Special values of the Riemann
zeta function via several
random variables
Takahiko Fujita
Faculty of Science and Engineering, Chuo University,
1-13-27, Kasuga, Bunkyo,Tokyo, 112-8551, Japan,
email:rankstatistics@gmail.com
1
Abstract: In this paper, using arcsine variables, we
π2
get a new elementary proof of ζ(2) = 6 , known as
the Basel problem and the Euler formula. Using
exponential variables, we get the formula like Euler.
We can also solve the Basel problem by using
Wigner’s semi-circle law and Legendre generating
function.
Keywords: arcsine random variable, Basel Problem,
Euler formula, Riemann’s Zeta Function, exponential
random variable, Wigner’s semi-circle law,
Legendre polynomial.
2
2000 AMS Subject classification: 11M06,
11M35, 60E05
1 Introduction
First we review our previous paper([1]).
Consider the Riemann zeta function
∞
∑
1
ζ(s) =
s
j
j=1
(for Re s > 1).
Using two independent Cauchy variables, we
obtain the following Euler’s formulae of the
Riemann zeta function, which is very classical
(see for example [9]):
Euler’s Formulae
1 ( π )2n+2
An
(1 − 2n+2 )ζ(2n + 2) =
.
2
2 2
Γ(2n + 2)
1
Here, the coefficients An are featured in the
series development
∞
∑
1
An 2n
=
θ
2
cos θ n=0 (2n)!
π
(|θ| < ).
2
We remark that many authors ([3, 4, 7, 8, 10]
π2
) have written elementary proofs of ζ(2) = 6 .
The problem of finding this value is known as
Basel problem ([2]).
We first review the density function of the
ratio of two independent random variables from
elementary probability theory.
Lemma1.1.
Consider two independent random variables
X, Y such that P (X > 0) = P (Y > 0) = 1 and
with density functions fX (x), fY (x).
Y
Then, f Y (x), the density function of X is
X
given by:
∫ ∞
f Y (x) =
fX (u)fY (ux)udu
X
0
.
Proof.
For x > 0, ∫ ∫
Y
fX (u)fY (v)dudv =
P ( X < x) =
u
<x
∫∞
∫ vx v
f
(v)dv
f
(u)du.
Then
differentiating
Y
X
0
0
both sides with respect to x, we get the result.
b) Applying Lemma 1.1. for
1
f|C1 | (x) = f|C2 | (y) = π2 1+x
2 1x>0 i.e.:
|C1 | ∼ |C2 | ∼ |C| where C is a Cauchy variable
1
with fC (x) = π1 1+x
2 , we get , for x > 0 :
∫ ∞
4
1
1
f Y (x) = 2
udu
2
2
X
π 0 (1 + u ) (1 + (ux) )
∫ ∞
1
2
= 2
du
π 0 (u + 1)(1 + ux2 )
∫ ∞
2
x
2
1
du
−
= 2
(
)
2
π 0 1 + u 1 + ux 1 − x2
∫ A
2
2
1
x
du
= lim 2
(
−
)
2
2
A→∞ π
1
+
u
1
+
ux
1
−
x
0
4 log x
= 2 2
c) Since 1 =
∫∞
0
f Y (x)dx, we have:
X
2
π
=
4
∫
∞
0
log x
dx
2
x −1
The righthand side R is equal to
∫
1
R=
0
∫
log x
dx
+
x2 − 1
∫
1
∞
log x
dx
x2 − 1
− log x
=2
dx
2
1
−
x
0
∫ 1
∞
∑
2k
x dx
(− log x)
=2
1
0
=2
∫
∞
∑ 1
k=0
0
k=0
(− log x)x2k dx
=2
=2
∫
∞
∑ ∞
k=0 0
∫
∞
∑ ∞
k=0
= 2Γ(2)
0
∞
∑
k=0
Thus, we obtained:
−2ku −u
ue
e
du
y
dy
−y
e
2k + 1
2k + 1
1
(2k + 1)2
∞
∑
k=0
1
π2
=
.
2
(2k + 1)
8
Noting that
∞
∞
∑
∑
1
1
1
ζ(2) =
=
+
ζ(2),
k2
(2k + 1)2
22
k=1
k=0
we obtain the desired result, i.e.:
ζ(2) =
∞
∑
4
3
k=0
1
π2 4
π2
=
=
.
