I am wondering if this integral can be evaluated using the "usual" integration techniques. I have written $| z - x | = \sqrt{(\operatorname{Re}({z}) - x)^2 + \operatorname{Im}(z)^2}$ but it doesn't seem to do much good..
According to the question Evaluate integral: $ \int_{-1}^{1} \frac{\log|z-x|}{\pi\sqrt{1-x^2}}dx$, it may require knowledge of conformal mappings, is this true?
Thanks in advance for your help.
No special knowledge required. Integrate by parts.
EDIT: If you want it in terms of $\log(z-1)$ and $\log(z+1)$, don't use real and imaginary parts. Instead, write $$\log |z - x| = \log \sqrt{(z - x)(\bar{z} - x)} = \dfrac{1}{2} \log (z-x) + \dfrac{1}{2} \log(\bar{z} - x)$$ For this to be valid you have to be a bit careful about which branch of the logarithm of complex numbers you use: you want the one with imaginary part in $(-\pi, \pi]$, and avoid having $z - x$ on the negative real axis. Since $x$ will go from $-1$ to $1$ in your integral, you don't want $z$ in $(-\infty, 1]$.
Then
$$\eqalign{&\left. \int_{-1}^1 \log(z-x) \ dx = (x-z) \log(z-x) + z - x \right|_{x=-1}^1 \cr&= (1-z) \log(z-1) + (1+z) \log(z+1) - 2}$$
The integral of $\log(\bar{z} - x)$ is the complex conjugate of this, so you get $\text{Re} ( (1-z) \log(z-1) + (1+z) \log(z+1)) - 2$.