6 Formulae BBP in base 2 : s <img src="mathematiciens/huvent/cmsy10-32.gif" alt=" (- " class="10x-x-32" /> <img src="mathematiciens/huvent/msbm10-4e.gif" alt="N" class="10x-x-4e" />,v = p q, x = 1 _ 2n dans <img src="mathematiciens/huvent/cmr10-9.gif" alt="Y" class="10x-x-9" />

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6 Formulae BBP in base 2 : $s (- N$ , $v = pq$ , $x = 12n-$ in $Y$

6.1 The considered integrals

In the worry of decomposing our integrals $integral 1lnk(y)R (y) ----------dy 0 Q (y)$ in simpler to calculate integrals, we are interested to the following "prime" integrals :

V~ - V~ - 1 --3 1 --3 o`u s1 = 2 (1+ i)+ 2 (1- i), s2 = 2 (1+ i)- 2 (1- i)

(83)

1 V~ 3 1 V~ 3 s3 = 2 (1 - i)+ 2--(1 + i), s4 = 2 (1- i)--2-(1+ i)

(84)

In [3], Broadhurst establish some relation between the different sums of polylogarithm. However he does not seem to consider the link with integrals. We will try to do so. Which is why we introduce the following integrals :

The link between the $J(k) n$ and the is the following :

6.2 The method

The calculation of the integrals $I(nk)$ will give linear combination of constants of order $k + 1$ like $pk+1$ or $z(k + 1)$ thanks to their expression under polylogarithm form of order $k+ 1$ . But furthermore, we can obtain BBP formula with the $I(kn)$ by using what Gery Huvent calls the denomination tables and which are just the expressions $I(k) n$ in the form of integrals $integral 1 P(y)lnk(y) 0 yn- a$ whom we have seen the direct expression under BBP serie form with the formula (69). We just need to obtain BBP series for the precise constant that we are interested in, which boils down to use a certain linear combination of $(k) In$ . For this, we intoduce normally the linear form $11 L (a ,...,a ) = sum a I(k) k 1 11 n=1 n n$ . Then we impose some relation between the $a n$ so to cancel out the coefficients in fronts of the unwanted constants. We then get a BBP serie for the remainder, $p$ or $log(2)$ or a lot more other things! Here is the denominator table. The method is then detailled with an example.

where $P$ is a polynomial with integers coefficient which depends both of $n$ and of the chosen denominator.

6.3 Formulae for $p$ , $ln(2)$ , $ln(3)$ and $ln(5)$

In the case where $k = 0$ , we get, by a calculation of integrals

6.3.1 BBP formulae applications for $p$

So to get some formula for $p,$ we impose the following relations :
$(-a2 - a3- a4 - 2a7- 2 a8- 4a10 -a1) = (a2 + a7) = (a3 +a10)$
$= (2a6 - 2a11) = 0$ .
So we get

So to get some BBP formula with few terms, we can in a first of all fix $a = a = a = a = a = 0 11 10 9 8 7$ , which gives

a5-p = - a I(0)+ a I(0)+ a I(0) 2 4 1 4 4 5 5

(113)

We then consult the denominator table. To simplify a sum that calls $I1,I4$ and $I5,$ we can write each integral under the form $integral 1Pn8(y)4dy, 0 y -2$ $n = 1,4,5.$ Hence

a5- (0) (0) (0) integral 1-P (y) -1- ( 4 ) 2 p = -a4I1 + a4I4 + a5I5 = 0 y8- 24dy = 24 BBP0 2 ,8,P (y)

(114)

where $P$ is a polynomial whose coefficients depends on $a4$ and $a5$ . If we fix $a5 = 1$ and $a4 = - 4r- 1$ we get the formula by Adamchik-Wagon (cf [6] )

sum oo ( ) p = -1- 8r+4- -8r--- -4r-+ -8r-2+ -2r-1 + -2r-1 + -r-- i=016i 8i+1 8i+2 8i+3 8i+4 8i+5 8i+6 8i+7

(Adamchik-Wagon)

The choice of $a4 = -a5 = 1$ give the formula by Plouffe (61). We can now look for a denominator in $y24- 212$ . We take then the formula (111) and we simply fix $a11 = a10 = 0$ .

