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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.

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@TMM: I think some tags related to complex numbers are needed. –  Babak S. Apr 2 '13 at 18:23
@BabakS. Certainly you are able to add those tags as well :) –  TMM Apr 2 '13 at 18:26

2 Answers 2

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,



But I don't know how to find $C$.

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You forgot a $\pi$ factor before the $\ln\left(z + \sqrt{z^{2} - 1}\right)$. –  Felix Marin Aug 31 '14 at 22:43

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\int_{-1}^{1}{\ln\pars{\verts{z - x}} \over \root{1 - x^{2}}}\,\dd x: {\large ?}}$

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

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+1 Nice work here. $\partial_\mu$ the ghost –  Integrals Sep 1 '14 at 2:02

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