(2k + 1)2
8 3
6
This is a probabilistic solution of Basel Problem.
|C2 |
|C1 | ,
Considering even moments of log
we
prove the Euler’s formulae of the Riemann zeta
function.
Lemma 1.2.
(
)2n
|C2 |
8
1
E( log
) = 2 Γ(2n+2)(1− 2n+2 )ζ(2n+2)
|C1 |
π
2
Proof.
(
)2n
∫∞
|C2 |
E( log |C1 |
) = 0 (log x)2n f |C2 | (x)dx =
|C1 |
∫
∫
2n+1
∞
1 (log x)2n+1
(log
x)
4
8
dx = π2 0
dx =
π2 0
x2 −1
x2 −1
∞
∑
∫
1
2k
8
2n+1
= π2 0 (− log x)
x dx
=
8
π2
8
π2
∞
∑
k=0
u2n+1 e−2ku e−u du =
k=0
∫
∞
∑ ∞
k=0
0
u
2n+1 −u du
(
)
e
2k + 1
2k + 1
=
8
π 2 Γ(2n
+ 2)
∞
∑
k=0
1
(2k + 1)2n+2
1
+ 2)(1 − 22n+2
)ζ(2n + 2)
Using this Lemma, we obtain the following
Euler Formula:
Theorem 1.1.
(Euler’s Formulae)
=
8
π 2 Γ(2n
(1 −
1
(
)
1 π 2n+2
)ζ(2n
+
2)
=
22n+2
2
2
An
Γ(2n + 2)
where, the coefficients An are obtained in the
series development
∞
∑
An 2n
1
=
θ
2
cos θ n=0 (2n)!
π
(|θ| < ).
2
Proof.
We only need to prove that
E(|C1 |α ) = cos1π α (|α| < 1), because by this,
2
we can easily get that E(e
α log
|C2 |
|C1 |
)=
1
(cos
π
2
α)
2
,
which is equivalent to
π 2n
2n
E((log |C1 /C2 |) ) = ( 2 ) An .
N
′
Noting that C1 ∼ N ′ where N and N are
two independent standard normal random
γ1/2
N2
2
variables, we get that (C1 ) ∼ (N ′ )2 ∼ γ ′
1/2
′
where γ1/2 and γ1/2
are two independent
gamma variables with parameter 1/2, i.e. its
x−1/2
density fγ1/2 (x) = √π e−x (x > 0).
Then we get that
α
α
−
α
E(|C1 | ) = E((γ1/2 ) 2 )E((γ1/2 ) 2 ) =
1
α
Γ( 12 + α
)
Γ(
−
1
π
2
2
2)
Γ(1/2) Γ(1/2) = π sin π( 21 + α
2)
1
= cos π α (|α| < 1), where we used
2
Γ(s)Γ(1 − s) = sinππs and E((γa )b ) =
Γ(a+b)
xa−1 −x
(x > 0).
Γ(a) , fγa (x) = Γ(a) e
the fact :
The above is a review of our previous
papers([1], [5]).
In this paper, instead of Cauchy, we use
arcsine random variables X with
2
fX (x) = π√1−x
2
Euler formula.
(0 < x < 1) and can get the
2 Basel Problem via arcsine
Take X1 ∼ X2 ∼ X(independent), then for
0<x<1
∫
1
1
4
√
√
f X2 (x) =
udu
2
2
2
2
X1
π
1−u 1−u x
0
∫ 1
2
1
√
= 2
du
π 0
1+x2 2
x2 −1 2
2
x ((u − 2x2 ) − ( 2x2 ) )
2
= 2
π x
∫
x2 −1
2x2
du
√
1+x2
x2 −1 2
2
− 2x2
u − ( 2x2 )
√
x2 +1
2
2
x − 1 2 2x2
2
= 2 [log(u + u − (
) )] 1+x2
2
− 2x2
π x
2x
1+x
2
(0 < x < 1)
= 2 log
π x
1−x
5 x) = 1 −
For x > 1,
then f X2 (x) = −f X2 (x)( x1 )′ =
X2
P ( X1
2
X1
log
2
π x
x+1
x−1
X1
(x > 1).
