Computation of a Feynman integral

Computation of a F eynman in tegral and some iden tities of Clausen function v alues. Gert Almkvist In tro duction. F or the 3 - lo op tetr a hedral F eynman diag ram with non- adjacent lines carr y- ing masses a and b o ne g ets the integral ( see Broadhurst [1] ) C ( a, b ) = − 16 b 2+ b Z 2 dw w ( w + a ) √ w 2 + b 2 − 4 arctan h w 2 − 4 − 2 b w √ w 2 + b 2 − 4 − 16 b ∞ Z 2+ b dw w ( w + a ) √ w 2 + b 2 − 4 arctan h b √ w 2 + b 2 − 4 Broadhurst also made the conjectur e C (1 , 1) = 4 √ 2( C l 2 (4 α ) − C l 2 (2 α )) where sin( α ) = 1 3 and C l 2 ( x ) = − x Z 0 log(2    sin( ϕ 2 )    ) dϕ = ∞ X n =1 sin( nx ) n 2 is the Claus e n function. I computed the tw o integrals ( for a = b = 1 )by hand and found using the LLL-a lgorithm the following identities Conjecture 1. Let tan( α ) = 1 √ 2 and tan( β ) = √ 8 + √ 3 Then 1.1. C l 2 ( α ) + C l 2 ( π − α ) + C l 2 ( π 3 − α ) − C l 2 ( 2 π 3 − α ) = 7 4 C l 2 ( 2 π 3 ) 1.2. C l 2 (6 α − π ) + C l 2 ( π + 2 α ) − 2 C l 2 (2 α ) + 2 C l 2 ( π − 4 α ) = 0 1.3. C l 2 ( π − 2 β ) + C l 2 (2 β − 4 α ) + C l 2 (2 β − 2 α ) − C l 2 (2 β + 2 α − π ) − C l 2 (2 α ) − 2 C l 2 ( π − 4 α ) − 2 C l 2 ( π + 2 α ) = 0 1.4. − 12 C l 2 (2 β − 2 α ) + 4 C l 2 ( π − 4 α ) − 12 C l 2 ( π − 2 β ) − 18 C l 2 ( π + 2 α ) + 7 C l 2 (4 α ) = 0 The fir st three identities give Br oadhurst’s v alue of C (1 , 1) . But I co uld no t prov e any of the identities. In July 19 9 8 I met Br oadhurst in V ancouv er a nd he gav e me [2] where he finds the v alue of C ( a, b ) for general a and b a nd a lso proves it is co rrect. But I wanted to prov e 1.1 -1.4 and decided to co mpute C ( a, b ) for a 2 + b 2 < 4 using elementary metho ds. The result con tained 32 differe nt Clausen v alue s . Putting a = 1 π and b = 1 e and us ing the LLL-algo rithm I found tw o genera l relations that sp ecializ e d to 1.1 and 1.3 when a = b = 1 . These tw o identities were easily prov ed by differentiation. 1 .2 follows fr om an identit y in [2] but 1.4 seems very difficult to pr ov e (it is not a consequenc e of any general identit y I found) 1 2. Computation of the i n tegrals. W e star t with t wo Lemmata. Lemma 1. If a 2 + b 2 ≥ c 2 and δ 1 = 2 arctan( a + c b + √ a 2 + b 2 − c 2 ) δ 2 = 2 ar ctan( a + c − b + √ a 2 + b 2 − c 2 ) then β Z α log( a cos( ϕ ) + b sin( ϕ ) + c ) dϕ = ( β − α ) log( √ a 2 + b 2 2 ) + C l 2 ( δ 2 − β ) − C l 2 ( δ 2 − α ) + C l 2 ( δ 1 + α ) − C l 2 ( δ 1 + β ) Pro of: W e hav e a cos( ϕ ) + b