14.10. THE GAMMA AND ZETA FUNCTIONS 363

By Theorem 14.5.3,

12πi

∫C(z0,r)

f ′ (z)f (z)

dz =p

∑k=1

12πi

∫C(zk,rk)

mk

(z− zk)

+1

2πi

∫C(zk,rk)

g′k (z)gk (z)

=p

∑k=1

mk = p ■

14.10 The Gamma and Zeta FunctionsUp to now we have considered the exponential function, logarithms, and polynomials. Twoother very important examples of analytic functions are the gamma and Zeta functions.It turns out these two functions are related. I am giving the obvious generalization ofthe Gamma function from the real variable version. However, more complete discussionsemploy infinite products to describe the gamma function, a topic not emphasized in thisbook. See [10] for a more complete discussion of these two functions.

Definition 14.10.1 For Re(z)> 0 define Γ(z)≡∫

0 e−ttz−1dz.

Theorem 14.10.2 For f , | f (t)| ≤ Ce−rt ,r > 0, G(z) ≡∫

0 f (t) tz−1dt is analyticfor Re(z) > 0. In case f (t) = e−t so G = Γ, the fundamental identity Γ(z+1) = zΓ(z)holds for all Rez > 0.

Proof: Formally differentiating under the integral sign, one would expect that G′ (z) =∫∞

0 f (t) ln(t) tz−1dt. This just needs to be shown.Is z→

∫∞

0 f (t) ln(t) tz−1dt continuous for Rez > 0? Let 0 < δ < Re(z)< ∆. If Re(ẑ) ∈(δ ,∆) , then

∣∣ f (t) ln(t) t ẑ−1∣∣≤ g(t) where

g(t)≡{

| f (t)| |ln(t)| t∆−1 for t ≥ 1| f (t)| |ln(t)| tδ−1 for 0 < t < 1

a function in L1. That this is in L1 is shown using integration by parts. Now if zn → z,then eventually Rezn ∈ (δ ,∆) and so f (t) ln(t) tzn−1 → f (t) ln(t) tz−1 and by the domi-nated convergence theorem, the integrals also converge. Thus z→

∫∞

0 f (t) ln(t) tz−1dt iscontinuous.

Now for Re(z)> 0, (G(z+h)−G(z))h−1 =

=∫

0f (t)h−1

(tz+h−1− tz−1

)dt =

∫∞

0f (t)h−1

(∫ 1

0h ln(t) tz+sh−1ds

)dt

=∫

0

∫ 1

0f (t) ln(t) tz+sh−1dsdt

The integrand is absolutely integrable and so we can use Fubini’s theorem to obtain

(G(z+h)−G(z))h−1 =∫ 1

0

∫∞

0f (t) ln(t) tz+sh−1dtds

By continuity, limh→0∫

0 f (t) ln(t) tz+sh−1dt =∫

0 f (t) ln(t) tz−1dt and so it follows thatlimh→0 (G(z+h)−G(z))h−1 =

∫∞

0 f (t) ln(t) tz−1dt.The last claim about the identity follows from Theorem 14.9.2 applied to the analytic

function Γ(z+1)−zΓ(z) . It is valid on (0,∞) so the identity continues to hold for Re(z)>0. All these positive real numbers are limit points of the set where Γ(z+1)− zΓ(z) = 0.You could probably prove this through a computation also. ■

14.10. THE GAMMA AND ZETA FUNCTIONS 363By Theorem 14.5.3,1 i f'(2) 1 i mPre dz = —,2Mi IC(<,r) f z) py 2Hi IC( ze rK) (z—Zx)(1 8, (2)+5— [ = m=p201 IC(zp,14) xk (2) d re14.10 The Gamma and Zeta FunctionsUp to now we have considered the exponential function, logarithms, and polynomials. Twoother very important examples of analytic functions are the gamma and Zeta functions.It turns out these two functions are related. I am giving the obvious generalization ofthe Gamma function from the real variable version. However, more complete discussionsemploy infinite products to describe the gamma function, a topic not emphasized in thisbook. See [10] for a more complete discussion of these two functions.Definition 14.10.1 For Re(z) > 0 define T(z) = ff ett! dz.Theorem 14.10.2 for f, |f(1)| <Ce~",r > 0, G(z) = Jo f(t)! dt is analyticfor Re(z) > 0. In case f(t) =e so G=T, the fundamental identity T (z+ 1) = 2 (z)holds for all Rez > 0.Proof: Formally differentiating under the integral sign, one would expect that G’ (z) =Jo f (t)In(t)t* ‘dt. This just needs to be shown.Isz— Jo f (t)In(¢)t* ‘dt continuous for Rez > 0? Let 0 < 6 < Re(z) <A. If Re(2) €(5,A), then | f(t) In(t)t*~!| < g(t) where_f [rl |hin()|e4—! fore > 1sO={ pinot! ford er eta function in L'. That this is in L' is shown using integration by parts. Now if z, — z,then eventually Rez, € (6,A) and so f(t)In(r)t*"—! — f(t)In(¢)*—! and by the domi-nated convergence theorem, the integrals also converge. Thus z > fo’ f (t)In(t)t®~'dr iscontinuous.Now for Re(z) > 0, (G(z+h)—G(z))h l==f for (er -e'ar= [pie ( [minine's"tas) dtco pl=| [ fein@easat0 JoThe integrand is absolutely integrable and so we can use Fubini’s theorem to obtain(G(z+h)—G(2))h = [ [rom ()*"ldtdsBy continuity, limp 0 Jo f (t)In(c) 2+" dt = Jy f (¢)In (1) 'dt and so it follows thatlimo (G(z+h) —G(z))h-! = Jy f (t)In (0) "dt.The last claim about the identity follows from Theorem 14.9.2 applied to the analyticfunction (z+ 1) —zI'(z) . Itis valid on (0, °°) so the identity continues to hold for Re (z) >0. All these positive real numbers are limit points of the set where (z+ 1) — 2I'(z) =0.You could probably prove this through a computation also. Hi