X1
P ( X2
5
1
x ),
Then we have the following proposition.
Proposition. 2.1
{
f X2 (x) =
X1
2
π2 x
2
π2 x
∫∞
log
log
1+x
1−x
x+1
x−1
(0 < x < 1)
(1 < x < +∞)
∫1
.
Since 1 = 0 f X2 (x)dx = 2 0 f X2 (x)dx =
X1
X1
∫
1 1
4
1+x
log
π2 0 x
1−x dx, then
∫1 1
π2
x2
x3
x4
4 = 0 x (x − 2 + 3 − 4 + · · · ) − (−x −
x2
2
2
−
∫1
0
x3
3
−
(1 +
therefore
x4
4
x2
3
− · · · ))dx =
+
∞
∑
j=0
x4
5
+ · · · )dx = 2
∞
∑
j=0
1
,
(2j + 1)2
π2
1
=
gives
ζ(2)
=
(2j + 1)2
8
π2
6
Remark.
√
We note that X ∼ β1/2,1/2 where
1 −1/2
(1 − x)−1/2 (0 < x < 1).
fβ1/2,1/2 (x) = π x
3 Euler formula via arcsine
In this section, we get the Euler formula for
ζ(2n) by using arcsine variables.
1
(0 < x < 1),
Take X with fX (x) = π2 √1−x
2
then
∫
α
1
2
x
α
For α > −1, E(X ) = π 0 √1−x2 dx =
α
∫
1
1
2
u
1
2
−
√
2 du
u
π 0
1−u 2
∫
1
1 1 α−1
1
1+α 1
−
2
2
=π 0 u
(1 − u) du = π B( 2 , 2 ).
−α
X2 α
α
For |α| < 1, E(( X1 ) ) = E(X2 )E(X1 ) =
1
1+α 1 1
1−α 1
π B( 2 , 2 ) π B( 2 , 2 )
1+α
1−α
1
1
Γ(
)Γ(
)
Γ(
)Γ(
1
2
2
2
2)
= π2 Γ( α +1)
1−α
Γ(
2
2 )
1
π
1
1−α
α
α
π sin π( 1+α
)
Γ(
)Γ(
2
2
2
2 )
π
tan πα
1
1 sin 2 α
= cos π α α π = πα 2
2
2
2
=
|α| < 1, where
we used the fact
π
Γ(s)Γ(1 − s) = sin πs (|α| < 1).
Here we note that
∞
∑
An
2n
tan x
x
(∗) by integrating
x =
(2n
+
1)!
n=0
∞
∑
An 2n
1
from 0 to x
u =
.
2u
(2n)!
cos
n=0
Lemma 3.1.
E((log X2 − log X1 )2n )
8
1
= 2 Γ(2n + 1)(1 − 2n+2 )ζ(2n + 2).
π
2
Proof.
E((log X2 − log X1 )2n ) = E((log
X2 2n
X1 ) )
=
∫∞
0
(log x)2n f X2 (x)dx
∫1
X1
1+x
2n 2
(log
x)
log
π2 x
1−x dx
0
∞
2k
∑
∫
x
1
4
= π2 0 2
(log x)2n dx
2k + 1
k=0
∫ 1
∞
∑
1
2k
2n
8
= π2
x (log x) dx
2k + 1 0
k=0
∫
∞
∞
∑ 1
−2ku 2n −u
8
e
u e du
= π2
2k + 1 0
=2
k=0
∫
∞
∑ ∞
u
2n du
=
e (
)
2k + 1
2k + 1
0
k=0
∞
∑
1
8
= π2 Γ(2n + 1)
(2k + 1)2n+2
8
π2
−u
k=0
=
8
Γ(2n
2
π
+ 1)(1 −
1
22n+2
)ζ(2n + 2).
Comparing (*) and
X1 α
α(log X1 −log X2 )
)
)
=
E(e
)
=
E(( X
2
∞
∑
α2n
2n
E((log X1 − log X2 ) ) (∵
(2n)!
n=0
E((log X1 − log X2 )odd )) = 0),
we get that
1
An π 2n
8
Γ(2n+1)(1− 2n+2 )ζ(2n+2) =
( ) ,
2
π
2
2n + 1 2
Using arcsine varialbles, we also can get the
following Euler formula.
