I am reading this set of lecture notes: https://www.southampton.ac.uk/~doug/qft/aqft5.pdf and I would like to understand how to go from the relation (page 52 no Eq number)
$4^\epsilon \int (\sin \theta)^{1-2\epsilon}\frac{1}{(E-p\cos\theta)^2}$
to
$\frac{2}{E^2-p^2}\left[1-2\epsilon \frac{E}{p}\log\left(\frac{E-p}{E+p} \right) + \mathcal{O}(\epsilon^2)\right]$
where they expanded around $\epsilon =0$ and integrate over $\theta$. The boundaries are not included but I assume $\theta \in [0,\pi]$.
Any help much appreciated!
Reading the papeer, I had the feeling that they first compute
$$I=4^\epsilon \int_?^? \frac{\sin^{1-2\epsilon} (\theta)}{(E-p\cos\theta)^2}\,d\theta$$ since they claim that they arrived to an hypergeometric function (I have not been able to find it) which was later expanded as a series around $\epsilon=0$.
Trying your idea, we have $$\sin^{1-2\epsilon} (\theta)=\sin (\theta )-2 \sin (\theta ) \log (\sin (\theta ))\epsilon+2 \sin (\theta ) \log ^2(\sin (\theta ))\epsilon ^2+O\left(\epsilon ^3\right)$$ So, the first term gives $$\int \frac{\sin (\theta)}{(E-p\cos\theta)^2}\,d\theta=-\frac{1}{p (E-p \cos (\theta ))}$$ the second one $$\int \frac{\sin (\theta)\log (\sin (\theta ))}{(E-p\cos\theta)^2}\,d\theta={-\frac{E \log (E-p \cos (\theta ))}{p(E^2-p^2)}+\frac{\log \left(\sin \left(\frac{\theta }{2}\right)\right)}{p(E-p)}+\frac{\log \left(\cos \left(\frac{\theta }{2}\right)\right)}{p(E+p)}-\frac{\log (\sin (\theta ))}{p(E-p \cos (\theta ))}}$$ For the next term, I did not find any solution.
Now, I have a serious problem with the bounds $0$ and/or $\pi$ (because of the two logarithms).
Edit
Working later on Wolfram Cloud, I have been able to obtain the formal expression of the antiderivative. The problem is that it contains six different Appell hypergeometric functions of two variables (I shall not reproduce it here since it is really messy). For $\theta=0$, its value is $0$ (which is good sign). I obtained expressions for all angles except for $\theta=\pi$. However, reading carefully the paper, I do not have the feeling that $\pi$ is the upper bound.
Nevertheless, the expansion of Appell hypergeometric function of two variables as a function of the argument does not look trivial (at least to me).