A proof that Euler missed... 


SOME WONDERFUL FORMULAE

An introduction to polylogarithms
Kingston conference, July 1979
Macquarie Mathematics Reports


Alfred van der Poorten


1

In June 1978, at the Journees Arithmetiques at Marseille–Luminy, the claims of Roger Apéry, that he had settled a longstanding problem in number theory, were rightly greeted with derision. He purported to have proved that ζ(3) is irrational, a question that had defeated Euler. Yet the methods that his audience glimpsed were those of Euler! Moreover Apery's proof appeared to depend on a number of outrageous assertions. Indeed he began by claiming that writing x = n2 and ak = –k2 in
 
   a1a2ak–1

(x + a1)(x + a2)…(x + ak )

 =  1

 x

k=1 

yields
   
ζ(3) =    1

 n3

 =  5

2

  
(–1)n–1
n3 ( 2n
n
)
.
 n=1 n=1 

Incredibly, it soon became apparent that Apery's allegations were correct. The story of our discovery of the veracity of the proof is recounted elsewhere [9]. Here it suffices to say that what Apery proves is that
    n
ζ(3) = 6 1

 n3bn bn –1

    with     bn = (  n
 k
) 2 
 
(  n+k
 k
) 2 
 
,
  n=1   k=0
(1)

and that the partial sums of the series (1) are rational numbers with a quite small denominator
  n
6 1 

 m3bm bm –1

 =   pn

 qn

 ;
  m=1

pn, qn are integers and qn = [1, 2, ..., n]3bn (here [1, 2, ..., n] is the lowest common multiple). It is now easy to see that 0 ≠ |qnζ(3) – pn| < qnδ, with δ > 0; more precisely, one has bn ~ (1 + √2)4n; [1, 2, ..., n] ~ en by the prime number theorem; and then one may compute that as n → ∞, δ → (4 ln(1 + √2) – 3)/(4 ln(1 + √2) + 3). It follows that ζ(3) cannot be rational. The reader will recognize the series (1) as being equivalent to the continued fraction expansion

ζ(3) = 6
 5 – 1
 117 – 64
 535 – 729
 1436 – 4096
 3105 – ... – n6

34n3 + 51n2 + 27n + 5


or, what is the same thing, as being equivalent to the remarks that the recursion Un+1 = (34n3 + 51n2 + 27n + 5)Un n6Un–1. has independent solutions {An}, with A0 = 0, A1 = 6 and {Bn} with B0 = 1, B1 = 5 so that An/Bn → ζ(3) and bn = Bn/n!3 is an integer, whilst if an = An/n!3 then so is [1, 2, ..., n]3an. The explanations that Apery provides in the summary of his Luminy lecture [1], and in his Kingston talk [2] show that indeed he accelerates a continued fraction expansion of ζ(3) to obtain his proof. In a quite similar manner to that sketched one has
  
ζ(2) =  π2

6

 = 3  
1
n2 ( 2n
n
)
 = 5   (–1)n–1

 n2bnbn–1

,
n=1 n=1 
bn = (  n
 k
) 2
 
(  n+k
 k
)  ~  ( 1 + √5

2

) 5n
 
n=1

and {bn} satisfies the recursion (n+1)2un+1 = (11n2 + 11n + 3)un + n2un–1. These remarks yield a proof of the irrationality of ζ(2) with a startlingly good irrationality measure for π2

 π2 –   p

 q

 >  1

 q11.850782... + ε

.

2

Amongst the many intriguing questions that remain are some that are in sense irrelevant to the main flow of Apery's proof. These concern the intermediate formulas:
   
ζ(2) =    1

 n2

 = 3   
1
n2 ( 2n
n
)
.
 n=1 n=1 
(2)
   
ζ(3) =    1

 n3

 =  5

2

  
(–1)n–1
n3 ( 2n
n
)
.
 n=1 n=1 
(3)

At first even these claims hardly seemed believable. But the series of the right are readily summed to the accuracy of one's calculator and on dividing the values obtained into ζ(2) = π2/6 and ζ(3) = 1.2020569… respectively and does obtain the constants 3.000000... and 2.500000... . This does suggest that (2) and (3) are truths. In any event, Apery provides proofs. We notice that
a1a2ak–1

(x + a1)(x + a2)…(x + ak)

 = Dk–1Dk   with   Dk a1a2ak

(x + a1)(x + a2)…(x + ak)

.