s in( ϕ ) + c = 2 p a 2 + b 2 sin( δ 2 − ϕ 2 ) s in( δ 1 + ϕ 2 ) T aking logar ithms a nd in tegrating we a r e done. Lemma 2: β Z α log(tan( ϕ ) − tan( δ )) dϕ = 1 2 { C l 2 (2 α − 2 δ ) − C l 2 (2 β − 2 δ ) + C l 2 ( π − 2 α ) − C l 2 ( π − 2 β ) }− ( β − α ) log(cos( δ )) Pro of: W e hav e log(tan( ϕ ) − tan( δ )) = log(2 sin( ϕ − δ )) − log(2 c o s( ϕ )) − lo g(cos( δ )) Then use β Z α log(2 cos( ϕ )) dϕ = 1 2 { C l 2 ( π − 2 β ) − C l 2 ( π − 2 α ) } Using the par tial fra ction 1 w ( w + a ) = 1 a ( 1 w − 1 w + a ) we get tw o integrals no t containing a I 1 = ∞ Z 2+ b dw w √ w 2 + b 2 − 4 arctan h ( b √ w 2 + b 2 − 4 ) 2 I 2 = 2+ b Z 2 dw w √ w 2 + b 2 − 4 arctan h ( w 2 − 2 (2 + b ) w √ w 2 + b 2 − 4 ) Put c = √ 4 − b 2 and make the substitution w = c sin( ϕ ) and use the nota tio n sin( α 1 ) = r 2 − b 2 + b tan( α 2 ) = c b Then I 1 = 1 2 c α 1 Z 0 log( c cos( ϕ ) + b s in( ϕ ) c cos( ϕ ) − b s in( ϕ ) ) dϕ = 1 4 c {− C l 2 (2 α 1 + 2 α 2 ) + C l 2 (2 α 1 − 2 α 2 ) + 2 C l 2 (2 α 2 ) } = 1 4 c {− q 1 + q 2 + 2 q 3 } Similarly using Lemma 1 a nd 2 I 2 = 1 2 c α 2 Z α 1 ( log( cos( ϕ ) − b 2 cos( ϕ ) + b 2 ) − 2 log(tan( ϕ 2 )) ) = 1 2 c  − C l 2 ( α 2 − α 1 ) + C l 2 ( α 2 + α 1 ) − C l 2 (2 α 2 ) − C l 2 ( π − 2 α 2 ) + C l 2 ( π − α 1 − α 2 ) − C l 2 ( π + α 1 − α 2 ) + 2 C l 2 ( α 2 ) − 2 C l 2 ( α 1 ) + 2 C l 2 ( π − α 2 ) − 2 C l 2 ( π − α 1 )  = 1 2 c {− q 4 + q 5 − q 6 − q 7 + q 8 − q 9 + 2 q 10 − 2 q 11 + 2 q 12 − 2 q 13 } W e will s how that I 1 + I 2 = 0 Using the duplication for mula C l 2 (2 x ) = 2 C l 2 ( x ) − 2 C l 2 ( π − x ) we find q 1 = 2 q 5 − 2 q 8 q 2 = 2 q 9 − 2 q 4 and also q 3 = q 6 3 It follows that I 1 + I 2 = 0 is equiv alen t to 2 q 4 + q 7 − 2 q 8 − 2 q 10 + 2 q 11 − 2 q 12 + 2 q 13 = 0 which follows from the following result ( α = α 1 , β = α 2 ) Theorem 1: Assume sin( α ) = tan( β 2 ) Then C l 2 ( π − 2 β ) − 2 C l 2 ( β ) − 2 C l 2 ( π − β )+2 C l 2 ( α )+2 C l 2 ( π − α )+2 C l 2 ( β − α ) − 2 C l 2 ( π − α − β ) = 0 Pro of: Since C l 2 ( π ) = 0 we find that the iden tit y is true when α = β = 0 . Let t an( α 2 ) = t and co nsider α a nd β as functions of t. W e hav e d dα C l 2 ( α ) = − log(2 sin( α 2 )) d dα C l 2 ( π − α ) = log(cos( α 2 )) Let LH S = f ( t ). Then − 2 d f dt =  − 2 log(2 cos( β )) − 2 log(2 sin( β 2 )) + 2 log(2 cos( β 2 ))  β ′ + n 2 log(2 s in( α 2 )) − 2 log(2 cos( α 2 )) o α ′ + 2 log(2 sin( β − α 2 ))( β ′ − α ′ ) +2 log(2 cos( β + α 2 ))( β ′ + α ′ ) = log( 2 cos( β 2 ) s in( β − α 2 ) c