INTEGRAL FUNCTIONS DEFINED BY TAYLOR’S SERIES. 
277 
Again, where .9 is sufficiently large, | & r+s | < l' r+s , where l'> l; and therefore 
l! f (-Yb r+s v (r) (6) 
r —0 r(r+l) 
< KZ /S , 
where K is finite and independent of .9 for all values of R, however large provided 
!L>V. 
Hence, when R tends to infinity, 
| {log {-x )} s V (-) r 6 r+ ,r (r) (fl) 
s=o r (,s+s) ?•=o r (r+ 1 ) 
< K 2 
5=0 
i 1 ' log ( } 9 
r(/3+s) 
+ T, 
where L is finite ; and thus the series is convergent provided |log(— x) | is finite. 
We therefore have the asymptotic equality 
(x ; 6) = 
{<£ (log (“*)} + 
J3-1 
V 
S = 1 
[log(-;r)l * 
r(^-s) 
Z (- ) r+s b r r (?,+s) (0) 
r= o r(r + s+l) 
+ J (*) j (~ x )~ e {log (—a:)}* 3-1 .(B). 
This equality is valid provided E (x) < 0. The function J (x) is such that its 
modulus, even when multiplied by \x\ \ however large the finite quantity l may be, 
can be made as small as we please by taking | x | sufficiently large. And <£ ( x ) denotes 
the integral function 
S - - c ’ | (- r&,w<->(0 = ; aft, , 
*=o. r (/3+s) ,-=o r(r+i) ,= 0 r(/3+«) ^ ‘ 
The quantity log(— x) is such that the modulus of its imaginary part is less 
than ?r/2. 
We see then that the process which we have employed, even in the case where (3 is 
an integer, makes the asymptotic expansion of F^ (x ; 6), when E (x) < 0, depend on 
that of the integral function of log (—x) which we have written cp {log (—,r)}. 
§ 35. We proceed now to apply the theory of contour integration to this problem. 
We have 
F p (x ; 6) 
x\ (n + 6) 
»=o(7H-0) p r(n+l) ’ 
and y ( x ) admits the expansion 2 b n /x n outside a circle of finite radius l. 
71 = 0 
We may represent F p (x; 6) by the contour integral 
