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.
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 $$ f = s-\frac{\lambda\pi}{1+\lambda\pi}s=\frac{1}{1+\lambda\pi}s, \;\;\; \lambda\ne -1/\pi. $$