POLYLOGARITHMS, MULTIPLE ZETA VALUES, AND

THE SERIES OF HJORTNAES AND COMTET

Notes by Tim Jameson (December 2009)

(edited by Graham Jameson, with references added, March 2014)

Introduction

The polylogarithm function is defined for |x| < 1 and any real s by

Lis(x) =∞∑n=1

xn

ns. (1)

For particular values of s, it can then be extended analytically to a wider range of x (real

or complex). It is simultaneously a power series in x and a Dirichlet series in s. The

first published study of the function was by A. Jonquiere in 1889, and it is sometimes called

Jonquiere’s function. Note that “polylogarithmic functions” means something quite different!

These notes are a summary of some results on polylogarithms that I have seen stated

in various places, such as Wikipedia and [BBC], largely with proofs that I have worked out

for myself. The methods are mostly elementary. We restrict to integer s = k ≥ 0, and x

usually real. However, in places the reader is invited to accept that formulae established for

real x also apply for complex x; I can supply a rigorous justification if pressed.

We start with a few immediate facts. Firstly,

Li0(x) =x

1− x,

extending meromorphically to the whole plane, and

Li1(x) = − log(1− x),

extending to all real x < 1 (or complex x excluding real x ≥ 1).

For |x| < 1, or |x| ≤ 1 with k > 1, we have

Lik(x) + Lik(−x) = 21−kLik(x2). (2)

For k > 1,

Lik(1) = ζ(k), (3)

Lik(−1) = −(1− 21−k)ζ(k). (4)

In particular, Li2(−1) = −12ζ(2) and Li3(−1) = −3

4ζ(3).

1

For k ≥ 2 and 0 ≤ x ≤ 1, we have from the series x ≤ Lik(x) ≤ ζ(k)x and −x ≤Lik(x) ≤ 0. Further, xLi′k(x) = Lik−1(x) (also, Li′k(0) = 1) and

Lik(x) =

∫ x

0

Lik−1(t)

tdt. (5)

(Termwise integration of the series is justified by uniform convergence for |x| < 1, and then

by continuity of Lik(x) at 1 and −1.)

We will work out various special values of Li2 and Li3, which were found by Landen

as long ago as 1780. In particular, we evaluate Li3(φ−2), where φ = 1

2(1 +

√5) is the golden

ratio. This is used to prove Hjortnaes’ series expression (35) for ζ(3). My proof of Comtet’s

corresponding sum for ζ(4) involves quantities like Li4(eπi/3).

The dilogarithm

The dilogarithm Li2 is sometimes called Spence’s function, in tribute to a pioneering

study of it by W. Spence in 1809. Beware of the fact that some computer algebra systems

denote Li2(1− x) by dilog(x).

By termwise integration of the series

− log(1− t)t

=∞∑n=1

tn−1

n

we obtain the identity

Li2(x) = −∫ x

0

log(1− t)t

dt, (6)

which we now use to define Li2(x) for all x ≤ 1. Note that

Li2(1) = ζ(2) = −∫ 1

0

log(1− t)t

dt = −∫ 1

0

log u

1− udu

and for x > −1,

Li2(−x) = −∫ −x0

log(1− t)t

dt = −∫ x

0

log(1 + t)

tdt.

Clearly, Li′2(x) = − log(1− x)/x, and hence Li2(x) is strictly increasing for all x < 1.

For 0 < x < 1, we have

Li2(x) = ζ(2) +

∫ 1

x

log(1− u)

udu

= ζ(2) +

∫ 1−x

0

log t

1− tdt.

2

Replacing x by 1− x, this says

Li2(1− x)− ζ(2) =

∫ x

0

log t

1− tdt

=[− log t log(1− t)

]x0

+

∫ x

0

log(1− t)t

dt

= − log x log(1− x)− Li2(x),

so we have the following identity, known as Euler’s reflection formula:

Li2(x) + Li2(1− x) = ζ(2)− log x log(1− x). (7)

In particular, using the well-known identity ζ(2) = π2/6, we have

Li2(12) = 1

2(ζ(2)− log2 2) =

π2

12− 1

2log2 2. (8)

We can rewrite (8) as an expression for ζ(2):

ζ(2) = log2 2 + 2∞∑n=1

1

n22n.

Taking log 2 as known, this can be regarded as a series for ζ(2) that converges much more

rapidly than∑∞

n=1(1/n2).

We now give two further identities for Li2. Firstly, for x > 0, we have

Li2(−x) = Li2(−1) + F (x) = −12ζ(2) + F (x)

where

F (x) =

∫ 1

x

log(1 + t)

tdt.

Combined with the same statement for 1/x, this gives

Li2(−x) + Li2(− 1x) = −ζ(2) + F (x) + F ( 1

x).

