# 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

## 1 Answer

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.