I have a difficult integral to compute. I know the result by guessing the answer, but need to know the method of calculation. The integral is
$$ \int_{a}^{b}{\rm d}p\,{p \over p^{2} - 2\mu}\, \left[{2\arcsin\left({1 \over p}\,{p^{2} - ba \over b-a}\right) - \pi}\right] =\pi\ln\left({2\mu - a^{2} \over \mu - {ba \over 2} + \frac{1}{2} \sqrt{\,\left(2\mu - a^{2}\right)\left(2\mu - b^{2}\,\right)\,}}\right) $$
I am considering the case where $a<b$, $a^2<2\mu$, and $b^2<2\mu$, so that there is no divergence from the denominator. I suspect it can be solved from some contour integral trick, especially since, by using the properties of arcsin, I can rewrite it in the following form:
$ {\rm Re} \int_{a}^{b} dp\, \frac{p}{p^2-2\mu} \Big[-2i \ln \big( \frac{1}{2} i (p^2 -ab) +\frac{1}{2} \sqrt{(p^2 - a^2) (b^2 - p^2) }\big)-\pi\Big] = \pi \ln \frac{ (2\mu - a^2)} {\mu-\frac{1}{2}ba + \frac{1}{2}\sqrt{(2\mu - a^2)(2\mu-b^2)}}$,
where the Re takes the real part. With this form, I am pretty sure there is a contour integral trick since the log function on the left is so similar to the log function on the right. Unfortunately I cannot seem to figure it out, although I can verify it numerically. Any help would be greatly appreciated! Thanks, Dan