But

F (x) =

∫ 1/x

1

u log

(1 +

1

u

)1

u2du

=

∫ 1/x

1

log(1 + u)− log u

udu

= −F ( 1x)− 1

2log2 x,

so we have

Li2(−x) + Li2(− 1x) = −ζ(2)− 1

2log2 x. (9)

3

So for x > 1, we have the following series expression for Li2(−x) in powers of 1x:

Li2(−x) = −12

log2 x− ζ(2) +∞∑n=1

(−1)n−1

n2xn,

and we can deduce Li2(−x) = −12

log2 x− ζ(2) + r(x), where 0 < r(x) ≤ 1/x.

For 0 < x < 1, we have also

Li2(1− x) = −∫ 1

x

log t

1− tdt

=

∫ 1x

1

log u

1− 1/u

1

u2du

=

∫ 1x

1

log u

(1

u− 1− 1

u

)du

=

∫ 1x−1

0

log(1 + v)

vdv − 1

2log2 x

= −Li2(1− 1x)− 1

2log2 x.

Applying this also to 1/x, we have the following identity for all x > 0:

Li2(1− x) + Li2(1− 1x) = −1

2log2 x. (10)

Using (2), (7), (9) and (10), we can reduce the computation of Li2(x) for any x < 1 to

computation of values of at most a few values of Li2(y) with |y| ≤ 12.

Now consider φ. Note that the fundamental equation satisified by φ may be written:

φ2 = φ+ 1, φ = 1 + φ−1, 1 = φ−1 + φ−2.

Putting x = φ−1 in (2), (10), (7) respectively, we obtain three linear equations relating

Li2(φ−1), Li2(−φ−1), and Li2(φ

−2)):

Li2(φ−1) + Li2(−φ−1) = 1

2Li2(φ

−2), (11)

Li2(φ−2) + Li2(−φ−1) = −1

2log2 φ. (12)

Li2(φ−1) + Li2(φ

−2) = ζ(2)− 2 log2 φ. (13)

Subtracting (12) from (11) gives

Li2(φ−1)− 3

2Li2(φ

−2) = 12

log2 φ. (14)

Now subtracting (14) from (13), we have

52Li2(φ

−2) = ζ(2)− 52

log2 φ,

4

i.e.

Li2(φ−2) = 2

5ζ(2)− log2 φ =

π2

15− log2 φ. (15)

Now (13) or (14) gives

Li2(φ−1) = 3

5ζ(2)− log2 φ =

π2

10− log2 φ, (16)

while (12) gives

Li2(−φ−1) = −25ζ(2) + 1

2log2 φ = −π

2

15+ 1

2log2 φ, (17)

hence finally (9) gives

Li2(−φ) = −35ζ(2)− log2 φ = −π

2

10− log2 φ. (18)

If we include Li2(0) = 0, then we have found eight special values of Li2(x) for real x. I have

seen it stated that these are the only ones known to be expressible in this kind of way.

We now briefly consider complex arguments. Substitution in the series gives

Li2(i) = −18ζ(2) + iG, Li2(−i) = −1

8ζ(2)− iG,

where

G = 1− 1

32+

1

52− · · ·

is Catalan’s constant. Assuming (10) valid for complex arguments, take x = 1 + i, so that

1− x = −i and 1− x−1 = 12

+ 12i. We obtain

Li2

(1 + i

2

)=

1

8ζ(2) + iG− 1

2

(1

2log 2 +

πi

4

)2

=5π2

96− 1

8log2 2 + i

(G− π

8log 2

).

For eiθ, we have

<Li2(eiθ) =

∞∑n=1

cosnθ

n2= π2B2

(θ

2π

),

where B2 is the Bernoulli polynomial B2(x) = x2 − x+ 16

and B2(x) = B2({x}). Also,

=Li2(eiθ) =

∞∑n=1

sinnθ

n2,

which is called the ‘Clausen’ function in [BBC].

There are many further formulae involving Li2: see, for example, Wikipedia, [Lew]

and [Max]. One is ‘Abel’s identity’, which equates a combination of five Li2 values to

log(1− x) log(1− y).

5

The trilogarithm

For 0 < |x| ≤ 1, we have by (5)

Li3(x) =

∫ x

0

Li2(t)

tdt. (19)

Since Li2(x) has been defined by (6) for all x ≤ 1, we can use (19) to define Li3(x) for all

such x.

So we have

Li3(x) =

∫ x

0

Li2(u)

udu

= −∫ x

0

1

u

∫ u

0

log(1− t)t

dt du

= −∫ x

0

log(1− t)t

∫ x

t

1

udu dt

=

∫ x

0

log(1− t)t

logt

xdt, (20)

valid regardless of the sign of x. For 0 < x < 1, substitution of (6) now gives

Li3(x) =

∫ x

0

log t log(1− t)t

dt+ Li2(x) log x, (21)

which can also be derived directly from the series expression. Further, for x > 0,

Li3(−x) =

∫ −x0

log(1− u)

ulog

u

−xdu

=

∫ x

0

log(1 + t)

tlog

t

xdt

=

∫ x

0

log t log(1 + t)

tdt+ Li2(−x) log x. (22)

In particular,

ζ(3) = Li3(1) =

∫ 1

0

log t log(1− t)t

dt. (23)

Integrating by parts in (21), we obtain for 0 ≤ x ≤ 1,

Li3(x) =[12

log2 t log(1− t)]x0−∫ x

0

12

log2 t−1

1− tdt+ Li2(x) log x,

= 12

∫ x

0

log2 t

1− tdt+ Li2(x) log x+ 1

2log2 x log(1− x). (24)

In particular, ∫ 1

0

log2 t

1− tdt = 2ζ(3).