We choose the other coefficients so to cancel out the most coefficients of $P$ (by imposing $a5 /= 0$ ).
The best choice seems to be
$[ai] = (a1,...,a11) = (0,- 1,0,1,1,0,1,-1,-1,0,0)$ which gives

We can also look for a denominator in $y40- 220$ by fixing $a = a = a = 0, 9 8 7$ the best choice seems to be
$[a ] = (a ,...,a ) = (0,0,0,0,-2,1,0,0,0,0,1) i 1 11$ which gives

In fact it turns out that the last formula $-P40(y)20 y -2$ can be simplified to $-Q20(y)10 y +2$ , which gives the alternated formula :

p = -1BBP0 (- 210,20,2y17- 5y14- 8y13- 8y9- 128y5- 160y4 + 512y)
26
1- oo sum (-1)i( -512-- -160- -128- --8--- --8--- --5--- --2--)
= 64 210i 20i+2 - 20i+5 - 20i+6 - 20i+10 - 20i+14 - 20i+15 + 20i+18
i=0

(Bellard)

(119)

Finnaly if we considered the formula (111) in all generality, it can be written $1 (a + a )p = integral 1-P(y)-dy 2 11 5 0y120- 260$ , we just need to choose the right $(ai)i$ . The choice that seems to lead to a formula with the least terms possible is
$[ai] = (a1,...,a11) = (0,0,0,0,0,1,0,0,- 2,0,1)$ and gives

1 (60 ) p = 256BBP0 2 ,120,P (y) o`u P (y) a 42 coefficients non nuls

(120)

We can explain $P$ by writting that
$integral 1-P(y)-- L0 (0,0,0,0,0,1,0,0,-2,0,1) = 0 y120- 260dy$ .
To finish off, the choice of $[a] = (0,0,0,0,0,1,0,0,-2,0,1)$ leads to

And $[a] = (0,0,0,0,2,0,0,0,-2,0,0)$ gives

6.3.2 BBP formulae for $ln (2), ln (3)$ and $ln(5)$

We can apply the same method to obtain some BBP formula for $ln(2)$ , $ln (3)$ and $ln(5)$ .

6.4 Case of polylogarithms of order $2$ : Formulaes of order 2

We can notice that for order 1, that is for the BBP formulae giving $p$ or for example, we had to deal with integrals with rational fractions. If we now want to get BBP series giving $p2$ or $p *ln(2)$ or $ln(2)2$ , we need to introduce a logarithm to the numerator of the integral. More precisely, for a BBP formula of order $n$ , we need to consider integrals of type $integral ln(x)n-1P(x)dx Q(x)$ . This is due to the fact that we then obtain polylogarithm combination of order $n$ .

This is what was done once more by Gery Huvent [12], noting as well that the first series were found by Plouffe and that Broadhurst [3] provided a few as well. The interest in order 2 is to be able to find BBP series for a famous constant which is Catalan's constant defined by $G = sum o o -(-1)n- n=0(2n+1)2$ . This shows, if you are not convinced, that Catalan's constant is "homogenuous" to an order 2, that is that it is without doubt of the same nature as $2 p$ or $p *ln(2)$ concerning the spread of its digit in base 2 or 16.

6.4.1 The classical expression : $(1) (1) (1) (1) I1 ,I2 ,I3 ,I4$ and $(1) I5$

A classical result by Euler is

2 2 I(11) = p-- ln-(2)- 24 4

(123)

Which allows us to write that

( ) (1) p2- ln2(2) 1 1 I2 = -24 + 4 + 4L2 4

(124)

Kummer's equation for the polylogarithm of order $2$ is written (cf [4])

The inverse formula is

( ) 1 p2- ln2(-z) L2 z = - L2(z)- 6 - 2

(Formule d’inversion)

and finally the duplication formula in the general case :

L (z)+ L (- z) =--1-L (z2) k k 2k-1 k

(Formule de duplication)

By applying Kummer's equation for $x = 1-2-i, y = 12$ then $x = 1+2i, y = 12$ , we get two equality which added gives

By using the inverse formula for $z = - 1+ i$ and $z = - 1- i$ and the duplication formula , we get

By duplication, we also have

( ) ( ) ( ) ( ) ( ) 1+-i 1--i -1--i- --1+-i 1 1 L2 2 + L2 2 + L2 2 + L2 2 = 4L2 - 4

(128)

We therefore deduce

Similarly, Kummer's equation for $x = 1-2i, y = 12$ and $x = 1+2i, y = 12$ , gives two equality which when taken away fives a new equality. By then using the inverse formula for $z = - 1+ i$ and $z = - 1- i$ and the duplication formula for $1+2i$ and $1-2i,$ we obtain

( ) ( ) L 1-+-i - L 1---i - (L (i)- L (- i))+ ip-ln(2) = 0 2 2 2 2 2 2 4

(131)

But $p2- L2(i) = - 48 + iG$ where $G$ is Catalan's constant. Hence

(1) ( (1 - i) (1 + i)) p ln (2) I5 = i L2 -2-- - L2 -2-- = ---4-- - 2G

(132)

