Calculation with non-logarithmic accuracy

The phonon dispersion $\omega = c\vert k\vert$ relation which was used in the derivation above leads to infrared divergence in some of the integrals (1.166) and was resolved by truncation of the integral. This problem can be cured using a more precise Bogoliubov dispersion law:

$$\omega(k) = \sqrt{(kc)^2+\left(\frac{\hbar k^2}{2M}\right)^2}$$ (1.175)

It is easy to see that results for the new dispersion can be obtained by changing formally the speed of sound $c\vert k\vert\to c\vert k\vert\sqrt{1+(\hbar k/2Mc)^2}$ in definitions of hydrodynamic operators (1.169-1.170. This will lead to a converging value of the integral (1.166):

$$-2\int\limits_0^\infty \frac{\eta\pi2m^2(1-\cos kx)}{2k\sqrt{1+(\... ... m^2\int\limits_0^\infty\frac{1-\cos z}{\sqrt{1+\varepsilon^2z^2}}\frac{dz}{z},$$ (1.176)

Here we introduced the notation $z = xk$ and $\varepsilon = \hbar/2Mcx$. Let us split the integral (1.176) in two parts:

$$\int\limits_0^\infty\frac{1-\cos z}{\sqrt{1+\varepsilon^2z^2}} \f... ...cos z)\frac{dz}{z} +\int\limits_N^\infty \frac{dz}{z\sqrt{1+\varepsilon^2z^2}},$$ (1.177)

in such a way that $1\ll N \ll 1/\varepsilon$. The term $(1+\varepsilon^2z^2)$ can be neglected in the integration up to $N$ and oscillating term can be neglected at larger distances. In order to proceed further we shift the lower integration limit by short distance $\epsilon \to 0$. Then the first integral becomes equal to $\ln N/\epsilon-\mathop{\rm Ci}\nolimits N+%% \mathop{\rm Ci}\nolimits\epsilon$. The second integral can be easily calculated by using substitution $y^2=1+\varepsilon^2z^2$ and equals $\frac{1}{2}\ln \frac{\sqrt{1+\varepsilon^2N^2}+1}{\sqrt{1+\varepsilon^2N^2}-1}$. Collecting everything together we obtain

$$\int\limits_0^\infty\frac{1-\cos z}{\sqrt{1+\varepsilon^2z^2}} \f... ...pprox \gamma + \ln \frac{4Mcx}{\hbar} = \gamma + \ln \frac{8\pi\rho_0 x}{\eta},$$ (1.178)

where $\gamma \approx 0.577$ is Euler's constant.

The static density-density correlation function (1.168) (more precisely its $m \ne 0$ part) is equal to

$$\frac{\langle\hat\rho(x)\hat\rho(0)\rangle}{\rho_0^2} = 1+2\sum\l... ...}{8\pi C}\right)^{\eta m^2} \frac{\cos(2\pi m\rho_0 x)}{(\rho_0 x)^{\eta m^2}},$$ (1.179)

where $C = e^\gamma \approx 1.781$.

The time-dependent result differ from the stationary case (1.179) only by substitution $x \to \sqrt{x^2-c^2t^2}$ in the denominator, as it was already shown in the calculation with logarithmic accuracy (compare 1.168 and 1.174):

$$\frac{\langle\hat\rho(x,t)\hat\rho(0,0)\rangle}{\rho_0^2} = 1+2\s... ...)^{\eta m^2} \frac{\cos(2\pi m\rho_0 x)}{(\rho_0 \sqrt{x^2-c^2t^2})^{m^2\eta}},$$ (1.180)