How does one extract the universal part of entanglement entropy? I want to know how equation 2.11 (page 9) follows from 2.10 (page 8) in this paper. 

The two references mentioned just before 2.11 also seem to skip this crucial step. Unless I am missing something obvious everywhere! 

It would be great if someone can help with this! 
 A: A taste of answer, skipping numerical coefficients.
The integrand, in the integral$(2.10)$, is $ I(y)=  (y^2-1)^{\large \frac{d-3}{2}}$
We are interested at the behaviour as $y \rightarrow  +\infty$
So, the integrand is : $I(y)=y^{d-3}(1 - \frac{1}{y^2})^{\large \frac{d-3}{2}}$
So, you get terms, in the integrand, as : 
$I(y)=a_{d-3}y^{d-3} + a_{d-5}y^{d-5} + .....a_{d-2n}y^{d-2n}....\tag{1}$
Now, there is a difference between odd and even $d$, that is :
$I_{even}(y) = a_{d-3}y^{d-3} + a_{d-5}y^{d-5} + ..a_1 y + a_{-1}y^{-1} + O(y^{-2})\tag{2}$
$I_{odd}(y) = a_{d-3}y^{d-3} + a_{d-5}y^{d-5} + ..a_0  +O(y^{-2})\tag{3}$
Now, we have to integrate the integrand between $1$ and $\frac{R}{\delta}$, with $\delta \rightarrow 0$: $J = \int_1^{\frac{R}{\delta}} dy~I(y) $
For even dimensions, you integrate the term $a_{-1}y^{-1}$ between $1$ and $\frac{R}{\delta}$, adn this gives a term $ a_{-1} \ln (\frac{R}{\delta})$. The complete integral is ((the constant $K,K'$ above ,comes frome the integration at the point $1$)): 
$$J_{even} = b_{d-2}(\frac{R}{\delta})^{d-2} + b_{d-4}(\frac{R}{\delta})^{d-4} + ...+b_{2}(\frac{R}{\delta})^{2} + a_{-1} \ln (\frac{R}{\delta})  + 0(\frac{\delta}{R})\tag{4}+ K$$
$$J_{odd} = b_{d-2}(\frac{R}{\delta})^{d-2} + b_{d-4}(\frac{R}{\delta})^{d-4} + ...+b_{1}(\frac{R}{\delta})  + 0(\frac{\delta}{R})\tag{5}+ K'$$
The paper said that "Of course, the power-law divergences in
the R´enyi entropies are not universal, e.g., see [5, 6], however, a universal contribution can be extracted from the subleading terms."
So, we have, finally (and noting that $K$ is neglectible relatively to $ln (\frac{R}{\delta}$)): 
$J_{even} \sim a_{-1} \ln (\frac{R}{\delta}), J_{odd} \sim K'$
