325 
(1—argth hp) and (1—wx"9 Ahr) always exceed a fixed value, hence 
Ty («, p) remains finite as x tends to unity. Accordingly the difference 
L(« OP) == qh (wv) 
is always finite, therefore the point 62 must be a singular point of 
L(z) and the continuation of L(z) across the circumference must be 
regarded impossible. 
To examine the behaviour of L(w6r) if vl, it will be sufficient 
to deduce a suitable expansion of the function 7}, (2, p). 
Putting 
ENE ; Wi eis, 
: l d En 1 
P (uy = —_—__ = — —___—__-_, 
Seca gee aes a ey 
we have at once 
r= © 
Th (@,p) = hi (nqy +hy) — p (nqy) |, 
n= 
and from this expression it is seen, that the application of a suitable 
summation-formula will lead to the desired result. 
I consider the set of trigonometrical series 
et SU ATL 
9 (t) = — = KL 
TE 1 k 
1 ‘© cos2akt 
bE) == Bes —, 
SS eae = 
l © sin 2x kt 
Is (t) = == Ding? je ’ 
and the identity 
Ome he He, 0 Or @(taypdt « or oo. (6) 
{| ( = gak po 
0 
Integration by parts transforms the indefinite integral into the 
expression 
h k=2m—1 h 
9, ( ti | 9; (t) | 4 (ta) + Sr Vak | nail ze ‚aen (t) | pl (tqy) + 
== 
| h 
=e gem en | on (: = ~ — 92m | gem(tg y) dt 
and here we have to introduce the limits 0 and o. In doing so 
h 
we must take into account the discontinuities of OP (— = and of 
g 
