382 
MR. E. W. BARNES ON THE THEORY OF THE 
Ill the first pLice let tlie contour l)e taken to lie along the real axis, passing from 
+ CO to + 03, and cutting the axis lietween the points cr = 2 and cr = 3, as the 
contour 1 of the figure. This is equivalent to taking the integral round a contour 
enclosing the points 3 and + cc. 
When 1 s I < 1 the integral is, by the theorem of the preceding paragraph, finite ; 
and by Cauchy’s theorem it will be equal in value to the sum of the residues inside 
the contour. 
Now by § 53, when 5=2 + 1:, where k is an integer, 
= I 
Hence the value of the integral along the contour 1 is 
03 ^2 + i + i- 
/.• = o(b+2)! d(c- + ^ 
log r, (a) , 
and by Taylor’s theorem this expression is, under the assigned limitations, equal to 
log 
+ a) 
Ta («) 
(a) . 
Let us now make tlie contour expand until it becomes a straight line perpendicular 
to the axis of cr, cutting the axis between the points 2 and 3, and a half circle at 
infinity. The value of the integral will be unaltered, since the contour in expanding 
passes over no poles of the subject of integration. And by the theorem of the previous 
paragrapli the })art of the integral which is taken along the semicircle at infinity 
vanishes. Hence the integral along the perpendicular line (the contour numbered 2 
in the figure) is equal to 
log 
+ (() 
H(cc) 
— 2 + 3 ^’^ (n), when 
< 1 , 
But the integral and this expression both remain continuously finite when 1 2 j becomes 
greater than unity. They are therefore equal to one another for all values of 1 2 j. 