6

Using (9), we can derive a corresponding statement for Li3. [This has been added to

Tim’s original version, in which the following identity was deduced from (26) below.] For

x > 0, write

F (x) =

∫ 1

x

log t log(1 + t)

tdt.

By (22),

Li3(−x) = F (0)− F (x) + Li2(−x) log x.

With x replaced by 1x, this says

Li3(− 1x) = F (0)− F ( 1

x)− Li2(− 1

x) log x.

Combining these expressions and using (9), we have

Li3(− 1x)− Li3(−x) = F (x)− F ( 1

x)− log x

[Li2(−x) + Li2(− 1

x)]

= F (x)− F ( 1x) + ζ(2) log x+ 1

2log3 x.

Substituting t = 1/u, we have

F (x) =

∫ 1

1/x

log u log

(1 +

1

u

)1

udu

= −∫ 1/x

1

1

ulog u [log(u+ 1)− log u] du

= 13

log3 1x

+ F ( 1x)

= −13

log3 x+ F ( 1x),

and we conclude that

Li3(− 1x)− Li3(−x) = ζ(2) log x+ 1

6log3 x. (25)

It follows that for x > 1, Li3(−x) = −16

log3 x− ζ(2) log x− r(x), where 0 < r(x) ≤ 1x.

We now prove another identity relating Li3(x), Li3(1−x) and Li3(1− 1x). It again goes

back to Landen around 1780. [The proof given here was found among Tim’s handwritten

papers; it replaces a more complicated proof in Tim’s typed version.]

Let 0 < x < 1 and

S(x) = Li3(x) + Li3(1− x) + Li3(1− 1x).

We express everything in terms of integrals on [x, 1]. The details are kept simpler by ex-

pressing the integrands in terms of Li2 rather than log terms. Firstly, using (19) and (7), we

7

have

Li3(x) =

∫ x

0

Li2(t)

tdt

= ζ(3)−∫ 1

x

Li2(t)

tdt

= ζ(3)−∫ 1

x

1

t[ζ(2)− log t log(1− t)− Li2(1− t)] dt

= ζ(3) + ζ(2) log x+

∫ 1

x

log t log(1− t)t

dt+

∫ 1

x

Li2(1− t)t

dt.

Secondly,

Li3(1− x) =

∫ 1−x

0

Li2(u)

udu =

∫ 1

x

Li2(1− t)1− t

dt.

Thirdly, with obvious substitutions,

Li3(1− 1x) =

∫ 1/x−1

0

Li2(−u)

udu

=

∫ 1/x

1

Li2(1− v)

v − 1dv

=

∫ 1

x

Li2(1− 1t)

1t− 1

1

t2dt

=

∫ 1

x

Li2(1− 1t)

t(1− t)dt.

Combining the three terms, we have

S(x) = ζ(3) + ζ(2) log x+

∫ 1

x

G(t) dt,

where

G(t) =log t log(1− t)

t+

(1

t+

1

1− t

)[Li2(1− t) + Li2(1− 1

t)].

Now by (10), we have

G(t) =1

tlog t log(1− t)− 1

2log2 t

(1

t+

1

1− t

)=

d

dt

[12

log2 t log(1− t)− 16

log3 t],

so ∫ 1

x

G(t) dt = 16

log3 x− 12

log2 x log(1− x).

So for 0 < x < 1, we have the identity

Li3(x) + Li3(1− x) + Li3(1− 1x) = ζ(3) + ζ(2) log x+ 1

6log3 x− 1

2log2 x log(1− x). (26)

8

One can deduce (25) by writing this with 1− x in place of x, taking the difference and

putting x for 1x− 1.

Putting x = 12

in (26) gives

2Li3(12) = −Li3(−1) + ζ(3) + ζ(2) log 1

2− 1

2log2 2 log 1

2+ 1

6log3 1

2

= 34ζ(3) + ζ(3)− ζ(2) log 2 + 1

2log3 2− 1

6log3 2

= 74ζ(3)− ζ(2) log 2 + 1

3log3 2,

so that

Li3(12) = 7

8ζ(3)− π2

12log 2 + 1

6log3 2. (27)

This gives a series converging rapidly to ζ(3). A more direct proof of (27) is given in [Mel].

