also an eigen function with same eigen value as follows:
$$g(x) = \int e^{2k \times \cos(x-y)+\beta \times \cos x} f(-y) dy$$
$$g(x) = \int e^{2k \times \cos(x+y)+\beta \times \cos x} f(y) dy$$
$$g(-x) = \int e^{2k \times \cos(-x+y)+\beta \times \cos -x} f(y) dy$$
$$g(-x) = \int e^{2k \times \cos(x-y)+\beta \times \cos x} f(y) dy$$
$$g(-x) = \lambda f(x)$$
$$g(x) = \lambda f(-x)$$
So wlog one can assume $f$ to be either even function or an odd function.
