14.10. THE GAMMA AND ZETA FUNCTIONS 365

Now z→ 1kz is analytic and so z→ ∑

nk=1

1kz is also analytic. Since the series converges

uniformly for Rez ≥ 1l + 1, it follows that z→ ∑

∞k=1

1kz is also analytic on Rez > 1. This

applies Lemma 14.8.5 to a small circle centered at z with Rez > 1, the circle lying inRez ≥ 1+ 1

l for some l ∈ N. The uniform convergence implies the limit of the functions∑

nk=1

1kz as n→ ∞ is analytic.

Proposition 14.10.5 z→ ζ (z) is analytic on Rez > 1 and on this region the followingidentity holds. ζ (z)Γ(z) =

∫∞

0 (es−1)−1 sz−1ds. In particular, z→∫

0 (es−1)−1 sz−1ds isanalytic on Rez > 1.

Consider 1ez−1 . This is undefined when z = 0. Writing it in terms of a power series in

the bottom, it equals1

z+ z2

2 + p̂(z)=

1z

11+ z

2 + p(z)

where p(z) is a power series beginning with exponent 2. Then for small z this is of theform

1z

(1−( z

2+ p(z)

)+( z

2+ p(z)

)2−( z

2+ p(z)

)3+ · · ·

)and this equals 1/z−1/2+q(z) where q(z) is some power series which begins with expo-nent 1. Thus 1

ez−1 −1z equals a power series which will converge near z = 0.

Now consider ∫ 1

0

(1

et −1− 1

t

)tz−1dt

The function 1ez−1 −

1z is unbounded near 2kπi for k a nonzero integer but elsewhere is

analytic. In particular, there is a power series p(z) whose first nonzero term is of degree 2such that

1ez−1

− 1z=

1z+ p(z)

− 1z=

−p(z)(z+ p(z))z

which can be redefined at z = 0 to make it analytic. Thus if |z| ≤ 1, this function 1ez−1 −

1z

will be bounded. By Theorem 14.10.2, applied to f (t) the zero extension of 1et−1 −

1t off

[0,1] , z→∫ 1

0( 1

et−1 −1t

)tz−1dt is analytic. It follows that for all Rez > 1,

ζ (z)Γ(z) =∫

0

(et −1

)−1 tz−1dt

=∫ 1

0

(1

et −1− 1

t

)tz−1dt +

∫ 1

0

1t

tz−1dt +∫

1

(et −1

)−1 tz−1dt

=

analytic on Rez>0︷ ︸︸ ︷∫ 1

0

(1

et −1− 1

t

)tz−1dt +

analytic at z ̸=1︷ ︸︸ ︷1

z−1+

analytic on Rez>0︷ ︸︸ ︷∫∞

1

(et −1

)−1 tz−1dt

That last integral is analytic because you can let f (t) be 0 on [0,1] and 1et−1 ≤ 3e−t . Indeed,

it is clear that et

et−1 ≤ 3 for t ≥ 1. Now apply Theorem 14.10.2. Thus the right side extendsthe product ζ (z)Γ(z) to an analytic function valid for Rez ∈ (0,1). By Corollary 14.10.3,z→ ζ (z) can be extended to an analytic function on 0 < Rez < 1 given by

ζ (z) =1

Γ(z)

(∫ 1

0

(1

et −1− 1

t

)tz−1dt +

1z−1

+∫

1

tz−1

et −1dt)

14.10. THE GAMMA AND ZETA FUNCTIONS 365Now z > z is analytic and so z > Y7_, z is also analytic. Since the series convergesuniformly for Rez > $ +1, it follows that z+ Yr, z is also analytic on Rez > 1. Thisapplies Lemma 14.8.5 to a small circle centered at z with Rez > 1, the circle lying inRez >1+ + for some / € N. The uniform convergence implies the limit of the functionsLi_| £ asin @~ is analytic.Proposition 14.10.5 z— € (z) is analytic on Rez > 1 and on this region the followingidentity holds. € (z)V'(z) = fy’ (e’— 1) 7! s*!ds. In particular, z > Jy (e’ — 1) 7! s*"ds isanalytic on Rez > 1.Consider =: This is undefined when z = 0. Writing it in terms of a power series inthe bottom, it equals1 | 1z+ 5+ A(z) — 2 1+5+p(2)where p(z) is a power series beginning with exponent 2. Then for small z this is of theform£(1- Gena) (+90) (96)' +)and this equals 1/z— 1/2+q(z) where q(z) is some power series which begins with expo-nent 1. Thus 75 — i equals a power series which will converge near z = 0.Now consider| (1! etyo \e—-1 tft1 1The function => — ; is unbounded near 2kzi for k a nonzero integer but elsewhere isanalytic. In particular, there is a power series p(z) whose first nonzero term is of degree 2such that1 1 1 1 —p(z)e&-1 z ztp(z) z. (ztpl(2))zwhich can be redefined at z = 0 to make it analytic. Thus if |z| < 1, this function a — 1will be bounded. By Theorem 14.10.2, applied to f (t) the zero extension of a4 — + off[0,1], 2 fo (a4 — 41) ldr is analytic. It follows that for all Rez > 1,cera = [ (é-1) teatTf 1 1\ Ly Tq, i Ae= | —-}f* ar+ | —t ar+ | (e'—1) ‘dto \e—-1l tf 0 t 1analytic on Rez>0 analytic at zA1 analytic on Rez>0——= [ tot dt + = +[o (e! —1) | lato \e—-1 tft z-l 1That last integral is analytic because you can let f (t) be 0 on (0, 1] and x4 <3e™. Indeed,it is clear that a <3 fort > 1. Now apply Theorem 14.10.2. Thus the right side extendsthe product € (z)I’(z) to an analytic function valid for Rez € (0,1). By Corollary 14.10.3,z— €(z) can be extended to an analytic function on 0 < Rez < | given by6-5 (f (4 -t) ett te [fRaa)