Tricky Integral: Evaluating Renormalized Ultraviolet "Divergent" Integral

Question

I am trying to rederive the results presented in the paper, in particular equation (30). That is, I am trying to compute the correction to the ground-state energy of a dipolar condensate due to beyond-mean field effects. This involves evaluating the following integral:

$$\Delta E = \frac{1}{2}V \int \frac{d^3q}{(2\pi)^3} \left[ \varepsilon_{\textbf{q}} - \frac{\hbar^2\textbf{q}^2}{2M} - n\tilde{V}_{\text{int}}(\textbf{q}) + \frac{\tilde{V}_\text{int}^2(\textbf{q})}{q^2}\frac{2Mn^2}{\hbar^2}\right]\tag{29}$$

where $$\varepsilon_\textbf{q} = \sqrt{\frac{\hbar^2\textbf{q}^2}{2M}\left(\frac{\hbar^2\textbf{q}^2}{2M} + 2gn_0\left[1 + \epsilon_\text{dd}(3\cos^2{\theta} - 1)\right] \right)}$$ is the Boglibov spectrum and $$\tilde{V}_\text{int}(\textbf{q}) = g\left[1 + \epsilon_\text{dd}(3\cos^2{\theta} - 1)\right]$$ is the Fourier Transform of the dipole-dipole interaction potential.

The integral is the inverse Fourier transform and $\theta$ is the Fourier polar angle. Taking $n_0 = n$, the paper finds the result to be

$$\Delta E = V\frac{2\pi\hbar^2a_s n^2}{M}\frac{128}{15}\sqrt{\frac{a_s^3n}{\pi}}\mathcal{Q}_5(\epsilon_{\text{dd}})\tag{30}$$

where they have reexpressed the result in terms of $a_s$ by substitution $g=4\pi\hbar^2a_s/M$. Here,

$$\mathcal{Q}_5(x) = (1-x)^{5/2} {}_2F_1(-\frac{l}{2}, \frac{1}{2};\frac{3}{2};\frac{3x}{x-1})$$

where ${}_2F_1(\alpha,\beta;\gamma;z)$ represents the hypergeometric function.

The last term in the integral is supposed to remove the divergence, however when I try to evaluate the integral I get still get a divergent result.

I would appreciate any attempts to explain how they reach their result or how to go about dealing with the divergence.

Answer 1

Let's factor out $\frac{\hbar^2}{2M}$ from the integrand and take: $$\lambda\equiv\frac{g n\big(1+ \epsilon (3 \cos^2\theta-1)\big)}{\hbar^2/2M}$$ Then the integral is rewritten as $$\Delta E=\frac{1}{2(2\pi)^3}V\frac{\hbar^2}{2M}\int \sin\theta d\theta\int d\phi\int_{0}^{+\infty}q^2 dq\left[q\sqrt{q^2+2 \lambda}-q^2-\lambda+\frac{\lambda^2}{q^2}\right]$$ You can change the upper bound of integral to $M$, perform the integral and then make a series expansion over $M$ at $\infty$ (I did it by Mathematica). Then you find (I would write the contributions from the four terms in the brackets separately) $$\Delta E=\frac{1}{2(2\pi)^3}V\frac{\hbar^2}{2M}\int \sin\theta d\theta\int d\phi\times\lim_{M\rightarrow \infty}\left[\big(\frac{M^5}{5}+\frac{\lambda M^3}{3}-\frac{\lambda^2 M}{2}+\frac{8\sqrt{2}\lambda^{5/2}}{15}\big)+\big(-\frac{M^5}{5}\big)+\big(-\frac{\lambda M^3}{3}\big)+(\lambda^2 M)\right]+\mathcal{O}\big(\frac{1}{M}\big)$$ There is apparently a typo in the reference you sent. The coefficient of regulator term seems to be twice of what it must be to cancel out $-\frac{\lambda^2 M}{2}$. Accepting this, we arrive at $$\Delta E=\frac{1}{2(2\pi)^3}V\frac{\hbar^2}{2M}\int_{-1}^{1} d(\cos\theta)(2\pi)\frac{8\sqrt{2}}{15}\big(\frac{g n}{\hbar^2/2M}\big)^{5/2}\big(1+ \epsilon (3 \cos^2\theta-1)\big)^{5/2}\\ =\frac{\sqrt{2}}{15\pi^2}V\big(\frac{\hbar^2}{2M}\big)^{-3/2}(g n )^{5/2}2\int_{0}^{1} d(\cos\theta)\big(1+ \epsilon (3 \cos^2\theta-1)\big)^{5/2}$$ You can simply preform the following integral by Mathematica $$Q(\epsilon)=\int_{0}^{1}d x \big(1+\epsilon(3x^2-1)\big)^{5/2}$$ It is easy to check that $Q(0)=1$ and $Q(1)\approx 2.6$, as mentioned in the reference paper.

Hope this helps.