MR. E. W. BARNES ON INTEGRAL FUNCTIONS. 
439 
*/» = ~ 
a relation which may also be written 
(z+ l)/(z+ 1) = 
2 sin 7rs 
2 sin 7 rz 
— x) z 1 dx, 
[x d (y)] (— x) z 1 dx. 
We thus see that 
z(z + l - P i) . . . (z + 1 - p m )f(z) 
= (-) 
»»+i 
2 sin 
~ [ [> (3 + p x - I) . . . (S + p m - l)y] (- x) z 1 dx, 
and also 
( z + 1 - a )f( z + 1) = 2^ir' z f [* + a ) 2/] (— ®)* ^ 
Therefore, since y is a solution of the equation 
£ + Pi ~ 1) • • • C& + p m ~ 1) y = — x (5- + a) y. 
we have 
/(*+!) = (-) 
_ V«-l Pll • • • (? + 1 pm) 
(z + 1 — «) 
The general solution of this difference-equation is 
/(*)• 
r (z) 
r(a — z) 
r (pd ... r (p m ) 
r («) ’ r ( Pl — z )... r ( Pm — 
_ a ® (u a > Pi, • • • p*), 
where nr ( 2 , a, p x ,. . . p m ) is a simply periodic function of 2 of period unity. 
We have then established the identity 
2 sin irz 
| F m (a, Pl , . . . ppi, —x)(— x) z 1 dx 
r( 2 ) ro - 2) r( Pl )...r( Pm ) 
r («) 
r (pi - 2) • • • r (p m — z) 
nr (2^ oc, pj, • • • p^*)* 
When a = pj, the expression on the left-hand side, and therefore that on the right- 
hand side must involve p 3 . . . p m only. Thus, when « = p ]5 nr ( 2 , a, p l5 . . . p ffi ) involves 
p 2 ... p m only. It must therefore be a function of a — p x ,... a. — p m . Not only so, 
but it cannot involve these quantities at all; for when a = p l5 nr will be a function of 
Fi — p 25 • • • Pi — p m , and yet it is independent of p x ; and so on when a = p 2 , ... a = p OT . 
Thus nr is a function of 2 , simply periodic of period unity, independent of a, p l5 ... p OT 
and m. 
Let m = 1 , a = pj = 1 ; then (a, p l5 p 3 , . . . p m , -r x) becomes e~ x , and the 
integral becomes 
2 sin 
— j e x ( — x) z 1 dx — T ( 2 ). 
7T2 J 
