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We have the functional integral $(1)$ of viriable $\delta p_j$. In our case this intergral follows from RPA (random phase approximation) method assuming $\ p_j(r)=p_j^*+\delta p_j(r)$. The task is to prove that after calculating Gaussian integral $(1)$ we will gain formula $(2)$

$$I=\int\prod_j D[\delta p_j]exp\biggl(-\frac{1}{2}\int\frac{d^3k}{(2\pi)^3}\biggl[ {\sum_j}\sum_q\delta p_j(k)\Bigl(\frac{\delta_{jq}}{p_j^*}+V_kZ_jZ_q\Bigr)\delta p_q(k)+ \\+2\sum_jV_kaZ_AZ_j\delta p_A(k)\delta p_j(k)\biggr]\biggr)exp\biggl[-\frac{1}{2}\int \frac{d^3k}{(2\pi)^3}V_k a^2 Z_A^2 \delta p^2_A(k)\biggr] (1)$$ $$V_k=\frac{4\pi l_b}{k^2}, \delta_{mp}-Kronecker \ delta,\ q \ (and \ j)=c,+,-,respectively.$$ $$p_j^*=\frac{n_j}{\Omega}$$

By calculating the Gaussian integral, the value of $I$ was found to be $$I=\frac{(2\pi)^{3/2}}{(\sqrt{A_cA_+A_-})}exp\biggl[-\frac{\Omega}{2}\int \frac{d^3k}{(2\pi)^3}ln[1+V_kA_0]\biggr]exp\biggl[-\frac{1}{2}a^2Z_A^2\int \frac{d^3k}{(2\pi)^3}\frac{V_k}{1+V_kA_0}\delta p_a^2(k)\biggl] (2)$$ where $A_j=\frac{\Omega}{n_j}, j=c,+,- $ and $A_0$ is given by $$A_0=\frac{1}{\Omega}\sum_{j=1,2,3}Z_j^2n_j=\frac{z^2}{4\pi l_B}$$

I don't exactly know how to operate with functional integral and which formula use to gain equation $(2)$ above. Can anyone explain how to transform equation $(1)$?

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  • $\begingroup$ If you could mention where you found this problem then I may be able to help. $\endgroup$ Commented Dec 30, 2018 at 10:12

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