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Given the following integral operator:

$(Au)(y)=\int \limits_{\Gamma}u^*(x,y) u(x) \,ds_x$
with $\Gamma=\{x\in\mathbb{R}^2:x=r\begin{pmatrix}\cos(2\pi t) \\ \sin(2\pi t) \end{pmatrix},0 \leq t < 1 \}$ and $u^*(x,y)=-\frac{1}{2\pi}\ln(\lVert \mathbf{x-y} \rVert)$.

I want to show that given a parametrization $\Gamma$ the eigenfunctions of $A$ are
$v_k(x)=e^{i2\pi kt}$ and I want to find the corresponding eigenvalues.

What I did so far was calculating:
$u^*=-\frac{1}{2\pi}\ln(\lVert \mathbf{x-y} \rVert)=-\frac{1}{4\pi}\ln(r\sqrt{(\cos(2\pi t_x)-\cos(2\pi t_y))^2+(\sin(2\pi t_x)-\sin(2\pi t_y))^2}$
and with $v_k(x)=e^{i2\pi kt}=\cos(2\pi kt_x)+i \sin(2\pi kt_x)$.
After putting everything together i got: $A=\int \limits_0^1 -\frac{1}{4\pi}\ln(r\{(\cos(2\pi t_x)-\cos(2\pi t_y))^2+(\sin(2\pi t_x)-\sin(2\pi t_y))^2\}^\frac{1}{2}(\cos(2\pi kt_x)+i \sin(2\pi kt_x))dt_x$

I also got some hints for calculating the integrals: $\int\limits_0^1 \ln(\sin(\pi t))\sin(2\pi k t) dt=0$ and $\int\limits_0^1 \ln(\sin(\pi t))\cos(2\pi kt)= -\ln(2)$, for $k=0$ and $\int\limits_0^1 \ln(\sin(\pi t))\cos(2\pi kt)=-\frac{1}{2k}$, for $k \geq 1$

I know that after the integration i want to get an equation like $(Av_k)(y)=\lambda v_k(x)$ . But I am just lost right now, I think I am missing the right trigonometric identities.

Thank your for your help!

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