Putting x = φ−1 and using the identities relating φ, φ−1 and φ−2, we get

Li3(φ−1) + Li3(φ

−2) + Li3(−φ−1) = ζ(3)− ζ(2) log φ− 12

log2 φ log φ−2 + 16

log3 φ−1

= ζ(3)− ζ(2) log φ+ log3 φ− 1

6log3 φ

= ζ(3)− ζ(2) log φ+ 56

log3 φ.

But by (2), Li3(φ−1) + Li3(−φ−1) = 1

4Li3(φ

−2), hence

Li3(φ−2) = 4

5ζ(3)− 2π2

15log φ+ 2

3log3 φ. (28)

It seems that there are no known such special values involving φ for higher k, although there

are important relations between various values. This is to do with ‘polylogarithm ladders’

(introduced by Leonard Lewin), which are important in K-theory and algebraic geometry,

and can be used in conjunction with the BBP algorithm for computing various constants.

A proof of ζ(2) = π2/6 and some power series related to arcsinx

For 0 ≤ x < 1, the substutution t =√1−x2x

u gives

x

∫ 1

0

dt

1− x2 + x2t2=

1√1− x2

∫ x√1−x2

0

du

1 + u2

=1√

1− x2arctan

x√1− x2

=arcsinx√

1− x2. (29)

Since alsod

dx(arcsinx)2 =

2 arcsinx√1− x2

,

9

we now have

(arcsinx)2 =

∫ x

0

2 arcsin y√1− y2

dy

=

∫ x

0

∫ 1

0

2y

1− y2 + y2t2dt dy

=

∫ 1

0

∫ x

0

2y

1− y2 + y2t2dy dt

= −∫ 1

0

1

1− t2[log(1− y2 + y2t2)

]y=xy=0

dt

= −∫ 1

0

log(1− x2 + x2t2)

1− t2dt. (30)

In particular, to show that ζ(2) = π2/6, we have

π2

4= (arcsin 1)2

= −∫ 1

0

2 log t

1− t2dt

= −2∞∑n=0

∫ 1

0

t2n log t dt

= 2∞∑n=0

1

(2n+ 1)2

= 32ζ(2).

Alternatively, we can equate the integral to Li2(1)−Li2(−1) by (6). This proof has similarities

with the one given by Nick Lord [Lo]. [Tim’s proof has now appeared in Math. Gazette [Jam].]

To derive a power series from (29), observe that for |x| < 1 and 0 ≤ t ≤ 1,

x

1− x2 + x2t2=∞∑n=0

x2n+1(1− t2)n =∞∑n=1

x2n−1(1− t2)n−1

and ∫ 1

0

(1− t2)n−1 dt =

∫ π/2

0

cos2n−1 θ dθ =2.4. . . . (2n− 2)

1.3. . . . (2n− 1)=

22n−1

n(2nn

) .So we have for |x| < 1,

arcsinx√1− x2

=∞∑n=1

22n−1x2n−1

n(2nn

) . (31)

This can also be seen as a case of the hypergeometric series F (1, 1; 32;x2). For example, the

case x = 12

gives

π

3√

3=∞∑n=1

1

n(2nn

) .10

Formulae like this form the basis for various programs which use a spigot algorithm to

calculate π, with time roughly proportional to the square of the number of digits required

(e.g. one by Winter and Flimmenkamp).

Now by integrating, or by similar reasoning from (30), we have

(arcsinx)2 =∞∑n=1

22n−1x2n

n2(2nn

) , (32)

equivalently for |x| < 2, (arcsin

x

2

)2=∞∑n=1

x2n

2n2(2nn

) . (33)

The case x = 1 givesπ2

18=∞∑n=1

1

n2(2nn

) .Either by the identity sinh−1 y = −i sin−1 iy, or by similar reasoning with x2 replaced

by −x2, we have also (sinh−1

x

2

)2=∞∑n=1

(−1)n−1x2n

2n2(2nn

) . (34)

Hjortnaes’ series for ζ(3)

We give a proof of this identity using (34) and the values of Li2(φ−2) and Li3(φ

−2).

The connection with φ arises from the fact that sinh−1 12

= log φ. Let

S =∞∑n=1

(−1)n−1

n3(2nn

) .Then by (34),

S =

∫ 1

0

4

x

(sinh−1

x

2

)2dx

=

∫ sinh−1 12

0

4

2 sinhuu22 coshu du (x = 2 sinhu)

= 4

∫ log φ

0

u2 cothu du

= 4

∫ log φ

0

u2e2u + 1

e2u − 1du

= 4

∫ 2 log φ

0

(v2

)2 ev + 1

ev − 1

dv

2

=1

2

∫ 2 log φ

0

v2ev + 1

ev − 1dv

11

=1

2

∫ 2 log φ

0

v2(

1 +2

ev − 1

)dv

=4

3log3 φ+

∫ 2 log φ

0

v2

ev − 1dv

=4

3log3 φ+

∫ φ−2

1

log2 u1u− 1

−duu

(v = − log u)

=4

3log3 φ+

∫ 1

φ−2

log2 u

1− udu

= 2ζ(3)−∫ φ−2

0

log2 u

1− udu+

4

3log3 φ.