So
K   K  
   a1a2ak–1

(x + a1)(x + a2)…(x + ak)

 =    (Dk–1Dk) = D0DK 1

x

 –  a1a2aK

(x + a1)(x + a2)…(x + aK)

.
k=1   k=1  

The mechanism employed is splitting the summands so that the terms telescope. As Apery advices, one writes x = n2, ak = –k2, (k = 1, 2, ...) being careful to take kn – 1. Apery [1] almost seems to insist that it is more decent to operate with divergent series, but this need not bother us here. After a fair bit of tidying we obtain
n–1   n–1  
   (–1)k–1(k–1)!2n(nk – 1)!

(n + k)!

 =    xn,k 1

n2

 – 2 
(–1)n–1
n2 ( 2n
n
)
k=1   k=1  

which looks hopeful. To get hold of ζ(3) we divide throughout by n (which is inviting spare on the left) and we sum over n, 1 ≤ nN, so as to obtain
N n–1   N   N  
xn,k

n

 = 1 

n3

 – 2
(–1)n–1
n3 ( 2n
n
)
.
n=1 k=1   n=1   n=1  

We invert summation and flushed with the success of the earlier telescope we write
N–1 N   N–1 N   N  
xn,k

n

 =  (En,kEn–1,k ) =  (EN,kEk,k )
k=1 n=1   k=1 n=1   k=1  

(noting that the term with k = N vanishes). It remains only to quess En,k if, indeed, it exits. This could be done mechanically, see [5], but we just check

En,k (–1)k–1(k – 1)!2(nk)!

2k·(n + k)! 

 = 
(–1)k–1
2k3 (  n+k
k
)(  n
k
)
.

So
N   N   N   N  
1 

n3

 – 2
(–1)n–1
n3 ( 2n
n
)
 =  1

2

(–1)k–1
k3 (  N+k
k
)(  N
k
)
 +  1

2

(–1)k–1
k3 ( 2k
k
)
n=1   n=1   k=1   k=1  

and all is revealed on noting that 2 + 1/2 = 5/2. Quite similarly one sees that

(–1)n–1xn,k = Fn,kFn–1,k

with
Fn,k (–1)n+k(k – 1)!2(nk)!

2(n + k)! 

 = 
(–1)n+k
2k2 (  n+k
k
)(  n
k
)
.

Then
N   N   N   N  
(–1)n–1

n2

 – 2
1
n2 ( 2n
n
)
 =  1

2

(–1)N+k
k2 (  N+k
k
)(  N
k
)
 –  1

2

1
k2 ( 2k
k
)
n=1   n=1   k=1   k=1  

yields (2) on noting that 2·(2 – 1/2) = 3.

3

It seems natural to ask, whether there are further formulas, with c, c' rational:
   
ζ(m) = c
1
 nm ( 2n
n
)
    or    ζ(m) = c'
(–1)n–1
 nm ( 2n
n
)
 .
 n=1 n=1 
(4)

The methods above do not readily assist, though they can be applied to yield less attractive formulas of which
  
ζ(5) =  5

2

   ( 1 +  1

22

 +  1

32

 + .. +  1

(n – 1)2

 –   4

5n2

) 
(–1)n
n3 ( 2n
n
)
 n=2 

is a simple example. Numerical experimentation on the other hand on
  
ζ(4) =  π4

90

 = c
1
 n4 ( 2n
n
)
 n=2 

yields c = 2.117647058... and one wonders whether c is a decent, god-fearing, rational number. Of course a computation cannot yields other than a rational so that it is the decency of c that is in question. Expanding c as a continued fraction yields c = 2 + 1/(8 + ½) = 36/17 to the quoted accuracy. Applying the Law of Small Numbers one conjectures that indeed
  
ζ(4) =  36

17

  
1
 n4 ( 2n
n
)
 .
 n=1 
(5)

Similar computations suggest that there are no cases other than the cited formulas (2), (3) and (5) for which c or c' is rational in ζ(4). In fact the formula (2) for ζ(2) is well-known (I quickly remembered a reference; for example [8], p.85) and the formula (3) for ζ(3) is quite well-known (a number of people knew it, and it can be found in the literature, for example [6]).