o s( β + α 2 ) sin( β 2 ) c o s( β ) ) β ′ + lo g( sin( α 2 ) c o s( β + α 2 ) cos( α 2 ) s in( β − α 2 ) ) α ′ Using tan( α 2 ) = t w e obtain cos( α ) = 1 − t 2 1 + t 2 sin( α ) = tan( β 2 ) = 2 t 1 + t 2 cos( β ) = (1 − t 2 ) 2 1 + 6 t 2 + t 4 sin( β ) = 4 t (1 + t 2 ) 1 + 6 t 2 + t 4 Then 2 cos( β 2 ) s in( β − α 2 ) c o s( β + α 2 ) sin( β 2 ) c o s( β ) = sin( β ) − sin( α ) tan( β 2 ) c o s( β ) = 4 t (1+ t 2 ) 1+6 t 2 + t 4 − 2 t 1+ t 2 2 t 1+ t 2 · (1 − t 2 ) 2 1+6 t 2 + t 4 = 1 4 Similarly sin( α 2 ) c o s( β + α 2 ) cos( α 2 ) s in( β − α 2 ) = 1 Hence d f dt = 0 and since f (0) = 0 we hav e f ( t ) ≡ 0 Corollarium (Conjecture 1.1): Assume sin( α ) = 1 √ 3 . Then C l 2 ( α ) + C l 2 ( π − α ) + C l 2 ( π 3 − α ) − C l 2 ( 2 π 3 − α ) = 7 4 C l 2 ( 2 π 3 ) Pro of: W e get tan( β 2 ) = sin( α ) = 1 √ 3 and hence β = π 3 . Using C l 2 ( π 3 ) = 3 2 C l 2 ( 2 π 3 ) and the Theo rem we get the wan ted res ult. Since I 1 + I 2 = 0 we get C ( a, b ) = 16 ab { I 3 + I 4 } where I 3 = ∞ Z 2+ b dw ( w + a ) √ w 2 − c 2 arctan h ( b √ w 2 − c 2 ) I 4 = 2+ b Z 2 dw ( w + a ) √ w 2 − c 2 arctan h ( w 2 − 4 − 2 b w √ w 2 − c 2 ) In I 3 we substitute w + a = 1 u Let p = a + b + 2 d = p 4 − a 2 − b 2 Then we get I 3 = 1 /p Z 0 du √ 1 − 2 a u − d 2 u 2 arctan h ( bu √ 1 − 2 a u + d 2 u 2 ) = 1 2 d α 4 Z α 3 log( dc cos( ϕ ) + bc sin( ϕ ) − ab dc cos( ϕ ) − bc sin( ϕ ) + ab ) dϕ 5 after the substitution u = c sin( ϕ ) − a d 2 where sin( α 3 ) = a c sin( α 4 ) = a c + d 2 cp Let δ 1 = 2 arctan( cd − ab 2 d + bc ) δ 2 = 2 arctan( cd − ab 2 d − bc ) δ 3 = 2 arctan( cd + ab 2 d − bc ) δ 4 = 2 arctan( cd + ab 2 d + bc ) Using Lemma 2 we obtain I 3 = 1 2 d  C l 2 ( δ 2 − α 4 ) − C l 2 ( δ 2 − α 3 ) + C l 2 ( δ 1 + α 3 ) − C l 2 ( δ 1 + α 4 ) − C l 2 ( δ 4 − α 4 ) + C l 2 ( δ 4 − α 3 ) − C l 2 ( δ 3 + α 3 ) + C l 2 ( δ 3 + α 4 )  = 1 2 d { r 1 − r 2 + r 3 − r 4 − r 5 + r 6 − r 7 + r 8 } Finally we substitute cu = w + p w 2 − c 2 to get I 4 = 1 c u 2 Z u 1 du u 2 + 2 a c u + 1 log( u 2 − f 2 f 2 u 2 − 1 ) where f 2 = 2 + b 2 − b u 1 = f u 2 = f + r 2 b 2 − b Then put u = d ta n( ϕ ) − a c to get I 4 = 1 d α 7 Z α 6 log( d 2 (tan( ϕ ) − a d ) 2 (tan( ϕ ) − a + b +2 d )(tan( ϕ ) − a − b − 2 d ) c 2 f 2 (tan( ϕ ) − a − b +2 d )(tan( ϕ ) − a + b − 2 d ) ) dϕ 6 where tan( α 6 ) = p d tan( α 7 ) = p + √ 2 b 2 + 4 b d Define δ 7 = arctan( a d ) δ 8 = arcta n( a + b + 2 d ) = α 6 δ 9 = arctan( a − b − 2 d ) δ 10 = arcta n( a − b + 2 d ) δ 11 = arcta n( a + b − 2 d ) Using Lemma 2 