To save some reading, what I'd generally like to know is if the following is true:
given: $0\le x \le 1 \land 0 \le a < \infty \land$ $k$ is some constant value of a, is it true that:
$$\ln\Bigg(\frac{f(x);x \gtrless a}{g(x);x \gtrless \frac{k}{a}}\Bigg); x \gtrless a \land x \gtrless \frac{k}{a} = \bigg[\ln\Big(f(x)\Big);x \gtrless a\bigg] - \bigg[\ln\Big(g(x)\Big);x \gtrless \frac{k}{a}\bigg]$$ ?
In more detail specific of my specific problem:
given: $0\le r_0 \le 1 \land 0 \le r < \infty \land$ $a$ is a given value of r, we want to find
$$\int_0^1 f(r,r_0)r_0dr_0 = ?$$
where:
$$f(r,r_0) = \ln\Bigg(\frac{a^2}{r_0^2} \frac{r^2+r_0^2-2rr_0\cos\theta}{r^2+\frac{a^4}{r_0^2}-2a^2\frac{r}{r_0}\cos\theta}\Bigg) $$ We factor out from the numerator and denominator a term so that both conform to a general form $\ln(1+z^2-2xz);|z|<1 , -1\le x\le 1$, because $\int_0^{2\pi}\ln(1+z^2-2xz);|z|<1 dx = 0$, and that drastically reduces the complexity of the integration as integrals of functions in this this form evaluate to 0, however doing this produces piecewise definitions that we need to parse through before integrating what is leftover. For example, if I factored out $r^2$ from the numerator and $\frac{a^4}{r_0^2}$, then we would have
$$f(r,r_0) = \ln\Bigg(\frac{a^2r^2}{a^4}\frac{1+(\frac{r_0}{r})^2 - 2(\frac{r_0}{r})\cos\theta ; |\frac{r_0}{r}|<1}{1+\frac{r^2r_0^2}{a^4}-2\frac{rr_0}{a^2}\cos\theta;|\frac{rr_0}{a^2}|<1}\Bigg);|\frac{r_0}{r}|<1\land |\frac{rr_0}{a^2}|<1 \\ = \ln\Bigg(\frac{a^2r^2}{a^4}\frac{1+(\frac{r_0}{r})^2 - 2(\frac{r_0}{r})\cos\theta ; \frac{r_0}{r}<1}{1+\frac{r^2r_0^2}{a^4}-2\frac{rr_0}{a^2}\cos\theta;\frac{rr_0}{a^2}<1}\Bigg) ; r_0<r \land r_0 <\frac{a^2}{r}$$
So we understand that the numerator and denominator have their own discontinuities for which they are defined on, and together, the full natural log satisfies both. So the full piecewise definition of $f(r,r_0)$ is:
$$ f(r,r_0) = \cases{\ln \Bigg(\frac{a^2 r^2}{r_0^2r^2}\frac{1+(\frac{r_0}{r})^2-2(\frac{r_0}{r})\cos\theta; \frac{r_0}{r}<1}{1+\frac{a^4}{r^2r_0^2}-2\frac{a^2}{rr_0}\cos\theta;\frac{a^2}{rr_0}<1}\Bigg) & $r_0<r \land r_0>\frac{a^2}{r}$ \\ \ln \Bigg(\frac{a^2 r^2}{r_0^2\frac{a^4}{r_0^2}}\frac{1+(\frac{r_0}{r})^2-2(\frac{r_0}{r})\cos\theta; \frac{r_0}{r}<1}{1+\frac{r^2r_0^2}{a^4}-2\frac{rr_0}{a^2}\cos\theta;\frac{a^2}{rr_0}>1}\Bigg) & $r_0<r \land r_0<\frac{a^2}{r}$ \\ \ln \Bigg(\frac{a^2 