invariant nature of loop integrals. A simple example is as follows: Due to Lorentz invariance, the integral must be proportional to the external momentum q µ and a function of the Lorentz scalar q $^{2}$. Contracting both sides of (2.42) with q$_{µ}$ , the function f ( q $^{2}$) can be determined and hence the integral on the left hand side of (2.42) can be calculated. This technique can be easily generalized to accommodate more complicated tensor structures. Ref. [98] constructs an algorithm to solve this problem in general. However, we will use a far more powerful method to deal with tensor integrals such as (2.42). The method that we will utilize was developed in Ref. [117] and relates tensor integrals in d -dimensions to scalar integrals in d + 2 N -dimensions. In Ref. [118] the method was applied to two-point propagator-type integrals, that is, integrals of the type A , B , J , V and F that we have considered so far. This method can be used to find a truly minimal set of basis integrals for the family of two-loop two-point integrals that we have studied so far. The key idea of the method is that it is possible to express an arbitrary tensor integral as where we have omitted the renormalization scale for brevity. The tensor operator T µ$_{1}$...µ$_{r}$λ$_{1}$...λ$_{s}$ is a function of the external momentum q $^{µ}$, derivatives with respect to each mass m$_{i}$ , and an operator d + that increases the dimension of any given loop integral by two units, i.e. d $^{+}$I( d ) = I ( d $^{+2)}$. We will now construct the explicit form of this operator. We will follow the derivation given in Ref. [118], although we will retain the loop integral normalization convention that has been used in this chapter. Introducing the auxiliary vectors a$_{1}$ and a$_{2}$ , 66 ∫ d $^{d}$k$_{1}$ (2 π ) d ∫ d $^{d}$k$_{2}$ (2 π ) d k µ$_{1}$ 1 . . . k µ$_{r}$ 1 k λ$_{1}$ 2 . . . k µ$_{s}$ 2 c n$_{1}$ 1 c n$_{2}$ 2 c n$_{3}$ 3 c n$_{4}$ 4 c n$_{5}$ 5 ≡ T µ$_{1}$...µ$_{r}$λ$_{1}$...λ$_{s}$ ∫ d $^{d}$k$_{1}$ (2 π ) d ∫ d $^{d}$k$_{2}$ (2 π ) d 1 c n$_{1}$ 1 c n$_{2}$ 2 c n$_{3}$ 3 c n$_{4}$ 4 c n$_{5}$ 5 , (2.43) c$_{1}$ = k 2 1 − m 2 $_{1}$, c$_{2}$ = k 2 2 − m 2 $_{2}$, c$_{3}$ = ( k$_{1}$ − q ) 2 − m 2 $_{3}$, (2.44) c$_{4}$ = ( k$_{2}$ − q ) 2 − m 2 $_{4}$, c$_{5}$ = ( k$_{1}$ − k$_{2}$ ) 2 − m 2 $_{5}$. (2.45) 1 µ 2( d − 4) ∫ d $^{d}$k$_{1}$ (2 π ) d ∫ d $^{d}$k$_{2}$ (2 π ) d k µ 2 k 2 ( k$_{1}$ − q ) $^{2}$( k$_{1}$ − k$_{2}$ ) 2 ≡ q $^{µ}$f $^{(}$q 2 ) . (2.42)