Mr Dixon, An example of complex double integration. 391 
and there being no term on the left without x, the coefficients of 
2 / ,l_1 in <£ and y m ~ l in -yjr must be as b to — a ; suppose that they are 
bcmn and — acmn. Then 
cj) — cm dv/dy and y\r + cn du/dy 
contain the factor x ; suppose the other factors to be co. Thus 
/ dv\ ( du\ 
xw = u [xx + cm + v yxcn — cn J 
= ® ( u x + vo) + c i)> 
and W — £7% — Vco — cJ is of degree m + n — 3 at most, so that the 
lemma is proved. 
The special case when W = J is worthy of notice : the number 
(mn) of solutions of U= 0 = V is given as j jjfydydx* over 
the field specified. To verify this we note that since J is of degree 
m+n— 2, the integral need not tend to zero as R increases but 
that only the terms of the highest degree in U, V, J will influence 
the result. The result will then not be affected if we substitute 
for U, V products of linear factors, so long as the highest terms in 
U, V are unchanged. Thus we may take 
m 
U = a IT (y — y r ) where y r = l r x + f r , 
r=l 
V — b II (y — rjr) where g r = g r x+ h r . 
r=l 
Then it is not hard to prove that 
m n 
J/Ur=Z 2 (g. - l,)Ky - y.) (y - 
r= 1 s=l 
(see Camb. Phil. Proc. Vol. xn. pt. 6, p. 30). Each of these mn 
fractions contributes (2i7r) 2 to the integral, so that the result 
follows. 
* It is stated by Picard and Simart that this integral, which was studied by 
Didon, does not give the number of common zeros of the factors of the denominator 
enclosed by the field of integration. (Fonctions algdbriques de deux variables 
independantes, vol. i. ch. 3.) 
