Express the solution of the integral equation in the resolvent form Express the solution of the integral equation $$f(x) = \phi(x)+\lambda\int_0^{2\pi} \cos(x+t)f(t) \, dt$$ in the resolvent form $$f(x) = \phi(x)+\lambda\int_0^{2\pi} \Gamma(x,t;\lambda)\phi(t) \, dt,$$ where $\lambda$ is not an eigenvalue.  Obtain the general solution, if it exists, for $\phi(x)=\sin x$.
I believe the kernel is separable, but other than that I just don't know where to go.
I'm not sure how to go about this at all.  Any help anyone can give would be great.
 A: You can write the resolvent equation using the inner product $(\cdot,\cdot)$ on $L^2(0,2\pi)$:
$$
        f=\phi+\lambda(f,c)c-\lambda(f,s)s,
$$
where $c(x)=\cos(x)$ and $s(x)=\sin(x)$. Using $(c,c)=(s,s)=\pi$ and $(c,s)=0$, begin iterating
\begin{align}
    f=\phi&+\lambda\left(\phi+\lambda(f,c)c-\lambda(f,s)s,c\right)c \\
           &-\lambda(\phi+\lambda(f,c)c-\lambda(f,s)s,s)s \\
   =\phi&+\lambda(\phi,c)c-\pi\lambda^2(f,c)c\\
        &-\lambda(\phi,s)s+\pi\lambda^2(f,s)s\\
   =\phi&+\lambda(\phi,c)c-\pi\lambda^2(\phi+\lambda(f,c)c-\lambda(f,s)s,c)c \\
      &-\lambda(\phi,s)s+\pi\lambda^2(\phi+\lambda(f,c)c-\lambda(f,s)s,s)s \\
   =\phi&+\lambda(\phi,c)c-\pi\lambda^2(\phi,c)c+\pi^2\lambda^3(f,c)c \\
      &-\lambda(\phi,s)s+\pi\lambda^2(\phi,s)s-\pi^2\lambda^3(f,s)s \\
   =\cdots = \\
   =\phi&+\frac{\lambda}{1+\lambda\pi}(\phi,c)c-\frac{\lambda}{1+\lambda\pi}(\phi,s)s,\;\;\; \lambda\ne -1/\pi.
\end{align}
If $\phi=s$, then
$$
   f = s-\frac{\lambda\pi}{1+\lambda\pi}s=\frac{1}{1+\lambda\pi}s,
     \;\;\; \lambda\ne -1/\pi.
$$