Now substituting from (24), and then the values of Li2(φ−2) and Li3(φ

−2) from (15) and

(28), we have

S = 2ζ(3)− 2Li3(φ−2) + 2Li2(φ

−2) log φ−2 + log2 φ−2 log(1− φ−2) + 43

log3 φ

= 2ζ(3)− 2Li3(φ−2)− 4Li2(φ

−2) log φ+ 4 log2 φ log φ−1 + 43

log3 φ

= 2ζ(3)− 2Li3(φ−2)− 4Li2(φ

−2) log φ− 83

log3 φ

= 2ζ(3)− 2Li3(φ−2)− 4

(25ζ(2)− log2 φ

)log φ− 8

3log3 φ

= 2ζ(3)− 2Li3(φ−2)− 8

5ζ(2) log φ+ 4

3log3 φ

= 2ζ(3)− 2(45ζ(3)− 4

5ζ(2) log φ+ 2

3log3 φ

)− 8

5ζ(2) log φ+ 4

3log3 φ

= 25ζ(3),

by a neat cancellation. So

ζ(3) =5

2

∞∑n=1

(−1)n−1

n3(2nn

) . (35)

This series was found by Hjortnaes in 1953 [Hj]. It has sometimes been wrongly at-

tributed to Apery, because it was used in his proof that ζ(3) is irrational [Ap]. However, it

is not used in the simpler proof of Apery’s theorem by Beukers [Beu]. A proof of (35) that

does not involve polylogarithms is outlined in [VDP]; it is reproduced in [BB, p. 378–379].

A generalized version is proved in [AG].

Some series involving Hn

We denote the harmonic sum by Hn =∑n

r=11r. For |x| < 1, we have easily

− log(1− x)

1− x=∞∑j=0

xj∞∑k=1

xk

k=∞∑n=1

Hnxn.

12

Similarly, in the product log2(1 − x) = (∑∞

j=1 xj/j)(

∑∞k=1 x

k/k, the coefficient of xn, for

n ≥ 2, isn−1∑j=1

1

j(n− j)=

1

n

n−1∑j=1

(1

j+

1

n− j

)=

2Hn−1

n,

so

log2(1− x) =∞∑n=2

2Hn−1

nxn. (36)

Since Hn = Hn−1 + 1n

(also for n = 1 with H0 = 0), we deduce

∞∑n=1

Hn

nxn = 1

2log2(1− x) + Li2(x).

Integration of (36) gives∞∑n=2

2Hn−1

n2xn =

∫ x

0

log2(1− t)t

dt, (37)

hence∞∑n=1

Hn

n2xn = 1

2

∫ x

0

log2(1− t)t

dt+ Li3(x)

Integrating again, we obtain

∞∑n=2

2Hn−1

n3xn =

∫ x

0

1

t

∫ t

0

log2(1− u)

udu dt

=

∫ x

0

log2(1− u)

u

∫ x

u

dt

tdu

=

∫ x

0

log2(1− u)

ulog

x

udu, (38)

which we’ll use with x = eπi/3 later.

Some multiple zeta values

We define the ‘multiple zeta value’ function as

ζ(s1, . . . , sk) =∑

n1,...,nk0<n1<···<nk

n−s11 · · ·n−skk .

(Annoyingly this is sometimes written ζ(sk, . . . , s1) instead!) Here I will only consider ζ(s, t).

We have

ζ(s, t) =∞∑m=1

1

ms

∞∑n=m+1

1

nt=

∞∑m=1

1

ms

∞∑n=1

1

(m+ n)t.

13

Also, reversing the order in the first expression, we have

ζ(s, t) =∞∑n=2

1

nt

n−1∑m=1

1

ms.

in particular,

ζ(1, t) =∞∑n=2

Hn−1

nt. (39)

Most obviously, we have for <s > 1

ζ(s, s) =∑m,nm<n

(mn)−s

=1

2

∑m,nm 6=n

(mn)−s

=1

2

(∑m,n

(mn)−s −∑n

n−2s

)

=1

2(ζ(s)2 − ζ(2s)). (40)

The case s = 2 gives

ζ(2, 2) =π4

2

(1

36− 1

90

)=

π4

120=

3

4ζ(4). (41)

Taking x = 1 in (37) and applying (24), we have

ζ(1, 2) =∞∑n=2

Hn−1

n2

=1

2

∫ 1

0

log2(1− t)t

dt

= ζ(3).

an identity already known to Euler. Later I found the following direct proof avoiding inte-

grals. [Note added by Graham Jameson: this method can be seen, for example, in [BBr, p.

7–8], where it is attributed to R. Steinberg]. Since

ζ(1, 2) =∞∑m=1

∞∑n=1

1

m(m+ n)2.

we have

2ζ(1, 2) =∞∑m=1

∞∑n=1

1

(m+ n)2

(1

m+

1

n

)=

∞∑m=1

∞∑n=1

1

mn(m+ n).

