It should be noted that qt η − $^{1}$/ Γ( η ) and L ( t ) t ρ − $^{1}$/ Γ( ρ ) actually represent the first two terms of the long-time expansion of h ( t ) only if η > ρ . In the opposite case, when L (1 /s ) /s ρ is the leading term of the asymptotic expansion of h$_{s}$ , this may not be true. The reason is that in this case the second term of the asymptotic expansion of L − $^{1}${ L (1 /s ) /s $^{ρ}$} at t → ∞ may not be negligible in comparison with qt η − $^{1}$/ Γ( η ). To illustrate this fact, let us assume that L (1 /s ) = k ln(1 /s ) ( k is a scale factor) and ρ = 2, and consider the Laplace transform h$_{s}$ = q/s η + ( k/s $^{2}$) ln(1 /s ). Since L − $^{1}${ (1 /s $^{2}$) ln(1 /s ) } = t (ln t + γ − 1), where γ = 0 . 5772 is the Euler constant [19], for the inverse Laplace transform of h$_{s}$ we obtain an exact result h ( t ) = qt η − $^{1}$/ Γ( η ) + kt (ln t + γ − 1). If η > 2 and t → ∞ then, keeping in h ( t ) the two leading terms, in accordance with (4.4) we find h ( t ) − qt η − $^{1}$/ Γ( η ) ∼ kt ln t . However, if η ≤ 2 then the second term of the asymptotic expansion of L − $^{1}${ L (1 /s ) /s $^{ρ}$} , k ( γ − 1) t , is not negligible compared to qt η − $^{1}$/ Γ( η ) and, as a consequence, the asymptotic formula (4.4) does not hold. In this case only the leading term of h ( t ), kt ln t , is determined from the Tauberian theorem. Thus, while at η > ρ the first two terms of the asymptotic expansion of h ( t ) can be determined from the modified Tauberian theorem, Eqs. (4.3) and (4.4), to solve this problem in the opposite case it is necessary to go beyond the Tauberian theorem. Next, we use the Tauberian theorem, Eqs. (4.1) and (4.2), and its modified version, Eqs. (4.3) and (4.4), to find the long-time behavior of the first two moments, 〈 Y$_{t}$ 〉 and 〈 Y 2 t 〉 , and the variance σ $^{2}$( t ) in the case of anomalous diffusion, i.e., when the conditions f = g$_{0}$ and α ∈ (0 , 2] hold simultaneously. Since the asymptotic solution of the CTRW is different for different intervals of α [20], we consider the cases with α ∈ (0 , 1), α ∈ (1 , 2), α = 1, and α = 2 separately. B. α ∈ (0 , 1) In this case it is convenient to represent the Laplace transform of the waiting time prob- ability density p ( τ ) in the form which follows from the definition p$_{s}$ = ∫ ∞ τ$_{min}$ dτ e − $^{sτ}$p ( τ ) and the normalization condition ∫ ∞ τ$_{min}$ dτ p ( τ ) = 1. Introducing the new variable of integration x = sτ and using the asymp- p$_{s}$ = 1 − ∫ ∞ τ$_{min}$ dτ (1 − e − $^{sτ}$) p ( τ ) , (4.5) 10