Evaluating the integral $\int_{-1}^1 \frac{\ln|z-x|}{\sqrt{1-x^2}}\mathrm dx$ I don't know how to deal with this integral:
$$\int_{-1}^{1}{\ln\left(\,\left\vert\,z - x\,\right\vert\,\right)\over
                \,\sqrt{\vphantom{\large A}\,1 - x^{2}\,}\,}\,{\rm d}x\,,$$
where $z$ is a complex number.
 A: Let $f(z)=\int_{-1}^1\dfrac{1}{\sqrt{1-x^2}}\ln|z-x|~dx$ ,
Then $\dfrac{df(z)}{dz}=\int_{-1}^1\dfrac{1}{(z-x)\sqrt{1-x^2}}dx$
According to http://www.wolframalpha.com/input/?i=int1%2F%28%28z-x%29%281-x%5E2%29%5E%281%2F2%29%29%2Cx%2C-1%2C1,
$\dfrac{df(z)}{dz}=\dfrac{\pi}{\sqrt{z^2-1}}$
$f(z)=\int\dfrac{\pi}{\sqrt{z^2-1}}dz=\ln(z+\sqrt{z^2-1})+C$
But I don't know how to find $C$.
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
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 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
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$\ds{\int_{-1}^{1}{\ln\pars{\verts{z - x}} \over \root{1 - x^{2}}}\,\dd x:
     {\large ?}}$

From
  ${\tt \mbox{@paul2357paul answer}}$ 
  it's clear that
  $\ds{C
     =\fermi\pars{1}
     =\int_{-1}^{1}{\ln\pars{\verts{1 - x}} \over \root{1 - x^{2}}}\,\dd x}$. So,
  we'll evaluate: 

\begin{align}
\color{#66f}{\Large C}&
=\int_{-1}^{1}{\ln\pars{\verts{1 - x}} \over \root{1 - x^{2}}}\,\dd x
=\int_{0}^{1}{\ln\pars{1 - x^{2}} \over \root{1 - x^{2}}}\,\dd x
=\lim_{\mu\ \to\ -1/2}\partiald{}{\mu}
\int_{0}^{1}\pars{1 - x^{2}}^{\mu}\,\dd x
\\[3mm]&=\lim_{\mu\ \to\ -1/2}\partiald{}{\mu}
\int_{0}^{1}\pars{1 - x}^{\mu}\,\half\,x^{-1/2}\,\dd x
=\half\,\lim_{\mu\ \to\ -1/2}\partiald{}{\mu}
\int_{0}^{1}x^{-1/2}\pars{1 - x}^{\mu}\,\dd x
\\[3mm]&=\half\,\lim_{\mu\ \to\ -1/2}\partiald{}{\mu}
\bracks{%
\Gamma\pars{1/2}\Gamma\pars{\mu + 1} \over \Gamma\pars{\mu + 3/2}}
=\color{#66f}{\large -\ln\pars{2}\pi} \approx {\tt -2.1774}
\end{align}

Then,
  $$\color{#66f}{\large%
\int_{-1}^{1}{\ln\pars{\verts{z - x}} \over \root{1 - x^{2}}}\,\dd x
=\pi\ln\pars{z + \root{z^{2} - 1}} - \ln\pars{2}\pi}
$$