we get (all logarithms cancel) I 4 = 1 2 d    2 C l 2 (2 α 6 − 2 δ 7 ) − 2 C l 2 (2 α 7 − 2 δ 7 ) − C l 2 (2 α 7 − 2 δ 8 ) + C l 2 (2 α 6 − 2 δ 9 ) − C l 2 (2 α 7 − 2 δ 9 ) − C l 2 (2 α 6 − 2 δ 10 ) + C l 2 (2 α 6 − 2 δ 11 ) + C l 2 (2 α 7 − 2 α 11 ) +2 C l 2 ( π − 2 α 6 ) − 2 C l 2 ( π − 2 α 7 )    = 1 2 d { 2 r 9 − 2 r 10 − r 11 + r 12 − r 13 − r 14 + r 15 − r 16 + r 17 + 2 r 18 − 2 r 19 } Using LLL on { r 1 , r 2 , ..., r 19 } with the sp ecial v alues a = 1 π and b = 1 e we found r 2 = r 9 r 5 = r 11 r 4 = − r 13 r 1 = r 15 r 8 = − r 17 r 6 = r 18 These identities are v a lid for all a a nd b such that a 2 + b 2 < 4 . They are not trivial. E.g. r 5 = r 11 says that 1 2 ( δ 4 − α 4 ) = α 7 − δ 8 which is equiv a lent to (after taking tan o n b o th sides) du p 2 + d 2 + p u = ab + cd 2 d + bc − pc − du d 2 + ap 1 + ( ab + cd )( pc − du ) (2 d + bc )( d 2 + ap ) 7 where c = p 4 − b 2 d = p 4 − a 2 − b 2 p = a + b + 2 u = p 2 b 2 + 4 b Maple v erifies this identit y and the o ther relations a s well. Using these iden tities we r educe I 3 + I 4 to eleven Clause n v a lues. In or der to simplify this a nd to get the same a ns wer a s Broa dhurst go t in [2 ] we hav e to int ro duce some mor e notation. Let φ = arctan( d p ) φ a = arctan( d a ) φ b = arctan( d b ) s 1 = C l 2 (4 φ ) s 2 = C l 2 (2 φ a + 2 φ b − 2 φ ) s 3 = C l 2 (2 φ a − 2 φ ) s 4 = C l 2 (2 φ b − 2 φ ) s 5 = C l 2 (2 φ a + 2 φ b − 4 φ ) s 6 = C l 2 (2 φ a ) s 7 = C l 2 (2 φ b ) s 8 = C l 2 (2 φ ) Then Broa dhurst’s result is 2 d ( I 3 + I 4 ) = s 1 + s 2 + s 3 + s 4 − s 5 − s 6 − s 7 − s 8 Comparing with o ur v a lue for I 3 + I 4 this is equiv alent to the follo wing identities found by LLL − 2 r 10 − 2 r 11 − r 14 + 2 r 15 − 2 r 19 − s 1 + s 6 + 4 s 8 = 0 r 3 = s 4 r 7 = − s 2 r 9 = s 3 r 12 = − s 7 r 16 = s 5 r 18 = s 8 8 The la st six of thes e follow from α 3 = π 2 − φ a α 6 = π 2 − φ δ 1 = − 2 φ + φ a + 2 φ b − π 2 δ 3 = 2 φ − φ a − 2 φ b + 3 π 2 δ 7 = π 2 − φ a δ 9 = − φ + φ b − π 2 δ 11 = π 2 + φ − φ a − φ b which are easily verified by Ma ple. It remains to pr ov e the first ide ntit y . It follows from the following r esult. Prop ositi on 1: Let d = p 4 − a 2 − b 2 p = a + b + 2 γ = ar ctan( p + √ 2 b 2 + 4 b d ) φ = arctan( d p ) φ a = arctan( d a ) Then 2 C l 2 (2 γ + 2 φ a − π ) + 2 C l 2 (2 γ + 2 φ − π ) + C l 2 (2 φ a − 4 φ ) − 2 C l 2 (2 γ − 2 φ + 2 φ a − π ) +2 C l 2 ( π − 2 γ ) + C l 2 (4 φ ) − C l 2 (2 φ a ) − 4 C l 2 (2 φ ) = 0 Pro of: co nsider the left hand side a s a