A little more surprisingly, thanks to Shanks I was alerted to an exercise in [4], p.89, where, without any noticeable hint, the reader is asked to prove that
 
  
1
( 2n
n
)
 =  1

3

 +  2π√3

27

 ,
n=1 
 
 
  
1
n  ( 2n
n
)
 =  π√3

9

 ,
n=1 
 
  
1
n2 ( 2n
n
)
 =  π2

18

 ,
n=1 
 
 
  
1
n4 ( 2n
n
)
 =  17π4

3240

 .
n=1 
(6)

Thus the formula (5) was known. The first 3 formulas of (6) are quite easy to prove but the last formula (5) defied those of us that had met it experimentally.

4

Some years ago I had met a curious book entitled Dilogarithms and associated functions by an English engineer, Leonard Lewin. It was filled with strange formulas of ancient vintage. But on finding that the dilogarithm of x, L2(x) seemed to be little more than
  
L2(x) =     xn

n2

 n=1 

I had dismissed the volume as an amateur eccentricity. Recently however I met Professor Lewin at Boulder, Colorado and he alerted me to a formula ([7], p.139) equivalent to (3). With interest rearoused I endeavuored to acquire a copy of his book. When eventually I succeeded in doing so (it has long been out of print) my joy was great to find in preface the wonderful formula
π/3 
 x ln2 ( 2 sin   x

2

) dx 17π4

6480

 .
0 
(7)

Obviously I had found the genesis of our formula (5).

5

The dilogarithm L2(z) of z with z | < 1 is given by the series
  
L2(z) =     zn

n2

 n=1 

If we take the complex plane with a cut from 1 to ∞ then we readily obtain an analytic continuation by the integral
 z
L2(z) = –    ln(1 – t)

t

 dt
0

provided if understand ln(1 – t) to be the principal branch of the logarithm with argument in the interval (–π, π], and we adopt the convention that ln(1 – x) = πi + ln(x – 1) for x real and x > 1. A number of amusing identities now follow more or less directly from the functional equation  ln xy = ln x + ln y. We have
 z² z
L2(z2) = –    ln(1 – t)

t

 dt = –2    ln(1 – u2)

u

 du =
00
(8)
 zz
= –2    ln(1 – u)

u

 du – 2    ln(1 – v)

v

 dv = 2L2(z) + 2L2(–z).
00

–1/z z
L2(–1/z) = –    ln(1 – t)

t

 dt + L2(–1) =     ln(1 + 1/u)

u

 du + L2(–1) =
–11
(9)
z z
   ln(1 – v)

v

 dv –     ln u

u

 du + L2(–1) = –L2(–z) + 2L2(–1) –   ln2 z

2 

 .
–11

Writing  z = eπi = –1 in (9) yields 2L2(1) = 2L2(–1) + ½π2 whilst (8) gives L2(1) = –2L2(–1), so
L2(1) =    1

 n2

 = ζ(2) =  π2

6

 .
n=1

It is amusing to see π2 appear as the square of a logarithm. Further for z |, | w | < 1
 z
L2 ( z

1 – z

  w

1 – w

)  = –   ln ( 1 –  u

1 – u

  w

1 – w

)  du

u(1 – u)

 =    put  t u

1 – u

  w

1 – w

   =
0
 z z z z
= –   ln ( 1 –  u

1 – w

)  du

u

 +     ln(1 – u)

u

 du –    ln ( 1 –  w

1 – u

)  du

1 – u

 +     ln(1 – w)

1 – u

 du =
0000

= L2 ( z

1 – w

)  + L2 ( w

1 – z

)  – L2(z) – L2(w) – ln(1 – z)·ln(1 – w).

Taking  z + w = 1  yields
L2(z) + L2(1 – z) =  π2

6

 – ln(1 – z)·ln z,
(10)

and  z = w  yields
L2(z2 ) = 2L2(–z) – 2L2 ( –  z

1 – z

)  – ln2 (1 – z)

which by (8) is
 L2(z) = –L2 ( –  z

1 – z

)  –  1

2

 ln2 (1 – z).
(11)

Curiously (cf. [3]) the dilogarithm is generally enjoying a revival.

