MR. E. W. BARNES ON INTEGRAL FUNCTIONS. 
487 
<T+ l 
:■{(-*) o sin 
7r(a-+l) U+<T) 
+ T<J"(—O') + 
?/ (~)' S *£( — cr-T.s ) 
s=—p sz s 
Now, when — + - = j) = an integer, 
7T 
<X + 1 
Sill 
(o- + 1) 
« x becomes infinite. 
We have then to substitute a corresponding finite expression derived by proceedin, 
1 . 
to the limit in the infinite terms of the subsidiary Fourier series. When is not 
integral, the series and its equivalent value are given by 
+ 
•S'T 
1+CT 
+ 
( — ) s e sW 
ire 
•=' Wi 
l+o- 
sm 7 t 
<T + 1 1 + 
id. 
When - is integral, this series, omitting the term for which .s' = 1 + in the second 
T T 
summation, is equal to the finite part of 
cr+ 1 
T (1 + ei6 + . . .) 
1 + 0 " 
( —) t e 
1 + cr 
— id — 
1 +cr 
T (-W jg — 1 
—-- e t 
1+0- 
when e = 0 ; that is to say, it is equal to 
(-) 
s 
We thus replace 
+ 1 , _ <r+1 / . _ t 
’ e* v ( - 16 + 
1 + 0 -/ 1 + 0 - 
iO. 
cr+ 1 
7T 
sin 
7T (o- + ] ) 0 + 1 
'T+ 1 
i (—*)~ Jb 
by , {log 2 - 
1 + 0 " 
Again. since <x is a positive integer, the only term of the form 
£ (— cr — ts) s = — p, . . . —1,1,2,... 
which becomes infinite is that for which s == — p. 
This term is £ (+ 1), which arises from the Maclaurin series 
m— 1 ^ 
£ _ 
»=i n 
{(») + 
1 — s m s 
— ——h • • • = log m + y — -—b 
S=1 2 in & 1 / 2m * 
We have already taken account of the substitution of log m for _ . — 1 ; we need, 
therefore, only replace £(fi- 1) by y. 
We have then the asymptotic expansion 
