1814 CHAPTER 57. INFINITE PRODUCTS
Change the variable and conclude
Γ(z)1nz =
∫∞
0e−nssz−1ds.
Therefore, for Rez > 1,
ζ (z)Γ(z) =∞
∑n=1
∫∞
0e−nssz−1ds.
Now show that you can interchange the order of the sum and the integral. This is pos-sibly most easily done by using Fubini’s theorem. Show that ∑
∞n=1
∫∞
0
∣∣e−nssz−1∣∣ds<
∞ and then use Fubini’s theorem. I think you could do it other ways though. It ispossible to do it without any reference to Lebesgue integration. Thus
ζ (z)Γ(z) =∫
∞
0sz−1
∞
∑n=1
e−nsds
=∫
∞
0
sz−1e−s
1− e−s ds =∫
∞
0
sz−1
es−1ds
By 57.6.26,
I (z) =(ez2πi−1
)∫ ∞
0
e(z−1) ln(t)
et −1dt
=(ez2πi−1
)ζ (z)Γ(z)
=(e2πiz−1
)ζ (z)Γ(z)
whenever Rez > 1.
24. ↑ Now show there exists an entire function, h(z) such that
ζ (z) =1
z−1+h(z)
for Rez > 1. Conclude ζ (z) extends to a meromorphic function defined on all of Cwhich has a simple pole at z = 1, namely, the right side of the above formula. Hint:Use Problem 10 to observe that Γ(z) is never equal to zero but has simple poles atevery nonnegative integer. Then for Rez > 1,
ζ (z)≡ I (z)(e2πiz−1)Γ(z)
.
By 57.6.26 ζ has no poles for Rez > 1. The right side of the above equation isdefined for all z. There are no poles except possibly when z is a nonnegative integer.However, these points are not poles either because of Problem 10 which states that Γ
has simple poles at these points thus cancelling the simple zeros of(e2πiz−1
). The