1Consider integrating \,\phi(\alpha)=\int_0^\pi\,\ln(1-2\alpha\cos(x)+\alpha^2)\;dx\,
now,
\begin{align} \frac{d}{d\alpha}\,\phi(\alpha)\, &=\int_0^\pi \frac{-2\cos(x)+2\alpha}{1-2\alpha \cos(x)+\alpha^2}\;dx\, \\ &=\frac{1}{\alpha}\int_0^\pi\,\left(1-\frac{(1-\alpha)^2}{1-2\alpha \cos(x)+\alpha^2}\,\right)\,dx\, \\ &=\frac{\pi}{\alpha}-\frac{2}{\alpha}\left\{\,\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\right\}\,\bigg|_0^\pi. \end{align}
As x varies from 0 to \pi, \left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,varies through positive values from 0 to infinity when -1<\alpha <1 and \left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\, varies through negative values from 0 , -\infty when \alpha < -1,or,\alpha >1
Hence,
\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=\frac{\pi}{2}\, when -1<\alpha <1
\\arctan\left(\frac{1+\alpha}{1-\alpha}\cdot\tan\left(\frac{x}{2}\right)\right)\,\bigg|_0^\pi=-\frac{\pi}{2}\, when \alpha < -1,or,\alpha >1
Therefore ,
\frac{d}{d\alpha}\,\phi(\alpha)\,=0\, when -1<\alpha <1
\frac{d}{d\alpha}\,\phi(\alpha)\,=\frac{2\pi}{\alpha}\, when \alpha < -1,or,\alpha >1
Upon integrating both sides with respect to wrt \alpha we get \phi (\alpha ) = C_{1} when -1<\alpha <1
and \phi (\alpha ) = 2\pi ln\left|\alpha \right|+C_{2} when \alpha < -1,or,\alpha >1
C1 can be determined by setting \alpha = 0
we get C1 = 0
to determine C2 we substitute http://latex.codecogs.com/gif.latex?\alpha%20=\frac{1}{\beta%20} where -1<\beta <1
\begin{align} \phi(\alpha) &=\int_0^\pi\left(\ln(1-2\beta \cos(x)+\beta^2)-2\ln|\beta|\right)\;dx\, \\ &=0-2\pi\ln|\beta|\, \\ &=2\pi\ln|\alpha|\, \end{align}
hence we can conclude \phi (\alpha ) = 0,-1<\alpha <1 and
\phi (\alpha ) = 2\pi ln\left|\alpha \right| when,-1>\alpha ,\alpha >1
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