r^2}{2r_0^2r^2}\frac{1+(\frac{r_0}{r})^2-2(\frac{r_0}{r})\cos\theta; \frac{r_0}{r}<1}{1-\cos\theta;\frac{a^2}{rr_0}=1}\Bigg) & $r_0<r\land r_0 = \frac{a^2}{r}$ \\ \ln \Bigg(\frac{a^2 r_0^2}{r_0^2r^2}\frac{1+(\frac{r}{r_0})^2-2(\frac{r}{r_0})\cos\theta; \frac{r}{r_0}<1}{1+\frac{a^4}{r^2r_0^2}-2\frac{a^2}{rr_0}\cos\theta;\frac{a^2}{rr_0}<1}\Bigg) & $r_0>r \land r_0>\frac{a^2}{r}$ \\ \ln \Bigg(\frac{a^2 r_0^2}{r_0^2\frac{a^4}{r_0^2}}\frac{1+(\frac{r}{r_0})^2-2(\frac{r}{r_0})\cos\theta; \frac{r}{r_0}<1}{1+\frac{r^2r_0^2}{a^4}-2\frac{rr_0}{a^2}\cos\theta;\frac{a^2}{rr_0}>1}\Bigg) & $r_0>r \land r_0<\frac{a^2}{r}$ \\ \ln \Bigg(\frac{a^2 r_0^2}{2r_0^2r^2}\frac{1+(\frac{r}{r_0})^2-2(\frac{r}{r_0})\cos\theta; \frac{r}{r_0}<1}{1-\cos\theta;\frac{a^2}{rr_0}=1}\Bigg) & $r_0>r\land r_0 = \frac{a^2}{r}$ \\ \ln \Bigg(\frac{2 a^2 r_0^2}{r_0^2r^2}\frac{1-\cos\theta; \frac{r}{r_0}=1}{1+\frac{a^4}{r^2r_0^2}-2\frac{a^2}{rr_0}\cos\theta;\frac{a^2}{rr_0}<1}\Bigg) & $r_0=r \land r_0>\frac{a^2}{r}$ \\ \ln \Bigg(\frac{2a^2 r_0^2}{r_0^2\frac{a^4}{r_0^2}}\frac{1-\cos\theta; \frac{r}{r_0}=1}{1+\frac{r^2r_0^2}{a^4}-2\frac{rr_0}{a^2}\cos\theta;\frac{a^2}{rr_0}>1}\Bigg) & $r_0=r \land r_0<\frac{a^2}{r}$ \\ \ln \Bigg(\frac{2a^2 r_0^2}{2r_0^2r^2}\frac{1-\cos\theta; \frac{r}{r_0}=1}{1-\cos\theta;\frac{a^2}{rr_0}=1}\Bigg) & $r_0=r\land r_0 = \frac{a^2}{r}$ } $$
Now I could probably integrate this, but finding the bounds of integration given the pair of discontinuities that we have to satisfy simultaneously for each definition sounds a bit rigorous. So here's what I'd like to confirm:
can each piecewise definition, like say for example: $\ln \Bigg(\frac{a^2 r^2}{r_0^2r^2}\frac{1+(\frac{r_0}{r})^2-2(\frac{r_0}{r})\cos\theta; \frac{r_0}{r}<1}{1+\frac{a^4}{r^2r_0^2}-2\frac{a^2}{rr_0}\cos\theta;\frac{a^2}{rr_0}<1}\Bigg);r<r_0 \land r_0>\frac{a^2}{r}$, instead be separated into 2 natural logs such that they retain their respective discontinuities:
$\bigg[\ln(a^2 r^2) + \ln \Big(1+(\frac{r_0}{r})^2-2(\frac{r_0}{r})\cos\theta)\Big);r_0<r\bigg] - \bigg[\ln(r_0^2r^2)-\ln\Big(1+\frac{a^4}{r^2r_0^2}-2\frac{a^2}{rr_0}\cos\theta\Big);r_0>\frac{a^2}{r}\bigg]$?
If so, the piecewise integration would become significantly more simple, I just don't know about separating when definitions specific to different regions of the domain come into play. If not, I'd like to know how exactly I'd set up these bounds of integration given the pair of discontinuities.
