L ( h ; µ ) where R is an r × r matrix with Let us write det R as or equivalently as Using this representation, we have Since the contour K is chosen so that for any z ∈ K , Re z < ( ∑ r j =1 h$_{j}$ 1+ h$_{j}$ ) − 1 S, the integrand in the above multiple integral is absolutely integrable on [0 , ∞ ) × K × ... × K , and Fubini’s theorem justifies the interchange of the order of the integrals, 27 ∫ ∞ 0 y p ( n − r )+ r ( r +1) / 2 − $^{1}$e − ny det R ( y )d y = ∑ ρ sgn ρ (2 π i) r × ∫ ∞ 0 ∮ K ... ∮ K y p ( n − r )+ r ( r +1) / 2 − 1 exp { − ( n − r ∑ j =1 nh$_{ρ}$$_{(}$$_{j}$$_{)}$ 1 + h$_{ρ}$$_{(}$$_{j}$$_{)}$ z$_{j}$ S ) y } × r ∏ j $_{=1}${ z j − 1 j p ∏ s =1 ( z$_{j}$ − λ$_{s}$ ) − 1 } d z$_{r}$... d z$_{1}$ d y. det R = ∑ ρ sgn ρ (2 π i) r ∮ K ... ∮ K r ∏ j $_{=1}${ e$^{y}$ S nh ρ ( j ) 1+ h ρ ( j $^{$_{)}$z$_{j}$}$z j − 1 j p ∏ s =1 ( z$_{j}$ − λ$_{s}$ ) − 1 } d z$_{r}$... d z$_{1}$ . (41) det R = ∑ ρ sgn ρ r ∏ j =1 1 2 π i ∮ K e$^{y}$ S nh ρ ( j ) 1+ h ρ ( j $^{$_{)}$z}$z j − 1 p ∏ s =1 ( z − λ$_{s}$ ) − $^{1}$d z, R$_{ij}$ = 1 2 π i ∮ K e y S nh i 1+ h $^{$_{i}$z}$z j − 1 p ∏ s =1 ( z − λ$_{s}$ ) − $^{1}$d z. L ( h ; µ ) = k$_{1}$ n $^{np}$Spr − r (1+ r ) / 2 Γ ( np ) ∫ ∞ 0 y p ( n − r )+ r ( r +1) / 2 − $^{1}$e − ny det R d y, (40)