117 where µ is the chemical potential. Therefore we obtain (with τ =0 as µ =0 ) Substituting this into Eq. (A.13) gives Now, we can obtain Eq. (A.17) from Eq. (A.12) by simply changing the integration variable as We can make progress in trying to solve the integral of J (1) − as follows. J ( − 1) can be written as where K is the modified Bessel function of the second kind. There exists a recurrence relation J (1) − ( | m$_{φ,}$$_{eff}$ | T , 0 ) ≡ 2 ∫ ∞ | m φ, eff | T √ x 2 − ( | m$_{φ,}$$_{eff}$ | T ) 2 e x − 1 d x (A.16) 〈 φ 2 〉 T = T 2 2 π 2 ∫ ∞ | m φ, eff | T √ x 2 − ( | m$_{φ,}$$_{eff}$ | T ) 2 e x − 1 d x (A.17) J ( − 1) = 2 ∞ ∑ n =1 K$_{0}$ ( nκ ) (A.21) ∂J ( ν ) ∓ ∂κ = − νκJ ( ν − 2) ∓ (A.22) k → x ≡ ω$_{k}$ T (A.18) = √ k $^{2}$+ ∣ m 2 φ, eff ∣ T (A.19) d x = k ω$_{k}$T d k (A.20)