253 
1907-8.] Dr W. H. Young on a Test for Continuity. 
function of y, and its integral may be got by “ term-by-term integration, 
and therefore by integration under the integral sign. Thus, for example, 
f sin x. log [ U (y)dy . 
Jo X Jo 
Example 2 . — To prove that the infinite product 
cos x. cos *x . . 
1 1 
. . COS -X. COS -X 
n n + 1 
is convergent and defines a bounded continuous function of x for all 
finite values of x. 
Let 
F n (flj) = COS X . cos \x ... . COS -X , 
then F n (x), being the product of n continuous functions, is a continuous 
function of x for all finite values of x. Also, since the cosine of a quantity 
lying in the closed interval ^0 , > 0 and <_ 1 , F x , F 2 . . . . form a mono- 
tone decreasing sequence, whose limit is therefore an upper semi-continuous 
function of x in the closed interval ^0 , , lying between 0 and 1, both 
inclusive : this limit is, however, the infinite product in question. 
Again, by the product form of sin x, the infinite product 
/2 • TT# 
. ■■= v sm — — 
/( , )= ( i -f_X i - 4 lX i - i -). 
J2 
is a continuous function of x throughout the closed interval (0 , 1), there- 
fore the same is true of each of the functions 
f n (x) = cos * . cos jx cos . (l - ig)(l - ■■■ ■ 
which form a monotone increasing sequence, since 
cos y > 1 - J?/ 2 . 
The limit of this latter sequence is therefore a lower semi-continuous 
function of x in the closed interval ^0 , ^ ; but this limit is the same in- 
finite product as before, and is thus a continuous function of x in the 
closed interval ^0 , ^ , lying between 0 and 1 , a fortiori between — 1 and 
+ 1. In the interval the same argument applies when we omit the 
