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
∞  
∑  a_{1}a_{2} … a_{k–1} (x + a_{1})(x + a_{2})…(x + a_{k }) 
=  1 x 
k=1 
yields
∞  ∞  
ζ(3) =  ∑  1 n^{3} 
=  5 2 
∑ 

.  
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

(1) 
and that the partial sums of the series (1) are rational numbers with a quite small denominator
n  
6  ∑  1_{ } m^{3}b_{m }b_{m –1} 
=  p_{n} q_{n} 
; 
m=1 
p_{n}, q_{n} are integers and
ζ(3) =  6  
5 –  1  
117 –  64  
535 –  729  
1436 –  4096  
3105 – ... –  n^{6} 34n^{3} + 51n^{2} + 27n + 5 
or, what is the same thing, as being equivalent to the remarks that the recursion
∞  ∞  
ζ(2) =  π^{2} 6 
= 3  ∑ 

= 5  ∑  (–1)^{n–1} n^{2}b_{n}b_{n–1} 
,  
n=1  n=1 
∞  
b_{n} =  ∑  (  n k 
)  ^{2} 
(  n+k k 
)  ~  (  1 + √5 2 
)  ^{5n} 
n=1 
and
π^{2} –  p q 
>  1 q^{11.850782... + ε} 
. 
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:

(2)  

(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
a_{1}a_{2} … a_{k–1} (x + a_{1})(x + a_{2})…(x + a_{k}) 
= D_{k–1} – D_{k} with D_{k} =  a_{1}a_{2} … a_{k} (x + a_{1})(x + a_{2})…(x + a_{k}) 
. 
So
K  K  
∑  a_{1}a_{2} … a_{k–1} (x + a_{1})(x + a_{2})…(x + a_{k}) 
=  ∑  (D_{k–1} – D_{k}) = D_{0} – D_{K} =  1 x 
–  a_{1}a_{2} … a_{K} (x + a_{1})(x + a_{2})…(x + a_{K}) 
. 
k=1  k=1 
The mechanism employed is splitting the summands so that the terms telescope. As Apery advices, one writes
n–1  n–1  
∑  (–1)^{k–1}(k–1)!^{2}n(n – k – 1)! (n + k)! 
=  ∑  x_{n,k} =  1 n^{2} 
– 2_{ } 


k=1  k=1 
which looks hopeful. To get hold of
N  n–1  N  N  
∑  ∑  x_{n,k} n 
=  ∑  1^{ } n^{3} 
– 2  ∑ 

.  
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  
∑  ∑  x_{n,k} n 
=  ∑  ∑  (E_{n,k} – E_{n–1,k }) =  ∑  (E_{N,k} – E_{k,k }) 
k=1  n=1  k=1  n=1  k=1 
(noting that the term with
E_{n,k }=  (–1)^{k–1}(k – 1)!^{2}(n – k)! 2k·(n + k)!^{ } 
= 

. 
So
N  N  N  N  
∑  1^{ } n^{3} 
– 2  ∑ 

=  1 2 
∑ 

+  1 2 
∑ 


n=1  n=1  k=1  k=1 
and all is revealed on noting that
with
F_{n,k }=  (–1)^{n+k}(k – 1)!^{2}(n – k)! 2(n + k)!^{ } 
= 

. 
Then
N  N  N  N  
∑  (–1)^{n–1} n^{2} 
– 2  ∑ 

=  1 2 
∑ 

–  1 2 
∑ 


n=1  n=1  k=1  k=1 
yields (2) on noting that
It seems natural to ask, whether there are further formulas, with c, c' rational:

(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 2^{2} 
+  1 3^{2} 
+ .. +  1 (n – 1)^{2} 
–_{ }  4 5n^{2} 
) 


n=2 
is a simple example. Numerical experimentation on the other hand on
∞  
ζ(4) =  π^{4} 90 
= c  ∑ 


n=2 
yields

(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
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

(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.
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,
∞  
L_{2}(x) =  ∑  x^{n} n^{2} 
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

(7) 
Obviously I had found the genesis of our formula (5).
The dilogarithm L_{2}(z) of z with
∞  
L_{2}(z) =  ∑  z^{n} n^{2} 
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  
L_{2}(z) = –  ∫  ln(1 – t) t 
dt 
0 
provided if understand

(8)  


(9)  