function f ( a, b ) . Differentiating we get − 2 d f = 4 log (2 sin( γ + φ a − π 2 ))( dγ + dφ a ) + 4 log(2 sin( γ + φ − π 2 ))( dγ + dφ ) 2 log(2 s in( φ a − 2 φ ))( dφ a − 2 dφ ) − 4 log(2 s in( γ − φ + φ a − π 2 ))( dγ − dφ + dφ a ) − 4 log(2 sin( π 2 − γ )) dγ +4 lo g(2 sin(2 φ )) dφ − 2 log(2 sin( φ a )) dφ a − 8 log(2 sin( φ )) dφ = 4 log( sin( γ + φ a − π 2 ) s in( γ + φ − π 2 ) sin 2 ( γ − φ + φ a − π 2 ) s in( φ a ) ) dγ 9 +2 log( sin 2 ( γ + φ a − π 2 ) s in( φ a − 2 φ ) sin 2 ( γ − φ + φ a − π 2 ) s in( φ a ) ) dφ a +4 log( sin( γ + φ − π 2 ) s in( γ − φ + φ a − π 2 ) s in(2 φ ) sin( φ a − 2 φ ) s in 2 ( φ ) ) dφ = 4 log( T 1 ) dγ + 2 log( T 2 ) dφ a + 4 log( T 3 ) dφ = 0 since T 1 = T 2 = T 3 = 1 . W e show that T 3 = 1 which is equiv alent to cos( γ + φ ) cos( γ + φ a − φ ) sin (2 φ ) = sin( φ a − 2 φ ) s in 2 ( φ ) if and only if 2(1 − ta n( γ ) tan( φ ))(1 − tan( γ ) ta n( φ a ) + ta n( φ ) tan( γ ) + tan( φ ) tan( φ a )) = tan( φ )(1 + tan 2 ( γ ))(tan ( φ a )(1 − tan 2 ( φ )) − 2 tan( φ )) if and only if 2 p p 2 b 2 + 4 b ( p 2 − d 2 − 2 ap + ( b + 2) p 2 b 2 + 4 b ) = ( p 2 − d 2 − 2 ap )( d 2 + p 2 + 2 b 2 + 4 b + 2 p p 2 b 2 + 4 b ) which is easily verified. Similarly o ne verifies that T 1 = T 2 = 1. Hence d f = 0 which means that f ( a, b ) is co nstant. Putting a = b = 0 we get d = 2 , p = 2 , γ = φ = π 4 , φ a = π 2 which g ives f (0 , 0) = 0 . It follows that f ( a, b ) ≡ 0 and the pro of is finished. Corollary: W e have if a 2 + b 2 < 4 C ( a, b ) = 8 ab √ 4 − a 2 − b 2  C l 2 (4 φ ) + C l 2 (2 φ a + 2 φ b − 2 φ ) + C l 2 (2 φ a − 2 φ ) + C l 2 (2 φ b − 2 φ ) − C l 2 (2 φ a + 2 φ b − 4 φ ) − C l 2 (2 φ a ) − C l 2 (2 φ b ) − C l 2 (2 φ )  Prop ositi on 2: W e hav e 2 C l 2 (2 φ ) − 4 C l 2 (2 φ b ) + C l 2 (4 φ b ) + 2 C l 2 (2 φ b − 2 φ ) − 2 C l 2 (2 φ a − 2 φ ) + C l 2 (2 φ a − 4 φ ) + 2 C l 2 (2 φ a + 2 φ b − 2 φ ) − C l 2 (2 φ a + 4 φ b − 4 φ ) = 0 Pro of: Let f ( a, b ) b e the LHS. Differentiating we get d f = 4 log( sin( φ ) sin( φ a − φ ) sin( φ a + 2 φ b − 2 φ ) sin( φ b − φ ) sin( φ a − 2 φ ) s in( φ a + φ b − φ ) ) dφ +2 log( sin( φ a − 2 φ ) s in 2 ( φ a + φ b − φ ) sin 2 ( φ b − φ ) sin( φ a + 2 φ b − 2 φ ) ) dφ a +4 log( sin(2 φ b ) s in( φ b − φ ) sin( φ a + φ b − φ ) sin 2 ( φ b ) s in( φ a + 2 φ b − 2 φ ) ) dφ b = 0 since all lo garithms are zero (chec ked by Maple). Since f (0 , 0) = 0 we hav e f ( a, b ) ≡ 0 . 