Prof. Bromwich, On a Definite Integral. 421 
Introducing z x , z 2 , ..., z n in V in place of x x , x 2 , ..., x n , we 
shall have 
V = 2c r z r 2 + 2 xfif r z r + px 0 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 0 2 is the value of V correspond- 
ing to the values 
^i = 0, z 2 = 0, . . ., z n — 0, 
or to 
?=°- 
oz x oz 2 
d f=0. 
OZn 
But, since z x , z 2 , ...,z n are connected with x X) x 2} ...,x n by 
a linear substitution whose determinant M is not zero 1 , it follows 
that the last set of equations are equivalent to 
du 
dx x 
= 0 , 
sr°- 
If these are solved for x x , x 2 , . .., x n we find 
UfiC r — UrX Q = 0, (r = 1, 2, . . ., n) 
where u r is the minor (with proper sign) of a w in the deter- 
minant u. Substituting these values in V, we have 
px 2 =V(x 0 , u x xfu Q , u 2 x 0 ju 0 , ..., u n x 0 lu 0 ), 
or pu 0 2 = V (u 0 , u x , u 2 , ..., u n ) 
— %b rs u r u s . (/, s = 0, 1, 2, . n) 
We can now evaluate the integral proposed, for we have, from 
the definition of M, 
Ve~ u dx x ... dx n 
(Ve~ u IM)dz x ...dz n , 
where the sign of M is supposed to be positive ; and since, for 
our present purpose, M is given only as 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 £ is 
from — qo to + oo , and so the original integral becomes 
u 0 ~i (. %c r z r 2 + 2 x 0 Xf r z r +px 0 2 ) exp [- %z r 2 - lx 0 2 ] dz x ... dz n . 
J (n) 
1 For we have seen that M 2 =u 0 ; and u 0 is the determinant of a definite 
quadratic form, and cannot, therefore, vanish. 
30—2 