Writing
∞  
L_{2}(1) =^{ }  ∑_{ }  1 n^{2} 
= ζ(2) =  π^{2} 6 
. 
n=1 
It is amusing to see π^{2} appear as the square of a logarithm. Further for
z  
L_{2}  (  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 = 
0  0  0  0 
= L_{2}  (  z 1 – w 
)  + L_{2}  (  w 1 – z 
)  – L_{2}(z) – L_{2}(w) – ln(1 – z)·ln(1 – w). 
Taking

(10) 
and
L_{2}(z^{2 }) = 2L_{2}(–z) – 2L_{2}  (  –  z 1 – z 
)  – ln^{2 }(1 – z) 
which by (8) is

(11) 
Curiously (cf. [3]) the dilogarithm is generally enjoying a revival.
An underlying reason is the fact the function
z  
ε(α) = α ln(1 – α) + 2πi exp  (  –  1 2πi 
∫  ln(1 – t) t 
dt  ) 
0 
is welldefined (is independent of the part from 0 to α that determines both
The Riemann surface of the dilogarithm is rather witty and deserves at least passing comment: there is of course a singularity at
so we obtain a logarithmic singularity at
j  
l_{2}^{(m1; n1; m2; n2; ...)}(α) = L_{2}(α) – 2πi  ∑  m_{j} ln α – 4π^{2}  ∑  n_{j}  ∑  m_{i }. 
j  j  i=1 
Plainly there are nontrivial paths (always with
We may define the higher polylogarithms by
z  
L_{n+1}(z) = –  ∫  L_{n}(t) t 
dt, n ≥ 2. 
0 
For the trilogarithm there is an interesting functional equation involving only trilogarithms and logarithms:
–z/(1–z)  z  
L_{3}  (  –  z 1 – z 
)  = –  ∫  L_{2}(t) t 
dt =  ∫  L_{2}  (  –  u 1 – u 
)  du u(1 – u) 
= 
0  0 
z  z  
=  (  ln u – ln(1 – u)L_{2}  (  –  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 = L_{2}(u) ln u  –  ∫  L_{2}(u) u 
du,  ∫  ln^{2 }(1 – u) u 
du =  
0  0  0  0 
z  z  1–z  
= ln u ln^{2 }(1 – u)  + 2  ∫  ln u ln(1 – u) u 
du = ln z ln^{2 }(1 – z) – 2  ∫  ln u ln(1 – u) u 
du.  
0  0  0 
Collecting everything yields for
L_{3}  (  –  z 1 – z 
)  = L_{2}  (  –  z 1 – z 
)  ln  z 1 – z 
+ L_{2}(z) ln z + L_{2}(1 – z) ln(1 – z) + 
+ ln z ln^{2 }(1 – z) – L_{3}(z) – L_{3}(1 – z) + L_{3}(1) –  1 3 
_{ }ln^{3 }(1 – z). 
Now applying (8), (10), and (11) eliminates the dilogarithms, so

(12) 
Moreover we note that on integrating (8) we have
1 4 
L_{3}(z^{2 }) = L_{3}(z) + L_{3}(–z). 
All this is not particularly exciting. But now denote
5 4 
L_{3}  (  1 τ² 
)  = L_{3}  (  1 τ² 
)  + L_{3}  (  1 τ 
)  + L_{3}  (  –  1 τ 
)  =  5 6 
ln^{3 }τ –  1 6 
π^{2 }ln τ + L_{3}(1). 
Hence we have obtained the amazing identity

(13) 
We shall see that it is equivalent to the formula (3) for
For the higher polylogarithms there are no known functional equations (other that the trivial ones for
For n ≥ 2, integration by parts yields (put
z  n–1  z  z  
L_{n}(z) – L_{n}(w) =^{ }  ∫  L_{n–1}(t) t 
dt =  ∑  (–1)^{k–1} k! 
ln^{k }t L_{n–k}(t)  ^{ }+  (–1)^{n–1} (n–1)! 
∫  ln^{n–1 }t 1 – t 
dt. 
w  k=1  w  w 
In particular

(14) 
We recall that
Now notice that for

(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

(16) 
This yields the socalled logsine integrals [7]
θ  
Ls_{a+b,a }(θ) = –  ∫  x^{a} ln^{b–1 }  2 sin  x 2 
dx.  
0 
Many wonderful formulas may be obtained. Here we consider only
π/3  π/3  π/3  
π^{3} 3^{3}·3! 
=  ∫  (  –  (x – π)^{2} 4 
+ ln^{2 }  (  2 sin  x 2 
)  )  dx = –  (x – π)^{3} 12 
+  ∫  ln^{2 }  (  2 sin  x 2 
)  dx.  
0  0  0 
so