10 Corollarium 1: Identit y 1.2 is true. Pro of: Put a = b = 1 in Pr op osition 2. Let α = a rctan( 1 √ 2 ). Then φ = π 2 − 2 α and φ a = φ b = π 2 − α . Ins erting this into Prop ositio n 2 we obtain 2 C L 2 ( π − 4 α ) − 4 C l 2 ( π − 2 α ) − C l 2 (4 α ) + C l 2 (6 α + π ) − C l 2 ( π + 2 α ) = 0 Using the duplication for mula C l 2 (4 α ) = 2 C l 2 (2 α ) − 2 C l 2 ( π − 2 α ) we get 1.2 Corollarium 2. Identit y 1.3 is true Pro of: With a = b = 1 in P rop osition 1 we have tan( γ ) = √ 8 + √ 3 . Hence γ = β in 1 .3. Prop osition 1 gives 2 C l 2 (2 β − 2 α ) + 2 C l 2 (2 β − 4 α ) + C l 2 (6 α − π ) − 2 C l 2 (2 β + 2 α − π ) + 2 C l 2 ( π − 2 β ) + C l 2 (2 π − 8 α ) − C l 2 ( π − 2 α ) − 4 C l 2 ( π − 4 α ) = 0 Using 1.2 for C l 2 (6 α − π ) and the duplica tio n formula for C l 2 (8 α ) w e o btain 1.3 Ac knowledgemen ts. The b eginning of this work was done in 1 998 at CECM, Simon F razer University . I wan t to thank Jo nathan and Peter Borwein for many stimulating discussions. The fina l part of the pap er could not have bee n written without B roadhurst’s result in [2]- References: 1. D.Bro a dhurst, A diloga rithmic 3-dimensio nal Ising tetrahedr on, hep-th/98 0502 5 2. D.Broadhurst, Solving differen tial e q uations for 3-loop diagrams: relatio n to h yp erb olic geometr y and knot theory , hep-th/9 8061 74. App endix. Pro of of an identit y found b y Broadhurst Let sin( α ) = 1 3 . Then Broa dhurst in [1] found the following identit y using PSLQ 4 √ 2( C l 2 (4 α ) − C l 2 (2 α )) = ∞ X n =0 ( − 1 2 ) 3 n 1 n + 1 2 ( 1 n + 1 2 − 3 log(2)) − 3 ∞ X n =1 ( − 1 2 ) 3 n H n n + 1 2 where H n = n X k =1 1 k W e will pr ov e this here using several formulas for the diloga rithm Li 2 ( x ) = ∞ X n =1 x n n 2 W e star t by q uo ting some formulas in Lewin [2 ] pp.244 Lemma 1 11 1.1. Li 2 ( x ) + Li 2 ( − x ) = 1 2 Li 2 ( x 2 ) 1.2. Li 2 ( x ) + Li 2 ( − x 1 − x ) = − 1 2 log 2 (1 − x ) 1.3. Li 2 ( 1 1 + x ) − L i 2 ( − x ) = π 2 6 − 1 2 log(1 + x ) lo g( (1 + x ) x 2 ) 1.4. (Ab el) Li 2 ( x 1 − x · y 1 − y ) = Li 2 ( x 1 − y )+ Li 2 ( y 1 − x ) − Li 2 ( x ) − Li 2 ( y ) − log (1 − x ) log(1 − y ) 1.5. Li 2 ( x ) + Li 2 (1 − x ) = π 2 6 − lo g( x ) log(1 − x ) Lemma 2. W e have ∞ X n =1 H n 2 n + 1 z 2 n +1 = 1 2  1 2 log 2 (1 − z ) − 1 2 log 2 (1 + z ) + log(2 ) log( 1 − z 1 + z ) + L i 2 ( 1 + z 2 ) − L i 2 ( 1 − z 2 )  Pro of: W e hav e by [3 ], p.7 15 ∞ X n =1 H n x 2 n = − log(1 − x 2 ) 1 − x 2 Int egrate ∞ X N 01 H n 2 n + 1 x 2 n +1 = − 1 2 x Z 0 (log(1 − t ) + lo g(1 + t ))( 1 1 − t + 1 1 + t ) dt and we are done by formula 3 .2 p.266 in Lewin [2 ] x