Can anyone give any hints on how to rewrite this in terms of dilogarithms?
$$ \int_{0}^{\infty}{{\rm d}t \over t}\,{1 \over t - s - {\rm i}\epsilon}\, {1 \over \,\sqrt{\, 1 - a/t\,}\,}\, \ln\left(1 - \,\sqrt{\,1 - a/t\,}\, \over 1 + \,\sqrt{\, 1 - a/t\,}\,\right)\,, $$ with $\epsilon$ small.
It looks horrible and I really don't know how to proceed, I have tried many variable substitutions, separate $t$ and $t - s$ via partial fractions etc...nothing seems to work. Any help is appreciated.
EDIT: So I have split the integral up in the following way, can someone comment my logic and mistakes:
Since $1-a/t$ is negative for $t<a$ we let $$ (1-a/t)^{1/2} = \mathrm{i}\sqrt{a/t-1}$$ in this interval, in the rest of the integration range it the usual square root. Then one can write the integral as
$$ I = \int_0^a \mathrm{d}t\frac{1}{t}\frac{1}{t-s-\mathrm{i}\epsilon}\frac{1}{\mathrm{i}\sqrt{a/t-1}} \ln\frac{1-\mathrm{i}\sqrt{a/t-1}}{1+\mathrm{i}\sqrt{a/t-1}}\\[3mm]+\int_a^\infty \mathrm{d}t\frac{1}{t}\frac{1}{t-s-\mathrm{i}\epsilon}\frac{1}{\sqrt{1-a/t}} \ln\frac{1-\sqrt{1-a/t}}{1+\sqrt{1-a/t}}$$ Now substituting $u=1-a/t$ and $u=a/t-1$ we get $$I=-\mathrm{i}\int_0^\infty\mathrm{d}u\frac{1}{\sqrt{u}}\frac{1}{a-s(1+u)-\mathrm{i}\epsilon}\ln\frac{1-\sqrt{u}}{1+\sqrt{u}}\\[3mm]+\int_0^1\mathrm{d}u\frac{1}{\sqrt{u}}\frac{1}{a-s(1-u)-\mathrm{i}\epsilon}\ln\frac{1-\sqrt{u}}{1+\sqrt{u}}$$ Now there is an obvious substitution $u=v^2$ but I how would I treat the first log since the stuff inside its parenthesis becomes negative for $u>1$ and I rather not mess with absolute value bars (e.g. in $\log z = \log|z|+\mathrm{i}\arg z$). Is there any other way from here on?