Sequences, Series, and Special Functions 619for n > 0 yieldslira r(z) = 00
z-+-n(8) Suppose for the moment that 0 < Rez < 1. Then we may write
r(z) = 1
00e-ttz-l dt
r(l - z) = 1
00e-ss-z ds
Hencer(z)r(l - z) = 1
001
00
e-(t+s)tz-·ls-z dtds
0 0.
and letting t + s = x, t/ s = y, we find thatr(z)r(l - z) = roo roo e-xyz-1 ldx dy
lo lo + Y
= r(l) roo yz-1 dy
lo 1 + YBut the value of the .last integral is known to be 7r /sin 'TrZ. t Therefore,
'Tr
r(z)r(l - z) = -.-
Slll'TrZ(8.20-13)Now g(z) = r(z)r(l - z) - 'Tr CSC'TrZ is analytic in D = C-{O, ±1, ±2," ·}
and we have shown that g(z) = 0 in 0 < Rez < 1. Then, by the identity
principle, we have g(z) = 0 everywhere in D, so that (8.20-13) holds good
in D.
Remark The special value r(1/ 2 ) = .,/1i may be obtained from (8.20-13)
by letting z =^1 /2 and noting that r( x) > 0 for x > 0.
(9) In (8.20-1) let t = pt', where p is a complex constant such thatRep > 0. We obtain
r(z) = 100 e-pt'pzt'z-1 dt'
or dropping the primes,
r(z) = roo e-pte-1 dt
pZ loRep> 0, Rez > 0
tThis integral is evaluated in Section 9.11 (formula (9.11-32)].(8.20-14)