Z 0 log( a + bt ) c + et dt = 1 e  log( ae − b c e log( c + ex c ) − L i 2 ( b ( c + ex ) bc − ae ) + L i 2 ( bc bc − ae  Pro of of Broadh urst’s identit y . Int ro duce the notatio n x = 1 2 (1 + i √ 8 ) 12 y = 1 2 u = √ 8 + i 3 z = − i 8 Then we wan t to compute C l 2 (4 α ) − C l 2 (2 α ) = Im( L i 2 ( u 4 ) − L i 2 ( u 2 )) In Ab el’s identit y (Lemma 1.4) we hav e x 1 − x = u 2 , y 1 − y = 1 , x 1 − y = 1 − z , y 1 − x = 1 1 + z and hence (2.1) Li 2 ( u 2 ) = Li 2 (1 − z ) + Li 2 ( 1 1+ z ) − Li 2 ( x ) − Li 2 ( 1 2 ) + log(2) lo g(1 − x ) By Lemma 1.5 a nd 1 .3 we get (2.2) Li 2 (1 − z ) = − Li 2 ( z ) + π 2 6 − 1 2 log( z ) log (1 − z ) (2.3) Li 2 ( 1 1+ z ) = Li 2 ( − z ) + π 2 6 − 1 2 log(1 + z ) log( 1+ z z 2 ) Adding (2.1), (2.2 ) and (2 .3) we obtain (2.4) Li 2 ( u 2 ) = Li 2 ( z ) − L i 2 ( z ) − Li 2 ( x ) + π 2 3 − Li 2 ( 1 2 ) + lo g(2) log(1 − x ) − log( z ) lo g(1 − z ) − 1 2 log(1 + z ) log( 1+ z z 2 ) By Lemma 1.2 a nd 1 .1 (2.5) Li 2 ( x ) + Li 2 ( − u 2 ) = − 1 2 log 2 (1 − x ) (2.6) Li 2 ( u 2 ) + Li 2 ( − u 2 ) = 1 2 Li 2 ( u 4 ) (2.7) Li 2 ( u 4 ) = 2 Li 2 ( u 2 ) − 2 Li 2 ( x ) − log 2 (1 − x ) Adding (2.4) and (2.7 ) g ives (2.8) Li 2 ( u 4 ) − Li 2 ( u 2 ) = Li 2 ( z ) − Li 2 ( z ) − 3 Li 2 ( x ) + π 2 3 − Li 2 ( 1 2 ) + log(2) log(1 − x ) − lo g 2 (1 − x ) − log( z ) log(1 − z ) − 1 2 log(1 + z ) lo g( 1+ z z 2 ) W e wan t (2.9) Im( Li 2 ( u 4 ) − L i 2 ( u 2 )) = 1 i ( Li 2 ( z ) − Li 2 ( z )) − 3 2 i ( Li 2 ( x ) − Li 2 ( x )) + Im  log(2) log(1 − x ) − lo g 2 (1 − x ) − log( z ) log (1 − z ) − 1 2 log(1 + z ) log( 1+ z z 2 )  W e get 1 i ( Li 2 ( z ) − Li 2 ( z )) = 1 i ∞ X n =1 1 n 2  ( i √ 8 ) n − ( − i √ 8 ) n  = 1 √ 2 ∞ X n =0 1 (2 n + 1 ) 2 ( − 1 2 ) 3 n and using Lemma 2 1 2 i ( Li 2 ( x ) − Li 2 ( x )) = 1 2 i ( Li 2 ( 1 − z 2 ) − Li 2 ( 1 + z 2 )) = − z i ∞ X n =1 H n 2 n + 1 z 2 n + 1 4 i  Li 2 ( 1 − z 2 ) − Li 2 ( 1 + z 2 )  + log(2 ) 2 i log( 1 − z 1 + z ) 13 = 1 √ 8 ∞ X n =1 H n 2 n + 1 ( − 1 2 ) 3 n + α lo g( r 9 8 ) + log(2) √ 8 ∞ X n =0 1 2 n + 1 ( − 1 2 ) 3 n Putting everything tog e ther we obtain Broadhurst’s formula (all logarithms can- cel). References. 1. D.Bro a dhurst, A diloga rithmic 3-dimensio nal tetrahedr on, hep-th/98 0525 2. L.Lewin, Dilogar ithms and ass o ciated functions, Macdonald, London 1958 3. A.P .Prudnik ov, O .I.Marichev, Y u.A.Brychko v , Integrals and serie s I (Russian) Moskv a Nauk a, 1 986 Institute of Algebra ic Meditation F ogdar¨ od 20 8 S-2433 H¨ o¨ or Sweden gert@maths.lth.se 14