6.4.2 Calculation of $I(71),I(18)$ and $I(91)$

Proposition 2 We have

2 ( ) I(17)= p-+ 1 L2 1 72 4 4

(133)

Proof. Kummer's equation with $V~ x = 1-i2-3, y = 12$ gives the equality

which gives straight away the wanted result because
$L (1)= p2 - ln2(2) 2 2 12 2$ , $L (1)+ L (- 1)= 1L (1), 2 2 2 2 2 2 4$
$(1+i V~ 3) (1-i V~ 3) (-1+i V~ 3) (--1- i V~ 3) L2 2 + L2 2 + L2 2 +L2 2$
$( ( V~ -) ( V~ -)) = 1 L2 -1+i-3 + L2 -1-i-3 2 2 2$ by the duplication formula and Kummer formula for $V~ - x = y = 1-i-3 2$ gives $( V~ -) ( V~ -) 2 L2 -1+i2-3 +L2 -1-i2-3 = - p-. 9$ _

Proposition 3 We have

2 2 I(81)= 5p-- ln-(2)- 36 2

(134)

and

I(91) = -2G- 3

(135)

Proof. Kummer's equation for $1+i 1-i V~ 3 x = -2-, y =--2--$ gives, considering
$( ) L2 (1 - i) = p126- i G + p-ln4(2)$ $:$

( ) ( ) ( ) L (s )+ L (s ) = 17-p2- i G + p-ln-(2) - L e-ip6 - L e-5ip6- 2 1 2 2 144 4 2 2

(136)

similarly, Kummer's equation for $V~ x = 1-2i, y = 1-i2-3$ give

17 ( pln(2)) ( ip) ( 5ip) L2 (s3)+ L2 (s4) = ---p2 + i G+ ------ - L2 e 6 - L2 e 6 144 4

(137)

With the help of the inversion formula

Which allows us to conclude that

and

So we just now need to calculate the last sum of polylogarithm. But we have gain in simplicity because those logarithm uses roots of unity.
We therefore use the multiplication formula

q sum -1 ( 2ikp-) L2(zq) = q L2 e q z k=0

(Formule de multiplication)

which gives with $z = -i$ and $q = 3$

1 ( ip) ( 5ip) -L2 (i) = L2 e 6 + L2 e 6 + L2 (- i) 3

(144)

then with $z = i$ and $q = 3$

1 ( ip-) ( 5ip) -L2(- i) = L2 e- 6 + L2 e- 6 + L2(i) 3

(145)

and allows us to easily conclude. _

6.4.3 Calculating $(1) I10 ,$ relation between $(1) I6$ and $(1) I11$

Kummer's equation for $x = -1$ and $y = 1+ i$ and for $x = -1$ and $y = 1 -i$ gives two equality which when added gives

Which is

( 1) 1 p2 J(51)= 2J(13)+ L2 - - + -ln2(2)- --- 4 4 16

(147)

and allows us to confirm that

2 2 ( ) I1(10)= 2p- - ln-(2)+ 1L2 - 1 15 2 4 4

(148)

If, instead of adding them, we substract them, we get

which gives us

(1) ( (1) (1)) p-ln-(2) J6 = 2 J2 - J4 + 4

(150)

i.e.

I(1)+ I(1)= p-ln-2- 12G 6 11 4 5

(151)

6.4.4 Application to the determination of BBP formulae

We now consider the linear form $L (a ,a ,...,a ) = a I(1)+ ...+ a I(1) 1 1 2 11 11 1111$ . Taking into account the equalities (123), (124), (129), (130), (132), (133), (134), (135) and (148), we have

Formulae for $p2$ So to obtain BBP formulae for $p2,$ we fix the following equality :
$(- 1a + 1a + 1a - 1 a - 1 a - 1a )= (- 2a - 2a - 12-a ) 4 1 4 2 4 3 4 4 2 8 2 10 5 3 9 5 11$
$= (a + a ) = (a + a ) = (a + a ) = (a -a ) = 0 5 11 2 7 3 10 6 11$ so to obtain the equality

We now just need to use particular variables so to obtain simple formulae.

