57.6. EXERCISES 1815

only remaining possibility for a pole for ζ is at z = 1. Show it has a simple pole atthis point. You can use the formula for I (z)

I (z) ≡∫ 2π

0

e(z−1)(lnr+iθ)

ereiθ −1ireiθ dθ +

∫∞

r

e(z−1)(ln t+2πi)

et −1dt (57.6.27)

+∫ r

∞

e(z−1) ln t

et −1dt

Thus I (1) is given by

I (1)≡∫ 2π

0

1ereiθ −1

ireiθ dθ +∫

∞

r

1et −1

dt +∫ r

∞

1et −1

dt

=∫

γrdw

ew−1 where γr is the circle of radius r. This contour integral equals 2πi by theresidue theorem. Therefore,

I (z)(e2πiz−1)Γ(z)

=1

z−1+h(z)

where h(z) is an entire function. People worry a lot about where the zeros of ζ arelocated. In particular, the zeros for Rez ∈ (0,1) are of special interest. The Riemannhypothesis says they are all on the line Rez = 1/2. This is a good problem for youto do next.

25. There is an important relation between prime numbers and the zeta function due toEuler. Let {pn}∞

n=1 be the prime numbers. Then for Rez > 1,

∞

∏n=1

11− p−z

n= ζ (z) .

To see this, consider a partial product.

N

∏n=1

11− p−z

n=

N

∏n=1

∞

∑jn=1

(1pz

n

) jn.

Let SN denote all positive integers which use only p1, · · · , pN in their prime factoriza-tion. Then the above equals ∑n∈SN

1nz . Letting N→∞ and using the fact that Rez > 1

so that the order in which you sum is not important (See Theorem 58.0.1 or recalladvanced calculus. ) you obtain the desired equation. Show ∑

∞n=1

1pn

= ∞.