800 PROFESSOE BOOLE ON THE COMPARISON OF TRANSCENDENTS, WITH 
From 5=0 to s~— 
4a 
From s 
/W-0 (6.) 
= /W=/(«-i^) ('■) 
( 8 .) 
From to s=oo f((T)=0 
Let us now examine the corresponding values of the element ds under the sign of 
integration in (4.). 
1st. From 5=0 to s=^ that element is 0 by (6.). 
2ndly. At the point we have 
( 9 -) 
But O') ds is the increment ofy(ff) corresponding to an infinitesimal increment ds in 
the value of s. At the break, where S"^^f[(T) changes in value from 0 to /(O), the 
initial value of and the increment offia) is/^O), whatever the value o/ds may 
he^ provided only that it is infinitesimal. Hence at this point we have 
3rdly. From s=i to «=j(Al)> 
4thly. At the second break, where o'=l, we find ds 
5thly. From 5 = ^ -- ^^^ to s=oo we have ds Thus on recapitulation 
I 
comprises two finite elements whose sum is 
M /(i) 
4a^ 4(a — 1)®’ 
and a series of infinitesimal elements which give by integration 
4a 
whence (5.) becomes 
f /(I) 
/(O) 
t4(a-ir- 
4a^ 
Ja ' 
£‘““Vf«-s)} (19-) 
Now this result may be verified by integrating the first member of the equation by parts, 
and transforming the integral which remains by assuming x=a — 