A few simple alread known formulae for $p2$

We obtain those formulae by choosing the integrals who give an denominator of a degree less than in the correspndance table.
Let us one last time show details an example :
So to obtain a denominator of the form $yb- a$ of smaller degree (in that case $y24- 212$ ), we choose $a9 = a10 = 0$ such that we cancel down $I(11)0$ and $I(111)$ . OSo we then get

This equality allows to give the general formula with $3$ parameters

( ) -1 -1 -1 2 -1- ( 12 ) 16a4 + 72 a7 + 18a8 p = 211BBP1 2 ,24,P (y)

(160)

So not to put too much on this page only the formulae with parameters that correspond to the denominator in $y24 - 212$ will be given. There exist some who are assoiciated to other denominators (for example $y120- 260$ ).
Let us look again at the equality (159), if we choose to fix $a = 0 7$ and $a = 16, 4$ this equality becomes

(1) (1) integral 1 y ln(y) integral 1(2y- 2)ln(y) 2 -16I1 + 16I4 = - 16 2-y2--2dy+ 16 -y2--2y-+2--dy = p 0 0

(161)

We can look at this equuality under the form of the sum of BBP formula.

We can bring vack this integral to a denominator of the form $yb - a$ with the help of the corresponding table so to obtain

integral ( ) 32 1 ln(y)-y6--2y5--2-y4--8y3--4-y2--8y+-8-dy = p2 0 y8 -16

(163)

which gives the following equality

32 BBP1 (16,8,y6- 2 y5- 2y4- 8 y3- 4y2- 8y + 8)= p2

(164)

oo sum ( ) p2 = 1-- --16-2-- --16-2-- --8-2- -16-2- --4-2- ---4-2 +---2-2 i=0 24i (8i+1) (8i+2) (8i+3) (8i+4) (8i+5) (8i+6) (8i+7)

(165)

This equality was already mentioned by Plouffe in [1] $.$
The choice of $a4 = 0, a7 = 72$ gives

which gives the following formula thanks to Plouffe :

A few simple and new formulae

An other solution consist of keeping the integrals which gives denominator of the form $b y - a$ of high degree but adjust the parameters so to gave many nul coefficients in the BBP formulae.
A few good choice seems to be the following :
$[ai] = (a1,...,a11) = (1,1,0,0,0,0,- 1,0,0,0,0)$ which gives the formula to $10$ terms

The polynomial $P (y)$ having only odd powers, this formula can be simplified to give

$[ai] = (2,1,0,1,0,0,- 1,- 1,0,0,0)$ which gives the formula to $11$ terms

$[a] = (- 2,-1,-1,- 2,0,0,1,0,0,1,0) i$ which gives the formula to $43$ terms (notice the $260$ ) :

If we are interested in formula with the least term, let us state that other formulae with $50$ terms ( $[ai] = (- 1,- 1,0,0,0,0,1,0,0,0,0)$ ) and with $55$ terms ( $[ai] = (- 2,-1,0,-1,0,0,1,1,0,0,0)$ ) exists.
We can even look for alternating series, for example
$[ai] = (0,0,0,-2,0,0,0,,0,0,0)$ gives

9 ( ) p2 = - --BBP1 - 26,12,y10- 8y8- 12y7- 4y6 + 8y4 + 48y3 + 64y2- 32 20

(Huvent)

Notice

The integrals equality allows to write in different ways $2 p$ as the sum of BBP formula.
For example, the result by (Huvent) ( $[ai] = (-1,- 1,0,0,0,0,1,0,0,0,0)$ ) gives the integral equality

Which can be written

Formulae for the constants $G,p ln(2)$ and $2 ln (2)$ For $p ln (2)$

The same method leads to the general formula for $pln(2)$ (when we impose a denominator in $24 12 y - 2$ )

pln(2) =-1-BBP (212,24,P (y)) o`u 211 1

(Huvent)

By adjusting the coefficients $a4$ and $a8,$ we have formulae with $17$ termes of the form

pln(2) = BBP1 (212,24,P (y))

One of the simplest seems to be $[ai] = (2,- 6,0,4,1,0,6,- 6,-3,0,0)$

An other formula with $18$ termes is obtained for $[a] = (0,0,0,0,1,0,0,0,-3,0,0) i$

To finish $[ai] = (0,0,0,0,4,1,0,0,0,0,1)$ gives

For Catalan's constant $G$

Similarly, by imposing a denominator in $y24- 212,$ we obtain

The most interesting case is obtain when all the $ai$ are zero except $a9$ which we let be equal to $1$ . We then obtain

The interest in this formula resides in the coefficients of $P$ which are all powers of $2$ .
The choice of $[ai] = (1,- 3,0,2,0,0,3,-3,1,0,0)$ allows to write
$G = BBP1 (212,24,P (y))$ where $P (y)$ has $16$ non-zero coefficients.
To finish $[a ] = (0,0,0,0,- 1,1,0,0,-1,0,1) i$ gives $G = BBP (230,60,P (y)) 1$ where $P$ has $28$ non-nul coefficients.

