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Chapter 6
THE RIEMANN ZETA FUNCTION
6.1. Riemann s Memoir
In Riemann s only paper on number theory, published in 1860, he proved the following result.
THEOREM 6.1. The function ¶(s) can be continued analytically over the whole plane, and satisfies
the functional equation
s 1 - s
À-s/2“ ¶(s) =À-(1-s)/2“ ¶(1 - s), (1)
2 2
where “ denotes the gamma function. In particular, ¶(s) is analytic everywhere, except for a simple pole
at s =1 with residue 1.
Note that the functional equation (1) enables properties of ¶(s) for Ã
properties of ¶(s) for à >1. In particular, the only zeros of ¶(s) for Ã
other words, at the points s = -2, -4, -6, . . . . These are called the trivial zeros of ¶(s).
The part of the plane with 0 d" Ã d" 1 is called the critical strip.
Riemann s paper is particularly remarkable in the conjectures it contains. While most of the con-
jectures have been proved, the famous Riemann hypothesis has so far resisted all attempts to prove or
disprove it.
THEOREM 6.2. (Hadamard 1893) The function ¶(s) has infinitely many zeros in the critical strip.
It is easy to see that the zeros of ¶(s) in the critical strip are placed symmetrically with respect to
the line t = 0 as well as with respect to the line à =1/2, the latter observation being a consequence of
the functional equation (1).
This chapter was first used in lectures given by the author at Imperial College, University of London, in 1990.
6 2 W W L Chen : Elementary and Analytic Number Theory
THEOREM 6.3. (von Mangoldt 1905) Let N(T ) denote the number of zeros Á = ² +i³ of the
function ¶(s) in the critical strip with 0
T T T
N(T ) = log - + O(log T ). (2)
2À 2À 2À
THEOREM 6.4. (Hadamard 1893) The entire function
1 s
¾(s) = s(s - 1)À-s/2“ ¶(s)(3)
2 2
has the product representation
s
¾(s) =eA+Bs 1 - es/Á, (4)
Á
Á
where A and B are constants and where Á runs over all the zeros of ¶(s) in the critical strip.
We comment here that the product representation (4) plays an important role in the first proof of
the Prime number theorem.
The most remarkable of Riemann s conjectures is an explicit formula for the difference À(X)-li(X),
containing a term which is a sum over the zeros of ¶(s) in the critical strip. This shows that the zeros of
¶(s) plays a crucial role in the study of the distribution of primes. Here we state a result closely related
to this formula.
THEOREM 6.5. (von Mangoldt 1895) Let
È(X - 0) + È(X +0)
È(X) = ›(n) and È0(X) = .
2
nd"X
Then
XÁ ¶ (0) 1 1
È0(X) - X = - - - log 1 - ,
Á ¶(0) 2 X2
Á
where the terms in the sum arising from complex conjugates are taken together.
However, there remains one of Riemann s conjectures which is still unsolved today. The open
question below is arguably the most famous unsolved problem in the whole of mathematics.
CONJECTURE. (Riemann Hypothesis) The zeros of the function ¶(s) in the critical strip all lie
on the line à =1/2.
6.2. Riemann s Proof of Theorem 6.1
Suppose that à >0. Writing t = n2Àx, we have
" "
s 2
“ = ts/2-1e-t dt =(n2À)s/2 xs/2-1e-n Àx dx,
2
0 0
so that
"
s 2
À-s/2“ n-s = xs/2-1e-n Àx dx.
2
It follows that for à >1, we have
" "
" "
s 2 2
À-s/2“ ¶(s) = xs/2-1e-n Àx dx = xs/2-1 e-n Àx dx,
2
0 0
n=1 n=1
Chapter 6 : The Riemann Zeta Function 6 3
where the change of order of summation and integration is justified by the convergence of
"
"
2
xÃ/2-1e-n Àx dx.
n=1
Now write
"
2
É(x) = e-n Àx.
n=1
Then for à >1, we have
" 1
s
À-s/2“ ¶(s) = xs/2-1É(x)dx + ys/2-1É(y)dy
2
1 0
" "
= xs/2-1É(x)dx + x-s/2-1É(x-1)dx. (5)
1 1
We shall show that the function
"
2
¸(x) = e-n Àx =1 +2É(x)
n=- "
satisfies the functional equation
¸(x-1) =x1/2¸(x)(6)
for every x>0. It then follows that
2É(x-1) =¸(x-1) - 1 =x1/2¸(x) - 1 =-1+x1/2 +2x1/2É(x),
so that for à >1, we have
" "
1 1
x-s/2-1É(x-1)dx = x-s/2-1 - + x1/2 + x1/2É(x) dx
2 2
1 1
"
1
= + x-s/2-1/2É(x)dx. (7)
s(s - 1)
1
It follows on combining (5) and (7) that for à >1, we have
"
s 1
À-s/2“ ¶(s) = + xs/2-1 + x-s/2-1/2 É(x)dx. (8)
2 s(s - 1)
1
Note now that the integral on the right hand side of (8) converges absolutely for any s, and uniformly
in any bounded part of the plane, since É(x) =O(e-Àx) as x ’! + ". Hence the integral represents an
entire function of s, and the formula gives the analytic continuation of ¶(s) over the whole plane. Note
also that the right hand side of (8) remains unchanged when s is replaced by 1 - s, so that the functional
equation (1) follows immediately. Finally, note that the function
1 s
¾(s) = s(s - 1)À-s/2“ ¶(s)
2 2
is analytic everywhere. Since s“(s/2) has no zeros, the only possible pole of ¶(s) is at s = 1, and we
have already shown earlier that ¶(s) has a simple pole at s = 1 with residue 1.
It remains to establish (6) for every x >0. In other words, we need to prove that for every x >0,
we have
" "
2 2
e-n À/x = x1/2 e-n Àx.
n=- " n=- "
6 4 W W L Chen : Elementary and Analytic Number Theory
The starting point is the Poisson summation formula, that under certain conditions on a function f(t),
we have
"
B
f(n) = f(t)e2Ài½t dt, (9)
A
Ad"nd"B ½=- "
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