An underlying reason is the fact the function arg(1 – z)·ln | z | – Im l2(z) is single valued (here arg, log and l2 are the corresponding branches). In this spirit Bloch [3] notes that for any complex α Ï (0,1)
 z
ε(α) = α ln(1 – α) + 2πi exp ( –  1

i

   ln(1 – t)

t

 dt )
0

is well-defined (is independent of the part from 0 to α that determines both ln(1 – α) and the dilogarithm).

The Riemann surface of the dilogarithm is rather witty and deserves at least passing comment: there is of course a singularity at  z = 1. If we start on principal branch and go around 1 m ≠ 0 times the new branch is given by

l2(m)(α) = L2(α) – im ln α

so we obtain a logarithmic singularity at α = 0. Generally if we go around 1  m1 times, then around 0  n1 times, then around 1  m2 times, then around 0  n2 times, ... we obtain the branch
 j
l2(m1; n1; m2; n2; ...)(α) = L2(α) – 2πi mj ln α – 4π2 nj mi .
jji=1

Plainly there are non-trivial paths (always with mj = 0) which return one to the principal branch.

6

We may define the higher polylogarithms by
 z
Ln+1(z) = –    Ln(t)

t

 dt,       n ≥ 2.
0

For the trilogarithm there is an interesting functional equation involving only trilogarithms and logarithms:
z/(1–z) z
L3 ( –  z

1 – z

)  = –     L2(t)

t

 dt   L2 ( –  u

1 – u

)  du

u(1 – u)

 =
0
 z  z
( ln u – ln(1 – u)L2 ( –  u

1 – u

) )    –    ( ln u – ln(1 – u) )  ln(1 – u)

u(1 – u)

 du.
0 0

But
 z  z  z  z
  ln u ln(1 – u)

u

 du = L2(u) ln u    –    L2(u)

u

 du,         ln2 (1 – u)

u

 du =
0 0 0 0
 z  z 1–z
= ln u ln2 (1 – u   + 2   ln u ln(1 – u)

u

 du = ln z ln2 (1 – z) – 2   ln u ln(1 – u)

u

 du.
0 0  0

Collecting everything yields for z | < 1

L3 ( –  z

1 – z

)  = L2 ( –  z

1 – z

)  ln  z

1 – z

 + L2(z) ln z + L2(1 – z) ln(1 – z) +

+ ln z ln2 (1 – z) – L3(z) – L3(1 – z) + L3(1) –  1

3

 ln3 (1 – z).

Now applying (8), (10), and (11) eliminates the dilogarithms, so

L3 ( –  z

1 – z

)  + L3(1 – z) + L3(z) =  1

6

 ln3(1 – z) –  1

2

 ln z ln2(1 – z) +  1

6

 π2 ln (1 – z) + L3(1).
 (12)

Moreover we note that on integrating (8) we have

1

4

 L3(z2 ) = L3(z) + L3(–z).

All this is not particularly exciting. But now denote τ = ½(1 + √5) – golden ratio; so τ2 = τ + 1, and 1 – 1/τ2 = 1/τ. Put  z = 1/τ2 = ¼(√5 – 1) in (12) to obtain

5

4

 L3 (  1

τ²

 )  = L3 (  1

τ²

 )  + L3 (  1

τ

 )  + L3 (  –  1

τ

 )  =  5

6

 ln3 τ –  1

6

 π2 ln τ + L3(1).

Hence we have obtained the amazing identity

L3(1) = ζ(3) =  5

4

    L3 (  1

τ²

 )  +  2

15

 π2 ln τ –  2

3

 ln3 τ    .
 (13)

We shall see that it is equivalent to the formula (3) for ζ(3).

For the higher polylogarithms there are no known functional equations (other that the trivial ones for Ln(zm) in terms of Lnz), εm = 1) which involve only polylogarithms of that level and logarithms. In this sense the case n = 3 is small.

