AND ON DEFINITE INTEGRATION. 4907 
then [%)] = (1) x (2) x 0(3) xa(4)......... 
F\(#+1) 
TEI) ro) Ca a RC Oe i ee oT ee (28) 
F(e+)—-F@) 
and a =0(v+1). Take the logarithm 
of each side of this equation, and we shall have 
F(e+1) -F,(2) = ee 
Now if in equation (13) we put 2 =0 we shall 
have the following value of F\(x) between the 
limits a=y and a=2, 
consequently x 
5 : | . log. o(x) dx + (A+1) 
[ oc) ] a ae a y 
y 
Slog. o(x) —log. o(y)t +Bilog. ‘o(@) —log. ‘a(y)h 
~ D§log. “o(@) —log. “o(y)t +&c., wef 
Multiply both sides of this equation by [o(e) ], 
and we shall have 
1 
. y log. 
L (2) aL) |xa ‘ 
, { . log.o(a@)da+(A+1) 
J 
jlog. o(x) —log. o(y)t +Blog. ‘o(a@) —log. ‘o(y)t 
a D} log. ““o(a) — log. “o(y) + vel SR: (29) 
