which corresponds to the integral (2.36) with m = 0. This integral occurs in Chapter 6. Setting the mass to zero does not change the recurrence relation (2.40), and therefore the integral above can be written as The integrals on the right hand side can be calculated using (2.19) and (2.24). Setting d = 4 + 2 ϵ and expanding using the properties of the Gamma function to perform the expansion, we find where ζ denotes the Riemann Zeta function. This integral is calculated in Ref. [97] using position space methods [34], and is in complete agreement with (2.61). Note that loop integrals that do not involve massive propagators can always be expressed in terms of Gamma functions which can be expanded easily using any computer algebra system. As we have seen, integrals that involve an external momentum and massive propagators tend to lead to hypergeometric functions whose indices are d -dependent. Therefore, after setting d = 4 + 2 ϵ we are required to expand around ϵ = 0 in the indices of a hypergeometric function, which can often be a non-trivial task. For example, consider the integral V ( d ; 1 , m ; 1 , 0 ; 1 , 0 ; 1 , 0) = 1 µ 2( d − 4) ∫ d $^{d}$k$_{1}$ (2 π ) d ∫ d $^{d}$k$_{2}$ (2 π ) d 1 ( k 2 1 − m $^{2}$) ( k$_{1}$ − q ) $^{2}$( k$_{2}$ − q ) $^{2}$( k$_{1}$ − k$_{2}$ ) 2 (2.62) 71 F ( d ; 1 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0) = − 1 (4 π ) 4 1 q 2 [ − q 2 4 πµ 2 ] ϵ ζ (3) , (2.61) F ( d ; 1 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0) = 1 d − 4 [ F ( d ; 2 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0 ; 0 , 0) − F ( d ; 2 , 0 ; 0 , 0 ; 1 , 0 ; 1 , 0 ; 1 , 0) + F ( d ; 1 , 0 ; 1 , 0 ; 2 , 0 ; 1 , 0 ; 0 , 0) − F ( d ; 1 , 0 ; 1 , 0 ; 2 , 0 ; 0 , 0 ; 1 , 0)] . (2.60)