7

For n ≥ 2, integration by parts yields (put L1(t) = ln(1 – t) )
 z n–1  z   z
Ln(z) – Ln(w) =     Ln–1(t)

t

 dt =   (–1)k–1

k!

 lnk t Lnk(t     (–1)n–1

(n–1)!

    lnn–1 t

1 – t

 dt.
w k=1  w  w

In particular

 1
Ln(1) =  (–1)n–1

(n–1)!

    lnn–1 t

1 – t

 dt,       n ≥ 2.
0
(14)

We recall that

L2(1) = ζ(2) = π2/6;     L4(1) = ζ(4) = π4/90; ... .

Now notice that for 0 < θ ≤ π

eiθeiθeiθ
(–1)n–1     lnn–1 z

1 – z

 dz     lnn–1 z

1 – z

 dz     lnn–1 z

z

 dz
111
(15)
( because    z 1

w

    is     dz

1 – z

 =  dw

1 – w

 +  dw

w

).

Hence, according as n is odd or even we can readily find the imaginary, respectively the real part of the integral on the left. On the other hand observe that (after writing z = 1 – eix)
1–eiθ θ
    lnn–1 z

1 – z

 dz = –i    ( i(x – π)

2

 – ln ( 2 sin   x

2

) ) n–1
 
dx.
1 0
(16)

This yields the so-called logsine integrals [7]
θ
 Lsa+b,a (θ) = –  xa lnb–1  2 sin  x

2

 dx.
0

Many wonderful formulas may be obtained. Here we consider only θ = π/3, moved by the fact that 1 – eiθ = eiθ so ln(2 sin θ/2) = 0. Taking imaginary parts in (16) and recalling (15) we read for n = 3 that
π/3  π/3 π/3
π3

33·3!

 =  ( –  (x – π)2

4

 + ln2 ( 2 sin   x

2

) )  dx = –  (x – π)3

12

    +  ln2 ( 2 sin   x

2

)  dx.
0 0 0

so
π/3 
 ln2 ( 2 sin   x

2

) dx 3

108

 .
0 
(17)

Similarly for n = 4 applying (14) and (15) yields
π/3
–  π4

33·4!

 –  π4

15 

 =  ( (x – π)3

8

 –  3(x – π) 

2

 ln2 ( 2 sin   x

2

) )  dx
0
 π/3 π/3 π/3
(x – π)4

32

   

2

 ln2 ( 2 sin   x

2

)  dx –   3

 2

 x ln2 ( 2 sin   x

2

)  dx.
0 0 0

So we have (7)
π/3 
 x ln2 ( 2 sin   x

2

) dx 17π4

6480

 .
0 
(18)

It seems that one cannot disentangle any other cases. Thus, for example
π/3 π/3
2  ln4 ( 2 sin   x

2

)  dx – 3   x2 ln2 ( 2 sin   x

2

)  dx
0 0

is a rational multiple of π5, but a numerical check (in which I was kindly assisted by Mr.Denis Payne) suggests that neither integral is such a rational multiple. Thus it appears that (17) and (18) are not representative of a much larger class of similar formulas.

8

It is well, but apparently not widely, known that
 
  
(2x)2n
n2 ( 2n
n
)
 = 2 arcsin2 x
n=1  
(19)

as is easily checked, or discovered (see for example [8], p.108). The first three formulas (6) follow by differentiating appropriately, whilst
½  u
  
1
n4 ( 2n
n
)
 = 8     du

 u

    arcsin2 x

x

 dx = 8I.
n=1 0 0

Changing the integration order we obtain
½  ½ ½
 I = –    ln 2u   arcsin2 u

 u

 du = –   ln2 2u

 2

 arcsin2 u    +    ln2 2u   arcsin u

 1 – u²

 du
0  0 0

noting that the integrated terms vanish at the endpoints. Now put u = sin(x/2) and we obtain
π/3
8I = 2  x ln2 ( 2 sin   x

2

)  dx 17π4

3240

 =   
1
n4 ( 2n
n
)
0 n=1

which proves (5).

