where the momentum integral can be evaluated using (2.13). The result is Setting m = 0, the Feynman parameter integral can be evaluated in terms of the Beta function (B.11). The result is B ( d ; n$_{1}$ , 0 ; n$_{2}$ , m ) = i (4 π ) 2 $^{(}$q 2 ) 2 − n$_{1}$ − n$_{2}$ [ − q 2 4 πµ 2 ] d $_{2}$− 2 Γ $^{(}$n$_{1}$ + n$_{2}$ − d 2 ) Γ( n$_{1}$ )Γ( n$_{2}$ ) ∫ 1 0 dx x n$_{2}$ − $^{1}$(1 − x ) n$_{1}$ − 1 [ x (1 − x ) − xm 2 q 2 ] d $_{2}$− n$_{1}$ − n$_{2}$ . (2.18) B ( d ; n$_{1}$ , 0 ; n$_{2}$ , 0) = i (4 π ) 2 $^{(}$q 2 ) 2 − n$_{1}$ − n$_{2}$ [ − q 2 4 πµ 2 ] d $_{2}$− 2 Γ ( d 2 − n$_{1}$ $^{)}$Γ ( d 2 − n$_{2}$ $^{)}$Γ $^{(}$n$_{1}$ + n$_{2}$ − d 2 ) Γ ( n$_{1}$ ) Γ ( n$_{2}$ ) Γ ( d − n$_{1}$ − n$_{2}$ ) . (2.19) B ( d ; n$_{1}$ , 0 ; n$_{2}$ , m ) = i (4 π ) 2 $^{(}$q 2 ) 2 − n$_{1}$ − n$_{2}$ [ − q 2 4 πµ 2 ] d $_{2}$− 2 Γ $^{(}$n$_{1}$ + n$_{2}$ − d 2 ) Γ( n$_{1}$ )Γ( n$_{2}$ ) z n$_{1}$ + n$_{2}$ − d 2 ∫ 1 0 dx x d $_{2}$− n$_{1}$ − $^{1}$(1 − x ) n$_{1}$ − $^{1}$(1 − zx ) d $_{2}$− n$_{1}$ − n$_{2}$ , z = 1 1 − m 2 q 2 . (2.20) B ( d ; n$_{1}$ , 0 ; n$_{2}$ , m ) = i (4 π ) 2 $^{(}$q 2 ) 2 − n$_{1}$ − n$_{2}$ [ − q 2 4 πµ 2 ] d $_{2}$− 2 z n$_{1}$ + n$_{2}$ − d 2 Γ ( d 2 − n$_{1}$ $^{)}$Γ $^{(}$n$_{1}$ + n$_{2}$ − d 2 ) Γ( n$_{1}$ )Γ( d $_{2}$) $_{2}$F$_{1}$$^{[}$ n$_{1}$ + n$_{2}$ − d 2 , d 2 − n$_{1}$ ; d 2 ; z$^{]}$ . (2.21) 57 This result agrees with an expression given for this integral in Ref. [97]. Also note that the result requires n$_{1}$ ≥ 0 and n$_{2}$ ≥ 0; if either of these is not satisfied the integral is a massless tadpole of the form (2.5) and is identically zero. In order to evaluate (2.18) when the mass is non-zero, note that it can be written as The Feynman parameter integral can be evaluated in terms of the Gauss hypergeometric function (B.16). The result is