Prof. Bromwich, On a Definite Integral. 421 



Introducing z 1} z. 2) ..., z n in V in place of x u x 2 , ..., x n , we 

 shall have 



V = %c r z r 2 4- 2x Xf r z r + P^o 2 , (r =1,2,..., n) 



where the quantities f r , p are certain new constants, of which 

 we only need to find p. Now px 2 is the value of V correspond- 

 ing to the values 



z 1 = 0, z 2 = 0, ..., z n = 0, 



dU ft dU n dU ft 



or to 5— = 0, k— = — = 0. 



But, since #1, £ 2 > •••j^w are connected with x 1 , x 2 , ...,x n by 

 a linear substitution whose determinant M is wo£ zero 1 , it follows 

 that the last set of equations are equivalent to 



^ =0 ^=0 ... a -^ = o. 



3#! ' dx 2 ' 3^n 



If these are solved for x u x 2 , ..., x n we find 



u x r — UrX = 0, (r = 1, 2, . . ., n) 



where u r is the minor (with proper sign) of a or in the deter- 

 minant u. Substituting these values in V, we have 



PXq == V \Xq, UiXqJUq, U^q/Uq, ..., Un^o/^o)> 



or £>w 2 = ^Oo, Ui, u 2 , ..., u n ) 



= %b rs u r u s . (r, s = 0, 1,2, . . ., n) 



We can now evaluate the integral proposed, for we have, from 

 the definition of M, 



s 



Ve~ u dx 1 ... dx n = I ( Ve~ u /M) dz x ... dz n , 



J (n) 



(n) J (n) 



where the sign of M is supposed to be positive ; and since, for 

 our present purpose, M is given only as u^, this assumption 

 merely requires that the square-root shall be positive. As already 

 remarked, M cannot vanish and so no difficulty can arise on this 

 account. 



The range of the variation of each of the variables z is 

 from — oo to + 00 , and so the original integral becomes 



u ~l I (%c r z r 2 + 2x Xf r z r + £># 2 ) exp [— %z r 2 — lx Q 2 ] dz x ... dz n . 



J (n) 



1 For we have seen that M 2 =u ; and u is the determinant of a definite 

 quadratic form, and cannot, therefore, vanish, 



30—2 



