7 s integer, v = p q, x = 1 _ 3n : BBP in base 3

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7 $s$ fixed integers, $v = pq$ , $x = 31n$ : BBP in base 3

So many results already ! But it's not over yet... in fact, even if we can not BBP formula for all bases, an important class of BBP formula was obtained in base 3, which correspond to taking $x = 13n$ . Here again, Gery Huvent explained the best the relations in [12].

7.1 Formulae for $p V~ 3$

We take the same kind of integrals than that of base 2, but this time it's the 3 that will be omnipresent in the polynomials.

An elementary calculation shows that

By cancelling the coefficients of $ln(2)$ and $ln(3),$ we obtain a formula for $V~ p 3$ :

The following particular case is the most elegant $( ) b = 0,d = -e = 13$ :

V~ - 2- oo sum -1-(--34-- --34-- --33-- ---3-- --3--- ---1---) p 3 = - 33 36i 12i+ 2 + 12i+ 3 + 12i+ 4- 12i+ 8 - 12i+ 9- 12i+ 10 i=0

(Huvent)

7.2 Formulae of order : integrales with $ln(y)$

There exist reals $(p,q,r)$ such that

( the two members are linear form in $(a,b,c)$ decompose on two different base of the dual of $R3$ )
Plouffe showed that

which under integral form, give

integral integral integral 1 ln(y)y- 1 ln(y)y- 1ln(y)(2y--3)- p2- -2 0 2y2- 3dy + 0 2 y2 + 3dy + 0 y2 - 3y+ 3 dy = 18

(271)

and correspond to the equality $p2- I1(-2,1,1) = r = 18$ . We can deduce the value of $r$ .

But we have

where $sum oo k L2(z) = z2 k=1k$
We deduce from it

( ) 3p- 2q = 1 L2 1 4 9

(273)

But Ramanujan proved that

( ) ( ) 2 2 L2 1 - 1L2 1 = p-- ln-(3)- 3 6 9 18 6

(274)

By using

We deduce from it

(8 4 ) (1 ) 1 (1 ) I1 -,- -,0 = L2 - - -L2 - 6 6 3 6 9

(276)

and

2 2 -2q = p--- ln--(3)- 3 18 6

(277)

From (273) and (277) we get the values of $p$ and $q$ and the equality

( ) p2- ln2(3)- L2-19-- I1(a,b,c) = (a- b+ 5c)36 + (- a+ b- 3c) 12 + (a+ 2b) 12

(278)

Note 7 $( 4 2 ) 1 1 (1 ) 1 1 2 I1 - 3,3,0 = L2(- 3)- 3L2 9 = - 18p2 + 6 ln 3$ is another relation proved by Ramanujan .

Note 8

o`u P (y) = 4y11 - 9y10 + 81y9 + 117y7 + 81y6 + 972y5 + 243y4 + 1053y3 + 6561y- 2187

We deduce

Note 9 $( ) I (a,b,0) = (a- b) p2- ln2(3) + (a+ 2b)L2(19)= a integral 12 ln(y)ydy + b integral 12 ln(y)ydy 1 36 12 12 0 y2-3 0 y2+3$ , with $a = -2$ and $b = 1$ and changing variables $2 u = y$ , we get

integral 1 ln-(u)(u-+-9) 1 2 1 2 0 u2 - 9 du = 6 p - 2 ln (3)

(281)

and the equality

sum oo 1 oo sum 1 2 2 2 9ii2+ -i(---1)2 = 3p - 2 ln (3) i=1 i=0 9 i+ 2

(282)

Note 10

oo oo sum 1-+ sum --(-1--)=2 ln 3 i=1 9ii i=09i i+ 12

(Huvent)

7.3 Formulae of order 3 : Integrales with ln $2 (y)$

Here again, the order 3 introduce some formulae giving $3 p$ and other $3 ln(2)$ but also $z(3)$ .

There exist reals $(p2,q2,r2)$ such that

The calculation of $r2$ is elementary because

Proposition 11 We have

13 (1 ) p2ln(3) ( 1 ) ln3(3) -- z(3)- 6L3 - - -------+ 3L3 -- + ----- = 0 2 3 2 3 2

(285)

This relation is equivalent to

Proof. Kummer's equation for the polylogarithm of order $3$ is written

With $x = 2 3$ and $y = 1, 3$ we obtain

Landen's equation is

( ) -z--- p2- 1 2 1 3 L3(z)+ L3(1 - z) + L3 z- 1 = z(3)+ 6 ln (1- z)- 2 ln(z)ln (1- z)+ 6 ln (1 - z)

(290)

Applied to $z = 3 4$ and to $z = 2 3$ , it allows to express $L3(3) 4$ and $L3 (2) 3$ in functions of $L3 (1), L3(-2), L3(- 1) 3 3$ and $L3(1) 4$ . The relation

(1-) p2- ln3(-z) L3 z = L3(z)+ ln(-z) 6 + 6

(291)

applied to $3, 2,- 2 4 3$ and $- 3$ allows to simplify the left member of (289).
Then all there is left to do is use the two following equalities

to conclude. _

With $integral 1 ln2(y)y 1 (1) 02 y2-3 dy = - 2L3 3$ and $integral 1 ln2(y)y 1 ( 1) 02 y2+3 dy = -2L3 -3 ,$ (285) can be written

If we replace $p2$ by it's value compared to $q2,$ the right hand side of $I2(a,b,c)$ becomes a polynomial of degree $1$ in $q2$ whose coefficient of $q2$ is equal to $3 ln-(63)c$ . Hence $I2(a,b,0)$ does not depend on $q2$ . After calculation, we get

integral 1 2 integral 1 2 ( ) ( 2 ) a 2ln2(y)ydy+ b 2ln2-(y)ydy = - (a+-2b)L3 1 + (--a+-b) 13z(3)+ ln3(3) - p-ln(3)- 0 y - 3 0 y + 3 12 9 36 36

(294)

In particular, we have with $a = 1,b = - 1$

integral 1 2 3 (1) 2 12 ----yln-(y)----dy = - ln-(3) - 13z(3)+ L3-9-+ p-ln(3) 0 (y2- 3)(y2 + 3) 18 18 24 18

(295)

which gives the equality

oo oo ( 2 ) sum -1-+ sum ----1---- = 4-ln-(3)--p2-ln(3)+ 52z(3) i=1 9ii3 i=09i(i+ 1)3 3 3 2

(Huvent)

Note 12 We can also take $a = -2,$ $b = 1$ and to a change of variables $u = y2$ which leads to

integral 1 2 ln-(u)2(u+-9)du = -1 ln3 3+ 1p2 ln 3- 13z (3) 0 u - 9 3 3 3

(296)

and to the same equality as for the series.

We just need to calculate the exact value of $p2$ and $q2$ .

Proposition 13 We have the equality

integral 1ln2(y)(2y--3)- 26 5p2-ln-(3) ln3-(3)- 0 y2- 3 y+ 3 dy = - 9 z(3)+ 36 - 12

(297)

Proof. Landen's equation applied to $z =-1 z1$ where $V~ - z1 = 3 + i-3 2 2$ is a root of $y2 - 3y+ 3$ (the other root is $z2$ such that $-1 = 1- -1 z2 z1$ ) give