INTEGRAL FUNCTIONS DEFINED BY TAYLOR’S SERIES. 
271 
§ 26. We will next show that , if s = u + lv and 3ft (/3 + s)>0, when \ v \ is very large, 
j S (s) | < Ke~ iu 1 v 1 , where K is a finite quantity independent of v. 
For, if 3ft(d)>0, 
S(*) 
V(-s) 
r(/3) 
yt-'e- 9 * (1 -e~ y Y + ‘ v dy 
< 
r(£) 
lu 
•p - 1 
(I —e~ y ) u dy < K, 
where K is a definite finite quantity independent of v. 
But, when |v| tends to infinity, | r (— s) | v 1 tends to a definite finite limit. 
We therefore have the given theorem if 3ft (d) > 0. 
When 3ft($)>0, we can, if \6\ be finite, find a finite number N such that 
3ft(# + N)> 0. We can write down a modified integral which shall express all but 
the first N terms of S ( s ). The argument used above will hold for the modified 
integral; the proposition to be proved is evidently true for the first N terms of S (s). 
And therefore we may establish that the theorem is true under the sole limitation 
3ft(/3 + s)> 0, 6 not being zero or a negative integer. 
§ 27. We will next show that, if C x be a contour in the finite part of the plane 
parallel to the imaginary axis .and cutting the reed axis in a point for which 
3ft (s)> —3ft (/3), and if 3ft ( x) > 0, e~ x G p (x ; 6) = — f S (s) x s ds. 
2m J C, 
In the first place we see that this integral is finite. For the series for S (s) is con¬ 
vergent if 3ft (s + /3) > 0, and as | v \ tends to infinity, | S (s)x s | < K | cc m(s) | e 1 vargx 1 1 v 1 . 
Thus the subject of integration tends exponentially to zero if |arg x I <i 7r - 
Let now C be a contour embracing the positive half of the real axis and cutting 
this axis in the same point as Ci. As we are ultimately to make 3ft (/3) very large, 
we shall assume that C x includes the origin. 
Evidently by Cauchy’s theory of residues 
1 
2m 
r (n—.s') x s ds 
00 /_\)H -.n + m 
= t i-^-4— = e~ x x\ 
m =o m! 
Hence if N G^ (x ; 6) denote the sum of the first (N + 1) terms of the series by which 
Gp (x ; 6) is defined, 
e“* M Gp (x ; 6) = - J— | % xS( l s - 
PV ’ 2mJc»=ow!(n+^ 
Now, if | arg x \ < ^ 7 r, | F ( n—s) x s j tends exponentially to zero as n tends to infinity 
if 3ft (s) > a finite negative quantity, and if s be not in the neighbourhood of the poles 
of T(-s). 
Therefore the previous integral will vanish when taken round an infinite contour 
for which 3ft (s) > a finite negative quantity. Hence 