Theorem. 3.1.
An
1 π 2n+2
1
ζ(2n + 2) = ( )
.
1
2 2
1 − 22n+2 Γ(2n + 2)
4 Euler like formula via
exponential
In this section, using exponential random
variables, we can get the Euler like formula for
special values of the Riemann zeta function.
Take e1 ∼ e2 ∼ e with fe (x) = e−x (x > 0),
then
∫
∞ α −x
α
For α > 0, E(e ) = 0 x e dx = Γ(1 + α).
e2 α
For |α| < 1, E(( e1 ) ) = Γ(1 + α)Γ(1 − α) =
αΓ(α)Γ(1 − α) =
πα
sin πα
Here we put that
x
sin x
(|α| < 1).
∞
∑
Cn 2n
=
x
(2n)!
n=0
(∗∗).
Lemma 4.1.
E((log e2 −log e1 ) ) = 2Γ(2n+1)(1−
2n
Proof.
Noting
f e2 (x) =
e1
1
(1+x)2
1
22n−1
)ζ(2n).
(x > 0) (Parato distribution),
for n = 1,
E(log e2 − log e1 )2n ) = E((log
∫∞
= 0 (log x)2n f e2 (x)dx
e1
∫∞
1
2n
= 0 (log x) (1+x)2 dx
∫1
1
= 2 0 (log x)2n (1+x)
2 dx
∞
∑
∫1
k−1
2n
k(−x)
(log x) dx
=2 0
e2 2n
e1 ) )
k=1
=2
∞
∑
k=1
∫
1
k
(−x)
0
k−1
2n
(log x) dx =
2
∞
∑
∫
∞
e
k
−(k−1)u 2n −u
0
k=1
∫
∞
∞
∑
=2
2
k
k=1
∞
∑
k=1
0
u e
du
u 2n −u du
( ) e
=
k
k
k−1
(−1)
k 2n
Γ(2n + 1)
= 2Γ(2n + 1)(
= 2Γ(2n +
∞
∑
−1
+
2n
(2k)
k=1
−1
1)( 22n ζ(2n)
∞
∑
k=0
+ (1 −
1
)
2n
(2k + 1)
1
22n )ζ(2n))
= 2Γ(2n + 1)(1 −
1
22n−1 )ζ(2n).
Comparing (**) and
E(( ee21 )α ) = E(eα(log e2 −log e1 ) )
∞
∑
α2n
2n
=
E((log e2 − log e1 ) )
(2n)!
n=0
E((log e2 − log e1 )
we get that
odd
2Γ(2n + 1)(1 −
(∵
)) = 0,
1
22n−1
2n
)ζ(2n) = Cn π ,
Theorem 4.1.
Cn π 2n
ζ(2n) =
2Γ(2n + 1)(1 −
Remark.
A decomposition
1
tan θ θ
=
cos θ
θ sin θ
is interesting for our discussion.
1
22n−1 )
.
θ
From tan
θ , we have the Euler formula. From
1
θ
,
we
have
the
Euler
fomula,
too.
From
sin θ
cos θ ,
we can get Lχ4 (2n + 1) (:special values of
∞
∑
(−1)j
Lχ4 (s) =
) which was already
s
(2j
+
1)
j=0
discussed in [1]).
This decomposition has following Brownian
interpretation.
Let Bt be a Brownian motion. We define
gt = sup{s|Bs = 0, s 5 t} and
Ta∗ = inf{t||Bt | = a} for a > 0. Then we can
see gTa∗ and Ta∗ − gTa∗ are independent and for
1
−λTa∗
λ > 0, E(e
) = cosh √2λa ,
E(e
−λ(Ta∗ −gTa∗ )
E(e
−λgTa∗
)=
√
2λa
√
√sinh 2λa
tanh
√ 2λa .
2λa
)=
and
5 Basel problem via Wigner’s
semi-circle law
Let W be√a random variable with
fW (x) = π4 1 − x2 (0 < x < 1), called
Wigner’s semi-circle random variables.
