Application of Frullani integral Show that :
$$\begin{align}
  & \int_{0}^{\pi }{\frac{f\left( \alpha +{{\text{e}}^{xi}} \right)+f\left( \alpha +{{\text{e}}^{-xi}} \right)}{1-2p\cos x+{{p}^{2}}}}\text{d}x=\frac{2\pi }{1-{{p}^{2}}}f\left( \alpha +p \right)\ \ \ \ \ \left| p \right|<1 \\ 
 & \int_{0}^{\pi }{\frac{1-p\cos x}{1-2p\cos x+{{p}^{2}}}}\left\{ f\left( \alpha +{{\text{e}}^{xi}} \right)+f\left( \alpha +{{\text{e}}^{-xi}} \right) \right\}\text{d}x=\pi \left\{ f\left( \alpha +p \right)+f\left( \alpha  \right) \right\} \\ 
\ \ \ \ \left| p \right|<1\end{align}$$
 A: Let $z = e^{ix}$ and $\mathscr{C}$ be the contour $|z| = 1$ in the complex plane.
$$\begin{align}
I(f; \alpha, p ) = & \int_{0}^{\pi} \frac{f( \alpha + e^{xi})+f( \alpha +e^{-xi} )}{1-2p\cos x+{p^2}}dx\\
= & \int_{-\pi}^{\pi}\frac{f( \alpha +e^{xi} )}{1-2p\cos x+{p^2}}dx\\
= & \int_{\mathscr{C}} \frac{f(\alpha + z) }{(z - p)(\bar{z} - p)} \frac{dz}{iz}\\
= & 2\pi\int_{\mathscr{C}} \frac{f(\alpha + z) }{(z - p)(1 - p z)}\frac{dz}{2\pi i}
\end{align}$$
If $f(z)$ is holomorphic over the closed unit ball centered at $\alpha$, then the only
singularity in the integrand is a pole at $z = p$. By Cauchy integral theorem, we have:
$$I(f; \alpha, p) = 2\pi \operatorname{Res}( \frac{f(\alpha+z)}{(z-p)(1-pz)}; p ) = \frac{2\pi}{1-p^2}f(\alpha+p)$$
Notice 
$$\begin{align} J(f;\alpha) 
=& \int_{0}^{\pi} ( f( \alpha + e^{xi})+f( \alpha +e^{-xi} ) ) dx\\
=& \int_{-\pi}^{\pi} f( \alpha + e^{xi} ) dx\\
=& \int_{\mathscr{C}} f(\alpha + z ) \frac{dz}{iz}\\
=& 2\pi \int_{\mathscr{C}} \frac{f(\alpha+z)}{z} \frac{dz}{2\pi i}
\end{align}$$
Under same assumption as before, the only signularity in the integrand is a pole at $z = 0$. By Cauchy integray theorem again, we have:
$$J(f; \alpha ) = 2\pi \operatorname{Res}( \frac{f(\alpha+z)}{z}; 0 ) = 2\pi f(\alpha)$$
From this, we get:
$$\begin{align}
  &\int_{0}^{\pi} \frac{1-p\cos x}{1-2p\cos x+{p^2}}( f( \alpha + e^{xi})+f( \alpha +e^{-xi} ))dx\\
= & \int_{0}^{\pi} \frac12 ( \frac{1-p^2}{1-2p\cos x+{p^2}} + 1 )( f( \alpha + e^{xi})+f( \alpha +e^{-xi} )dx\\
= & \frac12 ((1 - p^2) I(f;\alpha,p) + J(f;\alpha))\\
= & \pi (f(\alpha + p ) + f(\alpha))
\end{align}$$
