Ladder (Particle-particle bubble) of free electrons

$$$\begin{split} &\int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} T \sum_{i \omega_n} \frac{1}{i \omega_n+i \Omega_n-\frac{(\vec{k}+\vec{p})^2}{2 m}+\mu} \frac{1}{-i \omega_n-\frac{p^2}{2 m}+\mu} \\ =&\int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{f\left(\frac{(\vec{k}+\vec{p})^2}{2 m}-\mu\right)-f\left(-\frac{p^2}{2 m}+\mu\right)}{i \Omega_n-\frac{(k+\vec{p})^2}{2 m}-\frac{p^2}{2 m}+2 \mu} \end{split}$$$
• In the case $k>2k_F$

Define $\vec{k}+\vec{p}=-\vec{p}'$ for the first term, then rename it as $\vec{p}$,

$$$\begin{split} =& \int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{f\left(\frac{p^2}{2 m}-\mu\right)-f\left(-\frac{p^2}{2 m}+\mu\right)}{i \Omega_n-\frac{p^2}{2 m}-\frac{(\vec{p}+\vec{k})^2}{2 m}+2 \mu} \\ =& \int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{2f\left(\frac{p^2}{2 m}-\mu\right)-1}{i \Omega_n-\frac{p^2}{2 m}-\frac{(\vec{p}+\vec{k})^2}{2 m}+2 \mu} \\ =& -\int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{1}{i \Omega_n-\frac{p^2}{2 m}-\frac{(\vec{p}+\vec{k})^2}{2 m}+2 \mu} +\int \frac{d^3 \vec{p}}{(2 \pi)^3} \frac{2f\left(\frac{p^2}{2 m}-\mu\right)}{i \Omega_n-\frac{p^2}{2 m}-\frac{\left(\vec{p}+\vec{k}\right)^2}{2 m}+2 \mu} \end{split}$$$

Define $\vec{p}'=\vec{p}+\vec{k}/2$, the first term becomes

$$$\frac{1}{i \Omega_n-\frac{\left(\vec{p}’+\vec{k}/2\right)^2}{2 m}-\frac{\left(\vec{p}^{\prime}-\vec{k}/{2}\right)^2}{2 m}+2\mu}=\frac{1}{i \Omega_n-\frac{\vec{p}^{\prime 2}}{m}-\frac{k^2}{4 m}+2 \mu}$$$$$$\begin{split} =& -\int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{1}{i \Omega_n-\frac{p^2}{ m}-\frac{\vec{k}^2}{4 m}+2 \mu}\\ =&4 \pi \int \frac{d p}{(2 \pi)^3} \frac{p^2}{\frac{p^2}{m}+\frac{k^2}{4 m}-i \Omega_n-2 \mu} \\ =&4 \pi m^{3/2} \int \frac{d p/m^{1/2}}{(2 \pi)^3} \frac{p^2/m}{\frac{p^2}{m}+\frac{k^2}{4 m}-i \Omega_n-2 \mu} \\ =&4 \pi m^{3/2} \int \frac{d x}{(2 \pi)^3} \frac{x^2}{x^2+\frac{k^2}{4 m}-i \Omega_n-2 \mu} \end{split}$$$

Using $\int \frac{x^2 d x}{x^2+a}=x-\sqrt{a} \tan ^{-1}\left(\frac{x}{\sqrt{a}}\right)$,

$$$\begin{split} =&4 \pi m^{3/2} \int \frac{d x}{(2 \pi)^3} \frac{x^2}{x^2+\frac{k^2}{4 m}-i \Omega_n-2 \mu} \\ =& \frac{4\pi m^{3/2}}{(2\pi)^3}\frac{\Lambda}{\sqrt{m}}-\frac{4\pi m^{3/2}}{(2\pi)^3}\sqrt{\frac{k^2}{4 m}-i \Omega_n-2 \mu}\cdot \left.\tan^{-1}\left(\frac{x}{\sqrt{\frac{k^2}{4 m}-i \Omega_n-2 \mu}}\right)\right|^{\Lambda/\sqrt{m}}_0 \\ =& \frac{m}{2\pi^2}\Lambda-\frac{m^{3/2}}{4\pi}\sqrt{\frac{k^2}{4 m}-i \Omega_n-2 \mu} \end{split}$$$

We conclude

$$$\begin{split} &\int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} T \sum_{i \omega_n} \frac{1}{i \omega_n+i \Omega_n-\frac{(\vec{k}+\vec{p})^2}{2 m}+\mu} \frac{1}{-i \omega_n-\frac{p^2}{2 m}+\mu} \\ =&\frac{m \Lambda}{2\pi^2} -\frac{m^{3 / 2}}{4\pi} \sqrt{-i \Omega_n+\frac{k^2}{4 m}-2 \mu} +\int \frac{d^3 \vec{p}}{(2 \pi)^3} \frac{2f\left(\frac{p^2}{2 m}-\mu\right)}{i \Omega_n-\frac{p^2}{2 m}-\frac{\left(\vec{p}+\vec{k}\right)^2}{2 m}+2 \mu} \end{split}$$$
• Generic $k$
$$$\begin{split} =& \int \frac{d^3 \vec{p}}{\left(2\pi^3\right)} \frac{2f\left(\frac{p^2}{2 m}-\mu\right)-1}{i \Omega_n-\frac{p^2}{2 m}-\frac{(\vec{p}+\vec{k})^2}{2 m}+2 \mu} \\ =& \frac{1}{(2\pi)^2} \int_0^\pi \sin(\theta)d\theta \int_0^\infty p^2 dp \frac{2f\left(\frac{p^2}{2 m}-\mu\right)-1}{i \Omega_n-\frac{p^2}{m}-\frac{k^2}{2m}-\frac{kp\cos(\theta)}{m}+2 \mu} \end{split}$$$

The angle can be integrated explicitly,

$$$-\int_0^\pi \frac{d\cos\theta}{a+b\cos\theta} = \int_{-1}^{1} \frac{dx}{a+bx} = \frac{1}{b} \ln\frac{a+b}{a-b}$$$

where $a$ is not real-valued as long as $\Omega_n$ doesn't vanish, and $\ln(a/b) \equiv \ln(a)-\ln(b)$.

Therefore, for generic Matsubara-frequency,

$$$= \frac{m}{(2\pi)^2} \int_0^\infty dp \left[\frac{p}{k}\ln \frac{i\Omega_n -\frac{p^2}{m}-\frac{k^2}{2m}+\frac{pk}{m}+2\mu}{i\Omega_n -\frac{p^2}{m}-\frac{k^2}{2m}-\frac{pk}{m}+2\mu}\left(2f(\frac{p^2}{2m}-\mu)-1\right)-2\right]+\frac{m\Lambda}{2\pi^2}$$$

In the small momentum limit, it is simplifies to

$$$= \frac{m}{(2\pi)^2} \int_0^\infty dp \left[\frac{\frac{2p^2}{m}}{i\Omega_n -\frac{p^2}{m}+2\mu}\left(2f(\frac{p^2}{2m}-\mu)-1\right)-2\right]+\frac{m\Lambda}{2\pi^2}$$$