Take W1 ∼ W2 ∼ W (independent), then for
0 5 x 5 1,
√
∫ 1 16 √
f W2 (x) = 0 π2 1 − u2 1 − x2 u2 udu =
W1
√
∫
1
8
2 u)du.
(1
−
u)(1
−
x
2
π
0
2
= P (W2 5 W1 ) = P ( W
W1 5 1) =
√
2 u)dudx, and
(1
−
u)(1
−
x
05u,x51
Then
∫∫
8
π2
1
2
∫∫
√
=
√
1 − u( 1 − ux2 − 1 + 1)dudx
05u,x51
2
= −
3
π2
16
∫∫
√
05u,x51
√
1 − u(1 − 1 − ux2 )dudx
2
= −
3
∫∫
√
05u,x51
1−u
∞
∑
k=1
(
)
2k 2k k
1
x
u
du
22k (2k − 1) k
( )
1
2k
1
2
3
= −
B(k
+
1,
)
3
22k (2k − 1) k 2k + 1
2
k=1
( )
∞
3
∑
2
1
2k
1 Γ(k + 1)Γ( 2 )
= −
3
22k (2k − 1) k 2k + 1 Γ(k + 52 )
k=1
∞
∑
2
= −
3
∞
∑
1
(2k)! 1 22k+2 k!(k + 1)!
22k (2k − 1) k!k! 2k + 1
(2k + 3)!
k=1
∞
∑
2
= −2
3
k=1
1
(2k − 1)(2k + 1)2 (2k + 3)
, where we used that
(
)
∞
∑
√
1
2k k
1− 1−x=
x
.
22k (2k − 1) k
k=1
This solves the Basel problem.
Remark.
We remark that
2
α+1 3
α
E(|W | ) = π B( 2 , 2 ),
πα
tan
W2 α
1
2
,
E(| W1 | ) =
π
α2
1− 4
2
√
1 3
and W ∼ β( 2 , 2 ) and W ∼ U 1/2 X.
6 Two simple derivations of
Basel Problem
In this section, using some double integral
related to the Legendre Polynomials for the first
one, we get another ways to Basel Problem.
Theorem 6.1.
∫1
dt
−1
∫1
makes
1
dx
−1 1−2xt+t2
=
∫1
dx
−1
π2
ζ(2) =
.
6
∫1
1
dt
−1 1−2xt+t2
Proof.
∫1
1
For t(−1 < t < 1), −1 1−2xt+t2 dx =
1
1
[ −2t
log |1 − 2xt + t2 |]1−1 = −2t
(log(1 − t)2 −
1
2
log(1 + t) ) = t (log(1 + t) − log(1 − t)) =
1
t2
t3
t4
t5
t2
t3
t ((t − 2 + 3 − 4 + 5 + · · · ) − (−t − 2 − 3 −
t4
t5
t2
t4
t6
(∗).
4 − 5 + · · · ) = 2(1 + 3 + 5 + 7 + · · · )
Then
∞
∑
∫1
∫1
1
1
I = −1 dt −1 1−2xt+t2 dx = 4
.
2
(2k + 1)
k=0
Because each term is positive, the interchange
of sum
and
integral
is
clear.
By
Fubini
Theorem,
∫1
∫1
1
I = −1 dx −1 1−2xt+t2 dt =
∫1
∫1
1
dx −1 (t−x)2 +1−x2 dt =
−1
∫1
√ 1
√t−x ]1 =
dx[
arctan
2 −1
−1
1−x2
1−x
√
√
∫1
1−x
1+x
√ 1
(arctan 1+x + arctan 1−x )dx =
−1 1−x2
∫
π
π2
π 1 √ 1
1
dx
=
[arcsin
x]
=
,
where
−1
2
2 −1 1−x
2
2
we used the fact that 0 < α < π2 , 0 < β < π2
π
and tan α tan β = 1 gives α + β = 2 because
0 < α + β < π and
cos(α + β) = cos α cos β − sin α sin β =
cos α cos β(1 − tan α tan β) = 0. This makes
π
α + β = 2.
