256 
ME. E. W. BARNES ON THE ASYMPTOTIC EXPANSION OP 
By our original hypothesis the integral is convergent. It is finite for any assigned 
finite value of k, and when (x | is very large it tends to a finite limit. 
Hence |I 2 | tends exponentially to zero with l/\x\. 
Therefore for all finite values of k however large we may take \x\ so large that 
I— 2 c n /x p+n T (1 — /3—n) I | x^ +k ~ l | 
n= 0 
tends to zero as \x\ increases. 
Therefore I admits the asymptotic expansion 
oo 
t c n /xP +n r(l-P-n). 
= 0 
co 
Inasmuch as f(z ) admits 2 c n z n as a summable divergent series on the dissected 
72=0 
plane, we may say that, for our process for deriving an asymptotic expansion from an 
integral of the specified type to be possible, the contour C must be such that- for all 
points on and within it f(z ) must be representable by the summable divergent 
series. 
Part I. 
The Function G (x ; 6) 
x 
n =0 r(n+ 1) (71 + 6 ) 
§ 6. The function G (x \ 6) is a particular case of the function G^ (x ; 6) which will 
be considered in Part III. It can be discussed by more simple methods than are 
used in the more general case, and some of the formulae can only be deduced from the 
latter by employing the calculus of limits. I give here a brief summary of results 
and refer the reader elsewhere* for detailed analysis. We assume that 6 is not zero 
or a negative integer. 
I. By considering the contour integral 
5 Lf r <- s P , & 
2ttl j s + 6 
we can prove that 
—x ; 6) = —e~ x 2 —— 
72 = 0 
II. If | arg x | < 7T, 
G (— x ; 6)—x~ 6 Y (6) = — x~ e j" e~ y y 6 ~ 1 dy, 
J x 
where the line of integration is straight, tends to infinity in the positive half of the 
2 /-plane, does not cut the negative half of the real axis, and avoids the origin. 
G( 
l) (6 n) + (0-lJ.J(6 k l 
“X, ' ^ lF ' 1 ' ^ ' "" 
* See the ‘Quarterly Journal of Mathematics,’ 1906, vol. 37, pp. 289-313. 