Now write x = –iy in (19) to obtain
 
2 arsh2 y = 2 ln2 ( y + √1 + y² ) = (–1)n–1 
(2y)2n
n2 ( 2n
n
)
.
n=1
(20)

As before put τ = ½(1 + √5). Differentiating strategically, we see that
  
(–1)n–1
 ( 2n
n
) 
 =  1

20

 +  ln τ

4√5 

 ,
n=1
 
  
(–1)n–1
n  ( 2n
n
)
 =  2 ln τ

5 

 ,
n=1
 
  
(–1)n–1
n2 ( 2n
n
)
 = 2 ln2 τ.
n=1

Integration yields
½ ½ 2 ln τ
  
(–1)n–1
n3 ( 2n
n
)
 = 4  arsh2 y

y

 dy = –8  ln 2y   arsh y

 1 + u²

 dy = –2  x ln ( 2 sh   x

2

)  dx = I
n=1 0 0 0

supplying an example of a logsh integral. Writing t = e–x we get

 1  1  1
 I = 2   ln t ln(1 – t)

t

 dt –    ln2 t

t

 dt ( –2 ln t L2(t) –  1

3

 ln3 t )    +
1/τ² 1/τ² 1/τ²
 1
+ 2   L2(t)

t

 dt = –4 ln τ L2 (  1

τ²

 )  –  8

3

 ln3 τ – 2L3 (  1

τ²

 )  + 2L3(1).
1/τ²

But the expressions (10), (8), and (11) yield:

 L2 (  1

τ²

 )  + L2 (  1

τ

 )  =  π2

6

 – 2 ln2 τ,
–2L2 (  –  1

τ

 )  + L2 (  1

τ²

 )  – 2L2 (  1

τ

 )  = 0,
 L2 (  –  1

τ

 )  + L2 (  1

τ²

 )  = –  1

2

 ln2 τ,
hence     5L2 (  1

τ²

 )  =  π2

3

 – 5 ln2 τ.

Thus

5

2

 I 10

3

 ln3 τ –  2

3

 ln τ – 5L3 (  1

τ²

 )  + 5ζ(3).

But (13) is

0 =  10

3

 ln3 τ –  2

3

 ln τ – 5L3 (  1

τ²

 )  + 4ζ(3).

So indeed we have (3)
ζ(3) =  5

2

 I 5

2

  
(–1)n–1
n3 ( 2n
n
)
.
n=1

There remains a gap in our collection of formulas. We have
½ π/3
  
1
n3 ( 2n
n
)
 = 4  arcsin2 y

y

 dy = –2  x ln ( 2 sin   x

2

)  dx
n=1 0 0

but now there does not seem to be an instructive closed evaluation of the integral.

9

Of course my remarks only begin the story, and I have told only of those formulas that are readily deduced from easily accesible sources; of these Lewin [7] has proved the most useful.

from Queen's Papers in Pure and Applied Mathematics,
(Proc. 1979 Queen's Number Theory Conference, ed. P. Ribenboim),
54 (1980), pp.269–286.


References
  1. Roger Apéry. Irrationalite de ζ(2) et ζ(3). Journees arithmetiques de Luminy. Asterisque 61 (1979), pp.11–13. назад к тексту

  2. Roger Apéry. Lecture at this conference. назад к тексту

  3. Spencer Bloch. Applications of the Dilogarithm Function in Algebraic K-theory and Algebraic Geometry. назад к тексту

  4. Louis Comtet, Advanced Combinatorial Analysis. D.Reidel, Dordrecht, 1974. назад к тексту

  5. R. William Gosper, Jr. Decision procedure for indefinite hypergeometric summation. Proc. Nat. Acad. Sci. USA 75 (1978), pp.40–42. назад к тексту

  6. Margrethe Munthe Hjortnaes. Overforing av rekken ∑ 1/k³ til et bestemt integral. Proc. 12th Cong. Scand. Maths, Lund 10–15 Aug. 1953 (Lund 1954). назад к тексту

  7. Leonard Lewin. Dilogarithms and associated functions. Macdonald, London, 1958. назад к тексту

  8. Z. R. Melzak. Companion to Concrete Mathematics. Wiley–Interscience, New York, 1973. назад к тексту

  9. Alfred van der Poorten. A proof that Euler missed... Apéry's proof of the irrationality of ζ(3). The Mathematical Intelligencer 1 (1979), pp.195–203. назад к тексту

  10. Alfred van der Poorten. Some wonderful formulae. Footnotes to Apéry's proof of the irrationality of ζ(3). Séminaire Delange–Pisot–Poitou (Théorie des nombres) 20e année 1978/1979, 29, pp.1–7. назад к тексту

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