Integral $ \int_{-1}^1 \frac{\ln|x-z|}{\sqrt{1-z^2}} dz$

Question

I am currently going through the paper Some observations on the Green function for the ball in the fractional Laplace framework, by Claudia Bucur.

In it, I discovered the following integral identity ((A.32) in lemma A.7 in the paper): $$ \int_{-1}^1 \frac{\ln|x-z|}{\sqrt{1-z^2}} dz = \left\{ \begin{matrix} - \pi \ln2 & \text{ for } |x| \leq 1 \\ \pi \ln\left( |x|+\sqrt{x^2-1} \right) - \pi \ln 2 & \text{ for } |x| \geq 1 \end{matrix} \right. $$ This identity is given without proof and, although they give a reference, the reference does not prove it either but instead calls it a 'standard result'.

I have managed to prove the second case in the identity above but the first case eludes me. I've tried splitting the integral and integrating by parts in hopes of some cancellation or some such, I tried various substitutions, I tried using complex analysis methods (but the $\ln|x-z|$ term is not very amenable to that approach I think), and I even tried interpreting the integral above as a function of $x$ and then try to show that its weak derivative must vanish, all to no avail. At this point I can't really think of any more approaches to try, maybe there's something I'm missing.

If someone could prove the above result for the $|x|\leq 1$ case or give me a hint it'd be much appreciated!

Answer 1

Case $|x|<1$: Let $z=\sin t $ and $x=\sin s$ to rewrite the integral as

\begin{align} & \int_{-\pi/2}^{\pi/2} \ln \left| \sin t - \sin s\right|\,dt\\ &=\int_0^{\pi/2} \ln \left| \sin^2 t - \sin^2 s\right|\,dt =\frac14\int_0^{2\pi} \ln \left| \sin^2 t - \sin^2 s\right|\,dt\\ &=\frac{1}{4}\int_0^{2\pi} \ln \left| \sin \left( t+s\right)\right|+\ln \left| \sin \left( t-s\right)\right|\,dt\\ &=\frac{1}{2}\int_0^{2\pi} \ln \left| \sin t\right|\,dt= 2 \int_0^{\pi/2} \ln \sin t\ dt=-\pi \ln2 \end{align}

Answer 2

HINT:

$$\frac{1}{\sqrt{1-z^2}}=\sum_{n=0}^\infty\frac{(2n!)}{(n!)^2}\left(\frac z2\right)^{2n}$$