How does one go about evaluating the behavior of the BCS gap Δ=Δ(T) for T→0+ under the weak coupling approximation Δ/ℏωD≪1?
In Fetter & Walecka, Quantum theory of Many-Particle Systems, Prob. 13.9 it is said that the starting point is lnΔ0Δ=2∫ℏωD0dξ√ξ2+Δ21eβ√ξ2+Δ2+1, which I have no problem deriving from the theory, but I can't find a way to actually evaluate this integral even under the approximations \hbar\omega_D \to \infty , \Delta \approx \Delta_0 (in the RHS) and \beta\Delta \to \infty . [Of course \beta = (k_BT)^{-1} and \Delta_0 = \Delta(T = 0).] I have tried several approaches, used different Taylor-expansions and changes of variables, but I am simply stuck.
For the record, the expected behavior is supposed to be \tag{2} \Delta(T) \sim \Delta_0\left(1 - \sqrt{\frac{2\pi}{\beta\Delta_0}}e^{-\beta\Delta_0}\right) .
EDIT: Just leaving it here for the posterity. I found a more complete way to tackle this integral; specifically, under the WC approximation one has \int_0^{+\infty}{\frac{\mathrm d x}{\sqrt{x^2 + 1}}\frac{e^{-\beta\Delta\sqrt{x^2 + 1}}}{1 + e^{-\beta\Delta\sqrt{x^2 + 1}}}} = \int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}}\frac{e^{-\beta\Delta y}}{1 + e^{-\beta\Delta y}}} = = \sum_{k=1}^{+\infty}\int_1^{+\infty}{\frac{\mathrm d y}{\sqrt{y^2 - 1}} (-1)^{k+1}e^{-k\beta\Delta y}} = \sum_{k=1}^{+\infty}(-1)^{k+1}\int_0^{+\infty}{\mathrm d t\; e^{-k\beta\Delta \cosh{t}}} = \sum_{k=1}^{+\infty}(-1)^{k+1} K_0(k\beta\Delta),
K_0 being the 0-order modified Bessel function of the second kind, whose asymptotic behavior is known and may be used to solve the problem in a relatively clean way (and even find the corrections at higher orders, which are \in O(e^{-\beta\Delta}(\beta\Delta)^{-k - 1/2}) . Cf. Abrikosov, Gorkov, Dzyaloshinski, Methods of Quantum Field Theory in Statistical Phyisics, 1963. Pagg. 303-304.
Answer
Hints:
Define difference \delta:=\Delta-\Delta_0. Deduce from |\delta|\ll |\Delta_0| that the lhs. of eq. (1) is \tag{A}\text{lhs}~\approx~ -\frac{\delta}{\Delta_0}.
Substitute \xi=x\Delta in the integral on the rhs. of eq. (1). Deduce using \hbar \omega_D \gg \Delta that the rhs. is \tag{B} \text{rhs}~\approx~ \int_{\mathbb{R}} \! \frac{dx}{\sqrt{1+x^2}} \frac{1}{e^{\beta\Delta \sqrt{1+x^2}}+1}.
Deduce from \beta\Delta\gg 1 that we can simplify the rhs. further to a Gaussian integral \tag{C} \text{rhs}~\approx~ \int_{\mathbb{R}} \! dx~ e^{-\beta\Delta (1+\frac{1}{2}x^2)}~=~\sqrt{\frac{2\pi}{\beta\Delta}}e^{-\beta\Delta} . Such arguments are closely related to the method of steepest descent.
Deduce eq. (2).
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