OF UNRESTRICTED DEGREE, ORDER, AND ARGUMENT. 
455 
r(+1+) 
— (t'— l) n t' s ~ i dt, 
Jo 
it (%) n (s i \ 
which can easily be shown to be equal to 2i sin n-n ■ - / ■— - - • 
1 n (n + s + i) 
The expression with which we commenced is now r reduced to the form 
f„ m . — . 2 l sin rnr. e (n+m+1)in (/x 3 — I)** t 
II (n + m + 2s) II (n) II (s — -J) 1 
=o II ( tl -T Vfi) II (2s) XI ( n + s + -g-) /x' 
n+m\ 2 s +1 
which is 
2t sin utt . e (n+m+1)in (fj 2 — l) im — ^ l 2 - 
/*" 
n ( n + ■§•) 
1 fn + vi , .. n + m + 1 
i —x-h I, 
R+Wi-f 1 
9 
J ^ + fj 0 
When is a positive integer, we have in accordance with the usual definition ot 
Q» (a 1 )’ 
{M = w' — ft r r~u w,Mt + i»~ w 
2« +1 n (?i + ^) /x b 
hence, in this case, if we take 
we have 
/.° = 
g— (»+l)iir 
Q«(/*) = 
4x sin ?i7r 
g—(»i+l)irr r(—1 + , 1—) 
4x sin ?X7r 
r(-i+> l—> 1 
J ± (& - i)* (t - r)—'dt. 
Defining Q/ 2 (/x) when m is a positive integer, by means of the equation 
Q.-M = (/**-! ) ! '*|t.Q*(A 
we have 
Q** w = £k‘d. u 3 - 1)*” r i c - an* - a*)-*— 1 *, 
n (») 
we should consequently, when m and n are positive integers, choose f n m equal to 
e- {n+1)m n (n + m) 
• q • 
4tsinw7r II ( n) 
We shall now assign this value to f™, whatever m and n are ; we thus obtain the 
formula 
Q.* M = 
4t sin mr II ( n ) 
(t 2 — 1)" (t — dt (9) 
