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I have a rather involved question regarding the weakly attractive limit of the BCS ground state. We know for exampel from The book of Pitajevski and Stringari (Bose–Einstein Condensation and Superfluidity) or from several previous works, e.g. https://journals.aps.org/pra/pdf/10.1103/PhysRevA.77.023626 or other contributions here (Ground state of BCS mean field Hamiltonian) how the thermodynamic potential and hence the free energy of the BCS state at $T=0$ looks like, namely $$\frac{\Omega}{N_0 \varepsilon_F} = -\frac{3\pi}{8} \frac{\widetilde{\Delta}_0^2}{a_s k_F} - \frac{3}{4} \int_0^{\widetilde{\Lambda}} \text{d}\widetilde{\varepsilon} \sqrt{\widetilde{\varepsilon}} \bigg[ \sqrt{(\widetilde{\varepsilon} - \widetilde{\mu})^2 + \widetilde{\Delta}_0^2} - (\widetilde{\varepsilon} - \widetilde{\mu}) - \frac{\widetilde{\Delta}_0^2}{2\widetilde{\varepsilon}} \bigg]$$ where $\widetilde{\Delta}_0$ is the superfluid order parameter in orders of the Fermi-energy, $\widetilde{\mu}$ the chemical potential in units of the Fermi-energy and $\widetilde{\Lambda}$ my momentum cutoff.

Now i want to find the weakly attractive, or BCS, limit of this expression by performing a Taylor series around $a_s k_f \rightarrow 0^-$ where i derived out of the well known Gap-equation in the BCS limit the following expressions, which are also approved by the above paper, book and several other papers. $$ \widetilde{\mu} \approx 1 \hspace{1cm} \text{and} \hspace{1cm} \widetilde{\Delta}_0 \approx \frac{8}{e^2} \exp(\frac{\pi}{2} \frac{1}{a_s k_F})$$ When setting $\widetilde{\Delta}_0 = 0$ we recover the non interacting Fermi-gas and the integral simply evaluates to $\frac{3}{5}$. But how do i perform the Taylor expansion of the integral and how do i also treat the constant offset term in the front? The integral is not continuously differentiable in $\widetilde{\Delta}_0$ and hence i would not apply Liebnitz-theorem of differentiation under the integration?

In the end i want to know how to recover the approximate solution from the book of Pitajevski and Stringari (page 326. formula 16.56) $$ \frac{E_0}{N} = \frac{\Omega_0}{N \varepsilon_F} \varepsilon_F + \widetilde{\mu} \varepsilon_F = \frac{3}{5} \bigg( 1 - \frac{40}{e^4} \exp(\pi \frac{1}{a_s k_F})\bigg)$$ Because even with largest affords i could not reproduce this. However i could reproduce the BEC limit in Formula 16.57 by performing a Taylor expansion inside of the integrand by rescaling with the large chemical potential, and hence Leibnitz rule was applicable.

So my final question would be, how do i mathematically corretly reproduce this BCS limit?

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