Q$_{1}$ = α$_{3}$α$_{5}$ + α$_{4}$α$_{5}$ + α$_{2}$α$_{3}$ + α$_{3}$α$_{4}$ , Q$_{2}$ = α$_{4}$α$_{5}$ + α$_{3}$α$_{5}$ + α$_{1}$α$_{4}$ + α$_{3}$α$_{4}$ , Q$_{11}$ = − 1 4 ( α$_{2}$ + α$_{4}$ + α$_{5}$ ) , Q$_{22}$ = − 1 4 ( α$_{1}$ + α$_{3}$ + α$_{5}$ ) , Q$_{12}$ = − 1 2 α$_{5}$ . (2.53) T$_{µ}$ = 1 i ∂ ∂a µ 1 G ( d ) $^{(}$q 2 ) ∣ ∣ a$_{1}$ = a$_{2}$ =0 = q$_{µ}$ [ α$_{3}$α$_{5}$ + α$_{4}$α$_{5}$ + α$_{2}$α$_{3}$ + α$_{3}$α$_{4}$ + . . . ] G ( d $^{)}$( q $^{2}$) D ( α ) ∣ ∣ a$_{1}$ = a$_{2}$ =0 , (2.54) T$_{µ}$ = − (4 π ) 2 q$_{µ}$ [( i ∂ ∂m 2 3 ) ( i ∂ ∂m 2 5 ) +$^{(}$ i ∂ ∂m 2 4 ) ( i ∂ ∂m 2 5 ) +$^{(}$ i ∂ ∂m 2 ) ( i ∂ ∂m 2 3 ) +$^{(}$ i ∂ ∂m 2 3 ) ( i ∂ ∂m 2 4 )] d + , (2.55) T$_{µ}$$_{1}$$_{...µ}$$_{r}$$_{λ}$$_{1}$$_{...λ}$$_{s}$ $^{(}$q µ , ∂$_{k}$ , d + ) = 1 i r + s r ∏ i =1 ∂ ∂a µ$_{i}$ 1 s ∏ j =1 ∂ ∂a λ$_{j}$ 2 exp$^{[}$ i $^{(}$Q$_{1}$ ( a$_{1}$ · q ) + Q$_{2}$ ( a$_{2}$ · q ) + Q$_{11}$a 2 1 + Q$_{22}$a 2 2 + Q$_{12}$ ( a$_{1}$ · a$_{2}$ ) $^{)}$ρ$^{]}$$^{∣}$ ∣ ∣ a$_{k}$ = 0 α$_{k}$ = i∂$_{k}$ , ∂$_{k}$ = ∂ ∂m 2 k , ρ = − (4 π ) $^{2}$d + . (2.56) 68 Using T operators, any d -dimensional tensor integral which is of the same form as (2.45) 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 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 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 $_{1}$: