1550251515-Classical_Complex_Analysis__Gonzalez_

Sequences, Series, and Special Functions 619

for n > 0 yields

lira r(z) = 00

z-+-n

(8) Suppose for the moment that 0 < Rez < 1. Then we may write

r(z) = 1

00

e-ttz-l dt

r(l - z) = 1

00

e-ss-z ds

Hence

r(z)r(l - z) = 1

00

1

00

e-(t+s)tz-·ls-z dtds

0 0.

and letting t + s = x, t/ s = y, we find that

r(z)r(l - z) = roo roo e-xyz-1 ldx dy

lo lo + Y

= r(l) roo yz-1 dy

lo 1 + Y

But 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 that

Rep > 0. We obtain

r(z) = 100 e-pt'pzt'z-1 dt'

or dropping the primes,

r(z) = roo e-pte-1 dt

pZ lo

Rep> 0, Rez > 0

tThis integral is evaluated in Section 9.11 (formula (9.11-32)].

(8.20-14)