Q:

UV divergence integral

Could anyone please explain how to calculate integral such as 
$$\frac{\Omega}{2}\int_{-\infty}^{+\infty} \frac{d^3k}{(2\pi)^3}\ln\left[{1+\frac{a^2}{k^2}}\right]=-\frac{\Omega a^3}{12\pi}+I_0~?$$ This integral doesn't converge.
I guess we have to use UV divergence cut-off, but I don't fully understand the technique of this method. 
Reference:
Double screening in polyelectrolyte solutions: Limiting laws and crossover formulas
M. Muthukumar
http://dx.doi.org/10.1063/1.472362
Page 5187, formula 2.13

A:

Have you tried differentiating wrt to $a$?  If $I=(2\pi)^3$ times your integral, I get
$$
\frac {dI}{da} = 8\pi a  \int_0^\infty k^2 d k  \frac{1}{k^2+a^2}\\=8\pi a  \int_0^\infty  d k  \frac{k^2+a^2}{k^2+a^2} -  8\pi a  \int_0^\infty  d k  \frac{a^2}{k^2+a^2},
$$
which is in the form of divergent constant together with  a convergent integral proprtional to $a^2$. Integrating back up suggests that his $I_0$ is propertional to $\Lambda a^2$ where $\Lambda$ is the large-$k$ cutoff.

