Showing that $e_k$ is an eigenvector Lev $V$ be the space $L^2[-\pi, \pi] $ with the inner product $$\langle f,g\rangle=\int_{-\pi}^\pi f(x) \overline {g(x)} \, dx$$For integers $k$, let $e_k(t)=e^{ikt}$. Consider the operator $K$ on $V$ by $$(Kf)(x)=\int_{-\pi}^\pi cos(x-t)f(t)dt$$ Show that each $e_k$ is an eigenvector for K and find the corresponding eigenvalues. I started by writing the cosine in terms of exponentials. After a bit of elbow grease, I am nowhere near to proving the eigenvalue function $(Kf)(x)=\mu f(x)$ using the $e_k(t)$ above. How do I go about this?

We know that $\cos(x-t)=\frac{e^{i(x-t)}+e^{-i(x-t)}}{2}$. So $$\begin{align} Ke_k(x) &= \int_{-\pi}^\pi \frac{e^{i(x-t)}+e^{-i(x-t)}}{2} e^{ikt}dt \\\ &= \frac{e^{ix}}{2}\int_{-\pi}^\pi e^{i(k-1)t}dt + \frac{e^{-ix}}{2}\int_{-\pi}^\pi e^{i(k+1)t}dt \end{align}$$

Now you can use that $\int_{-\pi}^{\pi} e^{ilt} dt = 0$ if $l\in \mathbb{Z}^*$ and $2\pi$ if $l=0$.