446 
ME. E. W. BARNES ON INTEGRAL FUNCTIONS. 
The given series is therefore the asymptotic expansion of the function 
/(*)= (L) 
X 
1 - 
t 
X 
clt. 
Suppose first that the real part of x is positive; then, putting t = xz we have 
r“ e xll ~ z) 
/(*) = 
J o 
1 - 
dz. 
the integral being taken along a line along which T\ ( 2 ) is positive, that is, along 
which Tv (1 — 2 ) is negative. [These two conditions only differ when we consider a 
line practically parallel to the imaginary axis and therefore initially excluded.] 
~ er xy 
dy , the integral being taken along 
■1 V 
Putting 1 — 2 = — y, we have fix) = — 
a line along which P (y) is positive; and therefore 
e~ z dz 
f( x ) = -! 
( 1 ), 
the integral being taken along a line still in the positive half of the 2 plane. Thus 
e z dz 
+ e P z 
dz — 
f( x ) = “ 
Hence, if y be Euler’s constant,* 
er z — 1 
dz 
, where we take e | to be very small. 
f(x) = log c + y - 
dz — log e + log (— x) + terms which vanish with e I. 
Finally, on making j e | =0, 
f{x) = y + log (— x) + f 
J 0 
e 2 — 1 
dz. 
It will be noticed that the integral (l) obtained for f (x) has a pole along the line 
of integration so that it has an infinite number of values, all differing by '2ttl , which 
are implicitly involved in the logarithmic term. 
We see then that, when the real part of x is positive, the given series is the 
« . 00 x r 
asymptotic expansion of the function y -f- log (— x) + 2 —-• 
r=1 1 
Take next the case when the real part of x is negative. As in the first case, the 
r 00 e*(i-*) 
of j 
series in the asymptotic expansion 
for which H ( 2 ) is negative. 
Jot 
dz, the integral being taken along a line 
Thus it is the expansion of 
-CO —Til 
e x y 
1 y 
r e~ z dz 
dy along a line for which H (y) is negative 
along a line for which Tv ( 2 ) is positive. 
* See the author’s paper, ‘ Messenger of Mathematics,’ vol. 29, pp. 98 and 99. 
