In working out certain particle-to-particle collisions and entropy, I did come up with a problem. Hence, I am going to ask it here. There are similar calculations that enter the picture in dealing with Feynman integrals. So, other people may find this interesting as well.
QUESTION. Is the following integral independent of $0<z<1$? $$\int_0^1\frac{\log\vert z-t\vert}{\sqrt{t-t^2}}dt.$$
Denoting $z=\sin^2 a;\,a\in[0;\frac\pi2]$ $$I(z)=\int_0^1\frac{\log\vert z-t\vert}{\sqrt{t-t^2}}dt\overset{x=\sqrt t}{=}2\int_0^1\frac{\ln|z-x^2|}{\sqrt{1-x^2}}dx$$ $$\overset{x=\sin\phi}{=}2\int_0^\frac\pi2\ln|\sin^2a-\sin^2\phi|d\phi=\frac14\int_0^{4\pi}\ln|\sin^2a-\sin^2\phi|d\phi$$ $$=\frac14\int_0^{4\pi}\left(\ln\Big|2\sin\frac{\phi-a}2\,\cos\frac{\phi+a}2\Big|+\ln\Big|2\sin\frac{\phi+a}2\,\cos\frac{\phi-a}2\Big|\right)d\phi$$ $$=\frac12\int_0^{2\pi}\Big(2\ln 2+\ln|\sin(\phi-a)|+\ln|\sin(\phi+a)|+\ln|\cos(\phi-a)|+\ln|\cos(\phi+a)|\Big)d\phi$$ We see that $a$ (or, initially $z$) does not contribute due to periodicity of the functions of the integrand.
Taking into consideration that $\displaystyle \int_0^\frac\pi2\ln(\cos\phi)d\phi=\int_0^\frac\pi2\ln(\sin\phi)d\phi=-\,\frac\pi 2\ln2$ $$I(z)=2\pi\ln2+\frac12\cdot4\cdot4\int_0^\frac\pi2\ln(\cos\phi)d\phi=-2\pi\ln2$$