Warning 4 It seems that I was the first to have experimentaly discovery a real formula $BBP$ for $G$ (Mai 2000) without being able to find proof. The discussion between me and David Broadhurst with no doubt, but that's not very important...

For $ln2(2)$

We obtain

(9a + 2a + 8a )ln2(2) = - 1-BBP (212,24,P (y)) 4 7 8 29 1

(173)

The simplest case is given by $[ai] = (- 4,- 3,0,0,0,0,3,0,0,0,0)$ which leads to

The choice of $[ai] = (- 42,-21,-20,- 40,0,0,21,0,0,20,0)$ leads to $2 ( 60 ) ln (2) = BBP1 2 ,120,P (y)$ where $P$ has $52$ non zero coefficients.

6.4.5 A few composite formulae

In the determination of BBP formulae, we have systematicly cancel down the coefficients of $(1) ( 1) L2 4 , L2 -4$ and $(1) I6$ . If we then decide to keep those term, we can obtain among those possible formulae, the following result :
$[a] = (1,1,0,0,0,0,2,0,0,0,0)$ give

1 ( ) p2 3 (1 ) 16-BP P1 26,12,3y11 + 29- 165 +32 = 36-+ 4L2 4

(174)

which can be simplify to

This formula is equivalent to the formula (167). If we apply this idea to Catalan's constant, we obtain with $[a] = (0,0,0,0,0,-1,0,0,1,0,0)$ the equality

Who under serie form gives the two following equality :

The same idea leads, with $[a] = (0,0,0,0,-4,8,0,0,- 12,0,0)$ to

and with $[a] = (4,8,0,0,- 4,4,12,0,0,0,0)$ to

This kind of formula has not been systematically researched.

6.5 Cases of polylogarithms of order 3

The order 3 introduce of course formulae giving $p3$ and other $ln(2)3$ but mostly the famous constant $z(3)$ proved irrational by Apéry in 1978 [14] by developping a continued fraction of a factorial formula which will be mentioned later on.

This constant stays never the less mysterious, and the BBP formula intuitively shows that this constan is very likely not to be very different from a point of vue of the orderning of it's decimals and hence it's complexity.

Kummer's equation for the triologarithm is

and the inverse formula

( ) L 1 = L (z) + p2ln(-z) + 1ln3(-z) 3 z 3 6 6

(181)

A classical formula allows to state that

(2) 7 p2ln(2) ln2(2)- I1 = 8z(3)- 12 + 6

(182)

and by definition

1 ( 1) I2(2)= - -L3 - - 2 2

(183)

6.5.1 Calculation of $(2) I4$

Landen's equation for the trilogarithm is (cf [4] )

Applied to $1+i z = 2 ,$ we get (with $--3z(3)+-ip3 L3(i) = 32$ )

( (1 +i) (1 -i)) 35 5p2ln(2) ln3(2) I(42)= -2 L3 ---- + L3 ---- = --- z(3) + --------- ------ 2 2 16 48 12

(185)

result implicitely contained in [3].
We deduce by the duplication formula that

2 3 ( ) I(32) = 35z(3)- 5p-ln(2)+ ln-(2)- 1L3 -1 16 48 12 8 4

(186)

6.5.2 Calculation of $I(82)$

As for the calculation of $I8(1),$ we use Kummer's equation with $V~ x = 1+2i, y = 1-i2-3$ then with $V~ x = 1-2i, y = 1-i2-3.$ We add then those two equation we obtain. We simplify those equations with the help of the valuer of $L3(i)$ and the equality $( 1+ i V~ 3) z(3) 5ip3 L3 ------- = ----+ ---- 2 3 162$ . This allows us to state that

2 3 I(28)= - 119z(3)+ 5p--ln-(2) - ln--(2)- 48 36 6

(187)

6.5.3 Relation between $I(52)$ and $I(92),$ values of $I(210)$

We take the first two equation for the calculation of $I(81)$ that we substract this time. We then use the inverse formula with $z = 1+ i$ and $z = 1 - i$ so that we make appear the term $(L2 (1+2i ) - L2(1-2i))$ . Finaly one last application of the inverse formula with $z = V~ 3-i 2$ and with $z = V~ 3+i 2$ leads to

Which prove that

3p3 p ln2 (2) 3I9(2)- I(25)= ---- ------- 32 8

(189)

If we use Kummer's equation with $x = -1$ and $y = 1+ i,$ then with $x = - 1$ and $y = 1 -i,$ we obtain two equality that we substract. We simplify the obtained result with the inversion formula applied to $- 1- i,- 1+ i,1+ i$ and $1- i$ so to obtain

This equality is equivalent with the help of integrals to

1 (2) (2) (2) (2) (2) 13 3 -7 2 - 2J6 +I6 = 2J4 - I5 - 2J2 + 64p - 16 pln (2)

