where their PP contributions take the form Therefore, the sum of all PP contributions of the left-hand side of Eq. (103) becomes Let us now look to the right-hand side of the identity (103). The diagrams are + 1 2 ( − i ) $^{2}$(2 e )( − 4 ie $^{2}$) ∫ d $^{4}$l (2 π ) 4 − ( g α α − 2)(2 l − k ) $^{λ}$gµν − ( l − 2 k ) $^{µ}$gνλ − ( l + k ) $^{ν}$gµλ l $^{2}$( p + q + l ) 2 × sin( l ∧ q ) sin( − l ∧ p − l ∧ q ) sin( − l ∧ p + p ∧ q ) + 1 2 ( − i ) $^{2}$(2 e )( − 4 ie $^{2}$) ∫ d $^{4}$l (2 π ) 4 3 k $^{ν}$gµλ − 3 k $^{µ}$gλν l $^{2}$( p + q + l ) 2 sin $^{2}$( l ∧ p + l ∧ q ) sin( p ∧ q ) , PP[ p$_{µ}$ Λ µνλ d 1 ,d 2 ,d $_{3}$( p, q, k )] = 9 e 2 32 π 2 1 ϵ [( k $^{2}$g λν − k $^{λ}$k$^{ν}$) − ( q $^{2}$g νλ − q $^{ν}$q$^{λ}$)] . (119) PP[ p$_{µ}$ ( γ $^{µνλ}$( p, q, k ) b ( p $^{2}$) + Λ µνλ b 1 ,b 2 ,c,d 1 ,d 2 ,d $_{3}$( p, q, k ))] = 7 e 2 48 π 2 1 ϵ [( k $^{2}$g λν − k $^{λ}$k$^{ν}$) − ( q $^{2}$g νλ − q $^{ν}$q$^{λ}$)] . (120) = i Π νλ b ( q ) (121) = − i $^{2}$(2 e ) 2 ∫ d $^{4}$l (2 π ) 4 l $^{λ}$( l + q ) ν l $^{2}$( q + l ) 2 sin $^{2}$( l ∧ q ) , = i Π νλ c ( q ) (122) = 1 2 ( − i ) $^{2}$(2 e ) 2 ∫ d $^{4}$l (2 π ) 4 γ $^{ανβ}$( l, q, − l − q ) γ $^{βνα}$( l + q, − q, − l ) l $^{2}$( q + l ) 2 sin $^{2}$( l ∧ q ) , = i Π νλ d ( q ) (123) = 1 2 ( − i )( − 4 ie $^{2}$) ∫ d $^{4}$l (2 π ) 4 2( g α α − 1) g νλ l 2 sin $^{2}$( l ∧ q ) , 20