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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

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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.

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