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The explicit form of the operator T µ$_{1}$...µ$_{r}$λ$_{1}$...λ$_{s}$ can be determined by taking derivatives of (2.50) with respect to a$_{1}$ , a$_{2}$ , and then setting a$_{1}$ = a$_{2}$ = 0. For instance, consider the operator corresponding to the tensor structure k µ $_{1}$: 

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where the ellipses indicate additional terms that vanish when a$_{1}$ and a$_{2}$ are set to zero. Notice that each factor of α$_{i}$ multiplying G ( d $^{)}$( q $^{2}$) is proportional to a derivative of G ( d $^{)}$( q $^{2}$) with respect to m 2 $_{i}$. In addition, we can absorb the factor of D ( α ) in (2.54) into G ( d $^{)}$( q $^{2}$), effectively sending d → d + 2 in (2.50). Finally, note that when a$_{1}$ and a$_{2}$ are set to zero in G ( d $^{)}$( q $^{2}$) ( i.e. in Eq., (2.47)), what remains is the loop integral that the T operator acts upon in Eq., (2.45). Therefore, the T operator corresponding to the tensor k µ 1 is given by 

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Generalizing this result, the T operator corresponding to the tensor k µ$_{1}$ 1 . . . k µ$_{r}$ 1 k λ$_{1}$ 2 . . . k λ$_{s}$ 2 is given by 

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Using T operators, any d -dimensional tensor integral which is of the same form as (2.45) 