14

Now1

mn(m+ n)=

1

m2

m

n(m+ n)=

1

m2

(1

n− 1

m+ n

)and by cancellation

∞∑n=1

(1

n− 1

m+ n

)= 1 +

1

2+ · · ·+ 1

m= Hm,

so

2ζ(1, 2) =∞∑m=1

Hm

m2

= ζ(3) +∞∑m=2

Hm−1

m2

= ζ(3) + ζ(1, 2).

A similar method delivers an identity for ζ(1, 3), which will be used for Comtet’s series.

[Tim’s original method was by manipulation of the series for Li2(x)2; the following is slightly

shorter and more like the previous proof.]

2ζ(1, 3) =∞∑m=1

∞∑n=1

1

(m+ n)3

(1

m+

1

n

)=

∞∑m=1

∞∑n=1

1

mn(m+ n)2

and

1

mn(m+ n)2=

1

m2(m+ n)

(1

n− 1

m+ n

)=

1

m2n(m+ n)− 1

m2(m+ n)2

=1

m3

(1

n− 1

m+ n

)− 1

m2(m+ n)2,

so, as before,

2ζ(1, 3) =∞∑m=1

Hm

m3−∞∑m=1

∞∑n=1

1

m2(m+ n)2

= ζ(4) +∞∑m=2

Hm−1

m3− ζ(2, 2)

= ζ(4) + ζ(1, 3)− ζ(2, 2).

Hence, by (41),

ζ(1, 3) = ζ(4)− ζ(2, 2) = 14ζ(4). (42)

15

Comtet’s series for ζ(4)

For this, [BBC] gives a reference to Comtet’s book [Com], but I have not seen this or

any other proof. Here is my proof (completed 4-8-07). Perhaps it could be substantially

simplified! Let

S =∞∑n=1

1

n4(2nn

) .From (33), we have

∞∑n=1

y2n

n3(2nn

) =

∫ y

0

4

z

(arcsin

z

2

)2dz

so

S = 2

∫ 1

0

1

y

∫ y

0

4

z

(arcsin

z

2

)2dz dy

= 8

∫ 1

0

1

z

(arcsin

z

2

)2 ∫ 1

z

dy

ydz

= −8

∫ 1

0

1

z

(arcsin

z

2

)2log z dz

= −8

∫ π6

0

1

2 sin θθ2 log(2 sin θ) · 2 cos θ dθ (z = 2 sin θ)

= −8

∫ π6

0

θ2 cot θ log(2 sin θ) dθ

= −4

∫ π6

0

θ2d

dθ(log2(2 sin θ)) dθ

= −4[θ2 log2(2 sin θ)

]π6

0+ 4

∫ π6

0

2θ log2(2 sin θ) dθ

= 8

∫ π6

0

θ log2(2 sin θ) dθ.

Now expand as follows (maybe something circular is happening here?):

log(2 sin θ) = log

(eiθ − e−iθ

i

)=(θ − π

2

)i+ log

(1− e−2iθ

).

so

S = 8

∫ π6

0

θ

(−(θ − π

2

)2+ 2i

(θ − π

2

)log(1− e−2iθ

)+ log2

(1− e−2iθ

))dθ

= −8I1 + 16I2 + 8I3,

where

I1 =

∫ π6

0

θ(θ − π

2

)2dθ

16

=

∫ π6

0

(θ3 − πθ2 +

π2

4θ

)dθ

= π4

(1

4 · 64− 1

3 · 63+

1

2 · 4 · 62

)=

π4

4 · 64(1− 8 + 18)

=11π4

4 · 64,

I2 = i

∫ π6

0

θ(θ − π

2

)log(1− e−2iθ) dθ

= −i∞∑n=1

1

n

∫ π6

0

(θ2 − πθ

2

)e−2inθ dθ

= −i∞∑n=1

1

n

([(θ2 − πθ

2

)e−2inθ

−2in

]π6

0

−∫ π

6

0

(2θ − π

2

) e−2inθ−2in

dθ

)

= −i∞∑n=1

1

n

((π2

36− π2

12

)e−πin/3

−2in−[(

2θ − π

2

) e−2inθ

(−2in)2

]π6

0

+

∫ π6

0

2e−2inθ

(−2in)2dθ

)

= −i∞∑n=1

1

n

((−π

2

18

)ie−πin/3

2n−(π

3− π

2

) e−πin/3−4n2

+(−π

2

) 1

−4n2+

[2e−2inθ

(−2in)3

]π6

0

)

= −i∞∑n=1

1

n

(− π

2i

36ne−πin/3 − π

24n2e−πin/3 +

π

8n2− i

4n3(e−πin/3 − 1)

)=

∞∑n=1

(− π2

36n2e−πin/3 +

πi

24n3e−πin/3 − πi

8n3− 1

4n4e−πin/3 +

1

4n4

),

and by (36)

I3 =

∫ π6

0

θ log2(1− e−2iθ) dθ

=∞∑n=2

2Hn−1

n

∫ π6

0

θe−2inθ dθ

=∞∑n=2

2Hn−1

n

([θe−2inθ

−2in

]π6

0

−∫ π

6

0

e−2inθ

−2indθ

)

=∞∑n=2

2Hn−1

n

(πi

12ne−πin/3 −

[e−2inθ

(−2in)2

]π6

0

)

=∞∑n=2

2Hn−1

n

(πi

12ne−πin/3 +

1

4n2(e−πin/3 − 1)

)=

∞∑n=2

Hn−1

(πi

6n2e−πin/3 +

1

2n3e−πin/3 − 1

2n3

).