(191)

and gives the relation

(2) (2) (2) 23p3 p-ln2-(2) -6I5 + 5I6 + 5I11 = 160 - 8

(192)

If, instead of substracting the equality obtained previously, we add them, we get :

which gives

1 (2) (2) 15 2 5 3 (2) (2) (2) - 2J5 - I3 = 32 p ln(2)- 8 ln (2)- 7z(3) -2J3 - I4 - 2J1

(194)

and furnish

( ) (2) 959 2p2-ln-(2) ln3-(2)- 1 1 I10 = - 400z(3)+ 15 - 6 - 8L3 - 4

(195)

6.5.4 Calculation fo $(2) I7$

Similarly to the calculation of $(1) I7 ,$ Kummer's equation for the polylogarithm of order $3$ with $(1-i V~ 3- 1) (x,y) = --2--,2$ , then the inverse formula with $V~ - z = 1- i 3$ leads immediatly to

6.5.5 Application to the determination of BBP formulae

Let us now consider the linear form $L2 (a1,a2,...,a11) = a1I(21)+ ...+ a11I(211)$ . Then the previous result allows us to state that

Formulae for $p3$ If we look for formulae for $p3$ we cancel down the coefficients of the constants
$z (3),p2ln(2)$ ... to obtain

We can not choose $a = 0, 9$ which mean we have to use a denominator in $y120- 260$ for the BBP formulae giving $p3$ . In all generality, we hence obtain a BBP formula for $p3$ with two parameters ( $a 4$ and $a 10$ ), formula that the reader can establish. The simplest formula is hence obtain for $[a ] = (0,0,0,0,5,-5,0,0,3,0,- 5) i$ and with $90$ termes. So to wrie it, we introduce the polynomials $TTk$ defined by $(0) integral 1 TTk(y) Ik = 0y120-260dy$ , for example $2y(y120-260) TT1 (y) = ---y2- 2---$ .
We then have

This rquality can be written differently.
In fact $L2 (a1,a2,...,a11) = L2 (a1,a2,a3 -a10,a4,a5,a6 -a11,a7,a8,a9,0,0)+ L2(0,0,a10,0,0,-a11,0,0,0,a10,a11)$
But $integral 1 L2(a1,a2,a3- a10,a4,a5,a6- a11,a7,a8,a9,0,0) = 0 lny(y2)4-P1(21y2)$ and
$L2 (0,0,a10,0,0,- a11,0,0,0,a10,a11) = integral 1ln(y4)0P2(2y0) 0 y - 2$ where $P2$ has $8$ non-zero coefficients in general, and only $6$ if $a10 = 0$ or $a11 = 0$ or $a10 + a11 = 0$ .
If we apply this here, we have
$L2 (0,0,0,0,5,- 5,0,0,3,0,-5) = L2(0,0,0,0,0,5,0,0,0,0,-5)+ L2 (0,0,0,0,5,-10,0,0,3,0,0)$ which gives

This formula can also be written as

Formula for $2 pln (2)$ To obtain a simple BBP formula for $2 pln (2)$ , we apply the same idea as that for $3 p$ (the problem is the same, we need to use $I11$ which gives a denominator in $120 60 y - 2$ and gives an expression with two parameters)
We then obtain with $[ai] = (0,0,0,0,67,-75,0,0,69,0,- 75)$

This formula can also be written

Compare with those obtain for $p3$ .

Formulae for $z(3)$ For $z(3),$ we obtain, if we look for a denominator in $24 12 y -2 ,$

Which give the general formula

( ) 91a8 21a4 -1- ( 12 ) 144 + 32 z(3) = 210BBP2 2 ,24,P (y)

(Huvent)

The simplest BBP formula is obtain for
$[a] = (-3,- 1,0,- 2,0,0,1,2,0,0,0)$

We can also use the following :
$L2 (a1,a2,a3,a4,a5,a6,a7,0,0,0,0,) = L2 (0,a7,0,0,0,0,a7,0,0,0,0,)+ L2(a1,a2- a7,a3,a4,a5,a6,0,0,0,0,0,)$
But $integral 1 ln2(u) ( ) L2(0,a7,0,0,0,0,a7,0,0,0,0,) = a367 0 u+8-du = - a187L3 - 18$ and
$L2 (a1 + b1,a2 + b2,a3,a4,a5,a6,0,0,0,0,0,) = integral 1P2(y)8ln2(4y)dy 0 y -2$
If we then impose $a = a 8 10$ in (203), we have $L (7a , 9a ,0,a ,0,0,- 9a ,0,0,0,0) = -21a z(3) 2 2 4 2 4 4 2 4 32 4$
$= L (7a , 18a ,0,a ,0,0,-9a ,0,0,0,0) + L (0,--9a ,0,0,0,0,-9a ,0,0,0,0) 22 4 2 4 4 2 4 2 2 4 2 4$ which gives

Finally the simplest formula for a denominator in $120 60 y - 2$ is obtained with
$[ai] = (8,4,5,9,0,0,- 4,1,0,- 5,0)$ and contains $67$ termes.