∞
∑
π2
1
Then
=
and
(2k + 1)2
8
k=0
∞
∞
∑
∑
1
1
ζ(2) =
+
=
(2k + 1)2
(2k)2
k=0
k=1
∞
∑
1
1
+ ζ(2). These make
2
(2k + 1)
4
k=0
ζ(2) =
4
3
∞
∑
k=0
1
π2
4 π2
=
=
.
(2k + 1)2
3 8
6
Remark.
(*) is interpretated as the Parseval formula for
the Legendre generating function
∞
∑
n
1
P
(x)t
, where
=
g(t, x) = √1−2xt+t
n
2
n=0
Pn (x) are the Legendre Polynomials which is
1
dn
defined by Pn (x) = 2n n! dxn (x2 − 1)n , because
√
2n + 1Pn (x)(n = 0, 1, 2, . . . ) are C.O.N.S. in
L2 [−1, 1].
Theorem
6.2.
∫1
∫ +∞
1
dt 0
(1+x2 )+t(1−x2 ) dx =
−1
∫∞ ∫1
1
dx
dt
0
−1 (1+x2 )+t(1−x2 )
makes
π2
ζ(2) =
.
6
Proof.
∫ +∞
1
For −1 < t < 1, 0
(1−t)+x2 (1+t) dx =
x
π√ 1
1 √1
∞
√
[arctan 1−t ]0 = 2 1−t2 . Then
1+t
1−t
1+t
1+t
∫1
π√ 1
π2
L.H.S. = −1 2 1−t2 .dt = 2 .
∫1
For −1 (1−t)+x12 (1+t)) dt =
log x2
1
2
2
1
[ x2 −1 log |(x − 1)t + (x + 1)|]−1 = x2 −1 .
∫ ∞ log x2
∫ 1 log x2
So, = 0 x2 −1 dx = 2 0 x2 −1 dx. because
∫ ∞ log x2
∫ 1 log x−2 1
∫ 1 log x2
dx = 0 x−2 −1 x2 dx = 0 x2 −1 dx.
1 x2 −1
∫ 1 − log x2
Then, R.H.S.= 2 0 1−x2 dx =
∞
∑
∫1
2k
2
x dx =
2 0 (− log x )
k=0
4
4
∫
∞
∑ 1
(− log x)x dx =
k=0 0
∫
∞
∑ ∞
k=0
0
2k
−2kx −u
ue
e
du = 4
∞
∑
k=0
Then L.H.S.=R.H.S. gives
∞
∑
1
π2
=
and
ζ(2)
=
(2k + 1)2
8
k=0
1
.
(2k + 1)2
π2
.
6
参考文献
[1] Bourgade, P., Fujita, T. and Yor, M. :
Euler’s formula for zeta(2n) and Cauchy
variables, Elect. Comm. in Probab. 12
(2007) 73-80.
[2] Castellanos, D. : The Ubiquitous Pi. Part I.
Math. Mag. 61, (1988) 67-98.
[3] Choe, B. R. :An elementary Proof of
∑∞ 1
π2
n=1 n2 = 6 . Amer. Math. Monthly 94,
(1987) 662-663.
[4] Holme, F. : Ein enkel bereegning av
∑∞ 1
k=1 k2 Nordisk Mat. Tidskr. 18, (1970)
91-92 and 120.
[5] Fujita, T. : A probabilistic approach to the
special values of the Riemann zeta function,
R.I.M.S. Kokyuroku, Number Theory and
Probability, 1590, 1-8 (2008).
[6] : Fujita, T, Yano, Y: Special values of the
Hurwitz zeta function via generalized
Cauchy variables, Kyoto J. of Math, Vol. 52,
No.3, (2012) 465-477.
[7] Matsuoka, Y. : An Elementary Proof of the
∑∞ 1
π2
Formula n=1 n2 = 6 . Amer. Math.
Monthly 68, (1961)486-487.
[8] Papadimitriou, I. : A Simple Proof of the
∑∞ 1
π2
Formula k=1 k2 = 6 . Amer. Math.
Monthly 80, (1973)424-425.
[9] Serre, J.P. :Cours d’arithmétique, Collection
SUP, P.U.F., Paris, (1970).
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∑∞ 1
π2
k=1 k2 = 6 . Amer. Math. Monthly 76,
(1969)552-553.