17

Hence

S = −22π4

64+∞∑n=1

(−4π2

9n2e−πin/3 +

2πi

3n3e−πin/3 − 2πi

n3− 4

n4e−πin/3 +

4

n4

)+∞∑n=2

Hn−1

(4πi

3n2e−πin/3 +

4

n3e−πin/3 − 4

n3

)= − 11π4

23 · 34− 2πiζ(3) + 4ζ(4)− 4ζ(1, 3)

+∞∑n=1

(−4π2

9n2+

2πi

3n3− 4

n4

)e−πin/3 +

∞∑n=2

Hn−1

(4πi

3n2+

4

n3

)e−πin/3.

By (42), 4ζ(1, 3) = ζ(4), so the first four terms equate to

− 11π4

23 · 34+ 3ζ(4)− 2πiζ(3) =

(1

2 · 3 · 5− 11

23 · 34

)π4 − 2πiζ(3)

=22 · 33 − 5 · 11

23 · 34 · 5π4 − 2πiζ(3)

=108− 55

23 · 34 · 5π4 − 2πiζ(3)

=53π4

23 · 34 · 5− 2πiζ(3).

So our formula becomes

S =53π4

23 · 34 · 5− 2πiζ(3) +

∞∑n=1

(−4π2

9n2+

2πi

3n3− 4

n4

)e−πin/3 +

∞∑n=2

Hn−1

(4πi

3n2+

4

n3

)e−πin/3.

The imaginary part of S is zero, so we have proved

2πζ(3) =∞∑n=1

(4π2

9n2sin

πn

3+

2π

3n3cos

πn

3+

4

n4sin

πn

3

)+∞∑n=2

Hn−1

(4π

3n2cos

πn

3− 4

n3sin

πn

3

).

Incidentally, I cannot evaluate the separate contributions of any of these terms. The real

terms give

S =53π4

23 · 34 · 5+∞∑n=1

(−4π2

9n2cos

πn

3+

2π

3n3sin

πn

3− 4

n4cos

πn

3

)+∞∑n=2

Hn−1

(4π

3n2sin

πn

3+

4

n3cos

πn

3

). (43)

The two sums here will turn out to give a total of −ζ(4). We now evaluate the first sum.

The contribution of each of the three terms is of the form CBk(16), by the Fourier series for

18

the periodified Bernoulli polynomial

Bk(x) = Bk({x}0)

= −k!∞∑

n=−∞n6=0

e2πinx

(2πin)k

= − k!

(2πi)k(F (k, x) + (−1)kF (k,−x)),

where

F (s, α) = Lis(e2πiα) =

∞∑n=1

e2πinα

ns.

is the Lerch zeta function. Splitting according to parity gives

∞∑n=1

cos 2πnx

n2k= (−1)k−1

(2π)2k

2(2k)!B2k(x),

and∞∑n=1

sin 2πnx

n2k+1= (−1)k−1

(2π)2k+1

2(2k + 1)!B2k+1(x).

The Bernoulli polynomials up to B4 are

B0(x) = 1,

B1(x) = x− 12,

B2(x) = x2 − x+ 16,

B3(x) = x3 − 32x2 + 1

2x,

B4(x) = x4 − 2x3 + x2 − 130,

so we find∞∑n=1

cos(πn/3)

n2= π2B2(

16) =

π2

36,

∞∑n=1

sin(πn/3)

n3=

(2π)3

2 · 3!B3(

16)

=2π3

3 · 63

(1− 3 · 6

2+

62

2

)=

2π3

3 · 63(1− 9 + 18)

=20π3

3 · 63,

∞∑n=1

cos(πn/3)

n4= −(2π)4

2 · 4!B4(

16)

19

= − π4

3 · 64

(1− 2 · 6 + 62 − 64

30

)= − π4

3 · 5 · 64(125− 216)

=91π4

3 · 5 · 64.