Note 5 Similarly here, I thought I was the first to give a formula for $z (3)$ whose proof, a lot less simple but on the same model, was finished in june 200 with the help of Raymond Manzoni [5].

Formulae for $p2ln(2)$ We obtain the following general formula

( ) -3 a4 +-13a8 p2ln(2) =---1-- BP P2(212,24,P (y)) 80 360 5 × 210

(Huvent)

With $[ai] = (-8,- 3,0,- 5,0,0,3,5,0,0,0)$ , we have the formula with $15$ termes

This relation is noteworthy, in fact the equality (204) is obtain for $[ai] = (- 3,-1,0,-2,0,0,1,2,0,0,0)$ .
This leads to fix

Then

Finaly the simplest formula with a denominator in $120 60 y - 2$ is obtain with
$[ai] = (42,21,25,46,0,0,- 21,4,0,-25,0)$ and has $67$ termes.

Formulae for $ln3 (2)$ We have the general formula

(27a + 26a )ln3(2) =-3BP P (212,24,P (y)) 4 8 28 2

(Huvent)

and the simplest formula is ontained for
$[ai] = (- 22,-9,0,-12,0,0,9,12,0,0,0)$

And the one with a denominator in $120 60 y - 2$ is given by
$[ai] = (- 202,-99,-100,- 200,0,0,99,0,0,100,0)$

6.6 Cases of polylogarithm of order 4

6.6.1 The relations

For the polylogarithm of order $4$ and $5,$ we only have one tool left, this is Kummer's equation. It is written, with $e = 1- x,$ $n = 1 -y,$

and the inverse formula

( ) 2 4 L4(z) = - L4 1 - p--ln2(-z)- 7p--- -1 ln4(-z) z 12 360 24

(214)

We then so apply the following method :
- Using Kummer's formula with a particualr couple $(x,y)$
- Eliminating the polylogarith with argument modulus greater than $1$ with the inverse formula
- Eventual simplification with the help of the duplication formula or some value of $L4$ in $1,-1,i$ and $-i$ .

With $(x,y) = (- 1,1+ i),$ we get

Which with the integrals and $(3) Ik$ and $(3) Jk$ becomes
With $(x,y) = (- 1,i),$ we get

We obtain a second equality with conjugating (or with $(x,y) = (-1,- i)$ ). We add, or substract the two equality obtained and by replacing $L4 (i)$ by $4 -7p--+ ib(4) 11520$ or $sum k b (4) = -(-1)-4- k>0 (2k + 1)$ , this gives

and
With $(1-i V~ 3 1) (x,y) = --2--,2 ,$ we obtain

(221)

We then use the multiplication formula

$( ( ) ( ) ) (3) -1 -i V~ 3 -1 + i V~ 3 L4 z = 27 L4 ---2----z + L4 ---2----z + L4(z)$ (222)

which with $z = 1$ allow us to confirm that

$( V~ -) ( V~ -) 4 L4 -1-+i--3 + L4 -1---i-3 = - 13p- 2 2 1215$ (223)

and that

$( ) ( ) 4 12I(3)- 5L4 1 - 9L4 1 - p--- 5-ln4(2)+ 1-p2ln2(2) = 0 7 2 8 4 54 24 12$ (224)
Finaly, with $( V~ ) (x,y) = 1-i2-3, 1+2i$ , we obtain

(225)

By considering the real part and the imaginary part of this expression, we have respectively

6.6.2 Application of the determination of the BBP formula

We now consider the linear form $(3) (3) L3(a1,a2,...,a11) = a1I1 + ...+ a11I11$ . With the previous formula giving

6.6.3 Formula for $p4, p2ln2 (2)$ and $ln4(2)$

For $p4.$ The simplest formula (and the only associated with a denominator in $y24- 212$ ) is obtained for $[ai] = (37,9,0,26,0,0,-9,- 27,0,0,0)$ and gives

There exist a formula with two parameters with a denominator in $120 60 y - 2$ , the simplest is given by $[ai] = (34,18,25,41,0,0,- 18,18,0,-25,0)$

For $2 2 p ln (2)$ The simplest formula is obtained with
$[ai] = (- 10381,-3303,0,- 6836,0,0,3303,6957,0,0,0)$ and gives

For $4 ln (2)$ Wiht $[ai] = (-18932,- 6849,0,- 11176,0,0,6849,11322,0,0,0),$ we get

The case of $3 b(4),pln (2)$ and $3 p ln (2)$ It is not possible to determine formulae for those constant, we can only find two independant relation which are :

This shows that we only need to find a $BP P$ formula for one of those three constant and then we can deduce one for the two others.