(17) 
Similarly for n = 4 applying (14) and (15) yields
π/3  
–  π^{4} 3^{3}·4! 
–  π^{4} 15^{ } 
=  ∫  (  (x – π)^{3} 8 
–  3(x – π)^{ } 2 
ln^{2 }  (  2 sin  x 2 
)  )  dx = 
0 
π/3  π/3  π/3  
=  (x – π)^{4} 32 
+  3π 2 
∫  ln^{2 }  (  2 sin  x 2 
)  dx –  ^{ }3 2 
∫  x ln^{2 }  (  2 sin  x 2 
)  dx.  
0  0  0 
So we have (7)

(18) 
It seems that one cannot disentangle any other cases. Thus, for example
π/3  π/3  
2  ∫  ln^{4 }  (  2 sin  x 2 
)  dx – 3  ∫  x^{2 }ln^{2 }  (  2 sin  x 2 
)  dx 
0  0 
is a rational multiple of
It is well, but apparently not widely, known that

(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  
∑ 

= 8  ∫  du u 
∫  arcsin^{2 }x x 
dx = 8I.  
n=1  0  0 
Changing the integration order we obtain
½  ½  ½  
I = –  ∫  ln 2u  arcsin^{2 }u u 
du = –  ln^{2 }2u 2 
arcsin^{2 }u  +  ∫  ln^{2 }2u  arcsin u ^{ }√1 – u² 
du  
0  0  0 
noting that the integrated terms vanish at the endpoints. Now put
π/3  ∞  
8I = 2  ∫  x ln^{2 }  (  2 sin  x 2 
)  dx =  17π^{4} 3240 
=  ∑ 


0  n=1 
which proves (5).
Now write x = –iy in (19) to obtain

(20) 
As before put



Integration yields
∞  ½  ½  2 ln τ  
∑ 

= 4  ∫  arsh^{2 }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
1  1  1  
I = 2  ∫  ln t ln(1 – t) t 
dt –  ∫  ln^{2 }t t 
dt =  (  –2 ln t L_{2}(t) –  1 3 
ln^{3 }t  )  +  
1/τ²  1/τ²  1/τ² 
1  
+ 2  ∫  L_{2}(t) t 
dt = –4 ln τ L_{2}  (  1 τ² 
)  –  8 3 
ln^{3 }τ – 2L_{3}  (  1 τ² 
)  + 2L_{3}(1). 
1/τ² 
But the expressions (10), (8), and (11) yield:





Thus
5 2 
I =  10 3 
ln^{3 }τ –  2π^{2} 3 
ln τ – 5L_{3}  (  1 τ² 
)  + 5ζ(3). 
But (13) is
0 =  10 3 
ln^{3 }τ –  2π^{2} 3 
ln τ – 5L_{3}  (  1 τ² 
)  + 4ζ(3). 
So indeed we have (3)
∞  
ζ(3) =  5 2 
I =  5 2 
∑ 

.  
n=1 
There remains a gap in our collection of formulas. We have
∞  ½  π/3  
∑ 

= 4  ∫  arcsin^{2 }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.
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.
Roger Apéry. Irrationalite de
Roger Apéry. Lecture at this conference. black
Spencer Bloch. Applications of the Dilogarithm Function in Algebraic Ktheory and Algebraic Geometry. black
Louis Comtet, Advanced Combinatorial Analysis. D.Reidel, Dordrecht, 1974. black
R. William Gosper, Jr. Decision procedure for indefinite hypergeometric summation. Proc. Nat. Acad. Sci. USA 75 (1978), pp.40–42. black
Margrethe Munthe Hjortnaes. Overforing av rekken
Leonard Lewin. Dilogarithms and associated functions. Macdonald, London, 1958. black
Z. R. Melzak. Companion to Concrete Mathematics. Wiley–Interscience, New York, 1973. black
Alfred van der Poorten. A proof that Euler missed... Apéry's proof of the irrationality of
Alfred van der Poorten. Some wonderful formulae. Footnotes to Apéry's proof of the irrationality of