So the first sum in (43) is

∞∑n=1

(−4π2

9· cos(πn/3)

n2+

2π

3· sin(πn/3)

n3− 4

cos(πn/3)

n4

)= −4π2

9· π

2

62+

2π

3· 20π3

3 · 63− 4

91π4

3 · 5 · 64

=π4

3 · 5 · 64

(−4

9· 3 · 5 · 62 +

2

3· 20 · 5 · 6− 4 · 91

)=

π4

3 · 5 · 64

(−20 · 62

3+

40 · 5 · 63

− 4 · 91

)=

π4

3 · 5 · 64(−240 + 400− 364)

= − 204π4

3 · 5 · 64

= − 34π4

3 · 5 · 63.

Since 3 · 5 · 63 = 23 · 34 · 5 and 53− 34 = 19, we have proved

S =19π4

23 · 34 · 5+ 4

∞∑n=2

Hn−1

(π sin(πn/3)

3n2+

cos(πn/3)

n3

). (44)

We now define

f(x) =

∫ x

0

log2(1− u) log u

udu. (45)

By (38), we have

f(x) =

(∞∑n=2

2Hn−1

n2xn

)log x−

∞∑n=2

2Hn−1

n3xn,

so that (44) becomes

S =19π4

23 · 34 · 5− 2<f(eπi/3). (46)

The obvious integration by parts gives a reflection relation between f(x) and f(1− x):

f(x) =[log2(1− u) · 1

2log2 u

]x0−∫ x

0

(−2 log(1− u)

1− u

)12

log2 u du

= 12

log2(1− x) log2 x+

∫ x

0

log2 u log(1− u)

1− udu

= 12

log2(1− x) log2 x+ f(1)−∫ 1

x

log2 u log(1− u)

1− udu.

20

By (39) and (42), f(1) = −2ζ(1, 3) = −12ζ(4). Also,∫ 1

x

log2 u log(1− u)

1− udu. =

∫ 1−x

0

log2(1− u) log u

udu = f(1− x),

so

f(x) + f(1− x) = −12ζ(4) + 1

2log2(1− x) log2 x. (47)

Since f(z) = f(z) (this occurs when f is analytic and real on the real axis), the case x = eπi/3

gives us the value we want:

2<f(eπi/3) = f(eπi/3) + f(e−πi/3)

= f(eπi/3) + f(1− eπi/3)

= −12ζ(4) + 1

2

(−πi

3

)2(πi

3

)2

= −12ζ(4) + 1

2

(π3

)4=

π4

2

(− 1

2 · 32 · 5+

1

34

)=

π4

22 · 34 · 5(−32 + 2 · 5)

=π4

22 · 34 · 5.

Inserting this into (46) gives

S =17π4

23 · 34 · 5=

17

36ζ(4),

so finally we have Comtet’s series

ζ(4) =36

17

∞∑n=1

1

n4(2nn

) . (48)

We have had two occurences of ζ(1, 3) = 14ζ(4) in our calculation of S. These did not simply

cancel out: the first occurrence was −4ζ(1, 3), while the second was −f(1) = 2ζ(1, 3).

References (added by Graham Jameson)

[AG] G. Almkvist and A. Granville, Borwein and Bradley’s Apery-like formula forζ(4n+ 3), Experimental Math. 8 (1999), 197–203.

[Ap] R. Apery, Irrationalite de ζ(2) et ζ(3), Asterisque 61 (1979), 11–13.

[Beu] F. Beukers, A note on the irrationality of ζ(2) and ζ(3), Bull. London Math. Soc.11 (1979), 268–272.

[BB] Jonathan M. Borwein and Peter B. Borwein, Pi and the AGM, John Wiley (1987).

21

[BBr] Jonathan M. Borwein and David M. Bradley, Thirty-two Goldbach variations,Int. J. Number Theory, 2 (2006), 65–103.

[BBC] Jonathan M. Borwein, David M. Bradley and Richard E. Crandall, Computationalstrategies for the Riemann zeta function, J. Comput. Appl. Math. 121 (2000),247–296.

[Com] Louis Comtet, Advanced Combinatorics, Reidel, Dordrecht (1974).

[Hj] M. M. Hjortnaes, Overføring av rekken∑∞

k=1 1/k3 til et bestemt integral, Proc.12th Cong. Scand. Math. (Lund, 1953), Lund (1954).

[Jam] Tim Jameson, Another proof that ζ(2) = π2/6 via double integration, Math.Gazette 97 (2013), 506–507.

[Lew] Leonard Lewin, Polylogarithms and associated functions, North Holland (1981).

[Lo] N. Lord, Yet another proof that∑

(1/n2) = π2/6, Math. Gazette 86 (2002),477–479.

[Max] Leonard Maximon, The dilogarithm function for complex argument, Proc. RoyalSoc. London 459 (2003), 2807–2819.

[Mel] John Melville, A simple series representation for Apery’s constant, Math. Gazette97 (2013), 455–460.

[VDP] Alfred van der Poorten, A proof that Euler missed: Apery’s proof of the irrat-ionality of ζ(3), Math. Intelligencer 1 (1979), 195–203.

[Zag] Don Zagier, The dilogarithm function, maths.dur.ac.uk/dma˜0hg/dilog.pdf

updated with minor corrections, November 2018

22