Using Poisson's integral formula The problem asks to prove the following equality using Poisson's integral formula (or Poisson kernel, if I understood correctly from Wikipedia): > $$\int_0^{2\pi} \frac{e^{\cos \phi}\cos(\sin\phi)}{5-4\cos(\theta-\phi)}d\phi=\frac{2\pi}{3}e^{\cos \theta}\cos(\sin\theta)$$ The Poisson formula that is on the book is: > $$f(re^{i\theta})=\frac1{2\pi}\int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos(\theta-\phi)+r^2}f(Re^{i\phi}) d\phi$$ If we consider the circule defined by $|z|=R$, and a point $z=re^{i\theta}$ inside of it. Onto the problem, I have no idea how to do it. I guess from the RHS we get: $$f(re^{i\theta})=\frac{e^{\cos \theta}\cos(\sin\theta)}{3}$$ and from the numerator in the LHS: $$(R^2-r^2)f(Re^{i\phi})=e^{\cos \phi}\cos(\sin\phi)$$ This problem has too many unknowns, how am I suppose to solve this? The problem doesn't give the value of $R$, and $r$ obviously depends on it.

You want $5-4\cos (\theta-\phi) = R^2 - 2Rr\cos (\theta-\phi) + r^2$, that is, $R^2 + r^2 = 5$ and $2Rr = 4$. That makes $(R+r)^2 = R^2 + r^2 + 2Rr = 5+4 = 9$ and $(R-r)^2 = R^2+r^2-2Rr = 5-4 = 1$, so $R+r = 3$ and $R-r = 1$.

Further, it might be helpful to consider

$$e^{\cos \theta} \cos (\sin\theta) = \operatorname{Re} \left(e^{\cos\theta}(\cos(\sin\theta) + i\sin(\sin\theta))\right).$$