6.7 Case of polylogarithm of order 5

6.7.1 The relations

Broadhurst in [3] shows relations between the integrals $I k$ and $J k$ with the help of Kummer's equation (relations $(65)$ to $(67)$ . Then discover two equality through numerical mean (relations $(68)$ and $(69)$ ). He then prove the relation $(68)$ by using the hypergeometric series and Euler's sums. He deduce from it four equalities for $z(5)$ (relation $(70)$ )
We can establish those four equalities directly with the help of Kummer's formula which is written with $n = 1- x, e = 1 - y, a = - x, b = - y n e$

We apply then the same method as that for polylogarithm of order 4.

With $(1 1-i) (x,y) = 2, 2 ,$ we obtain, taking into account
$5 L5(i) = - 15-z(5)+ 5ip-, 512 1236$

wich gives us

Which can be simplified to, by using $(4) (4) 3- ( 1) I3 + I4 = - 32L5 -4$
With $(x,y) = (- 1,1+ i)$ we get

which is equal to
With $( V~ -) (x,y) = 1+2i, 1-i2-3$ , we get

We simplify this equality with the help of $( V~ -) L5 1+i-3 = 25z (5) + 17-ip5 2 54 5832$ and the multiplication formula $( ) ( V~ ) ( V~ ) L5 z3 = 81L5 --1+i2-3z + 81L5 --1-2i3z + 81L5(z)$ , which with $z = i$ allows us to have $( V~ ) ( V~ ) L5 -3-2i + L5 ---32-i = 82654z(5))- 62205208ip5$ .
Hence
With $( ) (x,y) = -1, 1-i V~ 3 , 2$ we obtain

which gives us

Note 6 The relation proved by Broadhurst ([3]) correspond to the calculation of $(4) I4 .$ The calculation given here seems a lot more simple.

6.7.2 Application of the determination of the BBP formulae

We now consider the linear form $(4) (4) L4(a1,a2,...,a11) = a1I1 + ...+ a11I11$ . The previous results gives

L3(a1,a2,...,a11) =
(4) (4) (4) (4)
+ a5I5 + a6I6 + a9I9 + a11I11 +
( 56575 6417 6417 3931389 403 )
- 2304 a8 - 256 a4 + 256 a3- 160000 a10- 864 a7 z(5)+
( 2237 1 343 861 343 )
-----a8 + ---a7- ----a3 + -----a10 +----a4 p4ln(2)+
2(5920 324 3840 10000 3840)
- 5-a4 - -1-a7 + -5 a3- -7- a8- -19 a10 p2ln3(2)+
96 216 96 144 400
(-11 -87- -1- ) 5
240 a8 + 1/20a4 - 1/20a3 + 2000 a10 + 144 a7 ln (2)
3 ( 1) 3 ( 1)
- - (a2 + a7)L5 - - - --(a10 + a3)L5 - -
2 2 32( 4 ) ( )
+ 15 a3- 15 a4- 411 a10- 17 a8- 3 a1- 7 a7 L5 1
2 2 50 2 2 3 2

(266)

6.7.3 Formulae for $z(5),p4 ln(2),p2ln3(2)$ and $ln5(2)$

We deduce from this some formulae for the constants $z(5),p4ln(2),p2ln3(2)$ and $ln5(2)$ . We remember that the polynomials are defined by $Ik(0)= integral 1-TT12k0(y)60dy 0 y -2$ .
We hence obtain
with $[ai] = (311416,168912,239375,382643,0,0,- 168912,96489,0,-239375,0)$

with $[ai] = (22046536,11642616,15741875,26174255,0,0,-11642616,5323725,$

$0,-15741875,0)$

with $[ai] = (32556,17892,25625,40413,0,0,-17892,10899,0,-25625,0)$

with $[ai] = (12347068,6435126,8249375,14111819,0,0,- 6435126,2392497,$

$0,-8249375,0)$

6.7.4 Simplification of those formulae

Those formula can be simplified by making the term $I3(4)+ I(14)0$ appear. For example $p4ln (2) = 311416I(14)+ 168912I2(4)+ 2× 239375I(34)+ 382643I(44)$