Using the result (2.26), setting d = 4 + 2 ϵ , and working in the MS scheme, we find In order to expand the hypergeometric function, it is useful to write it in series form Now we must expand the sum around ϵ = 0, and because of the overall Γ ( − ϵ ) term we must expand the sum to O ( ϵ ) in order to expand the entire integral to O ( ϵ $^{0}$). First, note that for n ≥ 1 we may safely set ϵ = 0 in the sum. Extracting the n = 0 term, we have Using the properties of the Gamma function, this can be written as where ψ ( n ) is the Polygamma function (B.6). Evaluating the sums and performing the 72 ∞ ∑ n =0 f ( ϵ , n ) z n n ! = Γ ( − 2 ϵ ) Γ (1 + 2 ϵ ) Γ (2 + ϵ ) + ∞ ∑ n =1 z n n ( n + 1) + ϵ ∞ ∑ n =1 [ 1 n $^{2}$(1 + n ) 2 − ψ ( n ) n (1 + n )$^{]}$ z n , (2.66) ∞ ∑ n =0 f ( ϵ , n ) z n n ! = f ( ϵ , 0) + ∞ ∑ n =1 [ f (0 , n ) + ϵ d dϵ f ( ϵ , n )$^{∣}$ ∣ ∣ ϵ =0 ] z n n ! , f ( ϵ , n ) = Γ ( − 2 ϵ + n ) Γ (1 + 2 ϵ + n ) Γ (2 + ϵ + n ) . (2.65) V ( d ; 1 , m ; 1 , 0 ; 1 , 0 ; 1 , 0) = − 1 (4 π ) 4 exp$^{[}$2 ϵ$^{(}$ log$^{[}$ m 2 µ 2 ] − γ$_{E}$ − log [1 − z ]$^{)]}$ Γ $^{2}$(1 + ϵ ) Γ ( − ϵ ) Γ (2 + 2 ϵ ) ∞ ∑ n =0 Γ ( − 2 ϵ + n ) Γ (1 + 2 ϵ + n ) Γ (2 + ϵ + n ) z n n ! . (2.64) V ( d ; 1 , m ; 1 , 0 ; 1 , 0 ; 1 , 0) = − 1 (4 π ) 4 exp$^{[}$2 ϵ$^{(}$ log$^{[}$ m 2 µ 2 ] − γ$_{E}$ − log [1 − z ]$^{)]}$ Γ $^{2}$(1 + ϵ ) Γ ( − ϵ ) Γ (1 + 2 ϵ ) Γ ( − 2 ϵ ) Γ (2 + 2 ϵ ) Γ (2 + ϵ ) $_{2}$F$_{1}$ [ − 2 ϵ , 1 + 2 ϵ ; 2 + ϵ ; z ] , z = 1 1 − m 2 q 2 . (2.63)