How to find eigenvalues and eigenfunctions of operator $A$ in $L^2[-\pi;\pi]$ $Ax(t)=\int_{-\pi}^{\pi}\sum_{n\in\mathbb{Z}}3^{-|n|}e^{-in(t-s)}x(s)ds$
Here is my attempt:
I've written equation for eigenfunctions:
$\int_{-\pi}^{\pi}\sum_{n\in\mathbb{Z}}3^{-|n|}e^{-in(t-s)}x(s)ds=\lambda x(t)$
At first I was trying to create differential equation to guess how do eigenfunctions look like.
I know what'll happen if I differentiate this $k$ times:
$\int_{-\pi}^{\pi}\sum_{n\in\mathbb{Z}}(-in)^{k}3^{-|n|}e^{-in(t-s)}x(s)ds=\lambda x^{(k)}(t)$, but this doesn't seem to help.
I also noticed that if $x(t)$ is eigenfunction, for every $m\in\mathbb{Z}$ $x(t)e^{imt}$ is also eigenfunction.
What else can I do here?
 A: Consider isometric isomorphism $U:L_2[-\pi,\pi]\to l_2(\mathbb{Z})$, which assigns to any element $x\in L_2[-\pi,\pi]$ its Fourier coefficients of the expansion in an orthonormal basis $\dfrac{e^{ikt}}{\sqrt{2\pi}}$, $k\in \mathbb{Z}$. Then we have that $$(UAx)(k)=\left(Ax,\dfrac{e^{ikt}}{\sqrt{2\pi}}\right)=\int\limits_{-\pi}^\pi\left(\int\limits_{-\pi}^{\pi}\sum_{n\in\mathbb{Z}}3^{-|n|}e^{-in(t-s)}x(s)ds\right)\dfrac{e^{-ikt}}{\sqrt{2\pi}}dt=$$$$=\dfrac{1}{\sqrt{2\pi}}\int\limits_{-\pi}^\pi\sum_{n\in\mathbb{Z}}3^{-|n|}\left(\int\limits_{-\pi}^{\pi}e^{-int}e^{-ikt}dt\right)e^{ins}x(s)ds=$$$$=\sqrt{2\pi}\int\limits_{-\pi}^\pi 3^{-|k|}e^{-iks}x(s)ds=2\pi\cdot3^{-|k|}\left(x,\dfrac{e^{ikt}}{\sqrt{2\pi}}\right)=(TUx)(k),$$
where $Tx=(2\pi\cdot3^{-|k|}x(k))$, $k\in\mathbb{Z}$. Thus, the operator $A$ is equivalent to the diagonal operator $T$, whose eigenvalues are well known and equal to $\lambda_k=2\pi\cdot3^{-|k|}$, $k\in\mathbb{Z}$. The corresponding eigenfunctions are $x_k=(...,0,1,0,...)$, were $1$ stands in $k$-th place. Then $A$ has the same eigenvalues and its eigenfunctions must have Fourier coefficients equal to $x_k$. Thus, the eigenfunctions of the operator $A$ are equal to $e^{ikt}$.
