Computer Algebra Algorithms for Special Functions in Particle Physics.

Transcription

1 Computer Algebra Algorithms for Special Functions in Particle Physics. Jakob Ablinger RISC, J. Kepler Universität Linz, Austria joint work with J. Blümlein (DESY) and C. Schneider (RISC) 7. Mai 2012

2 Definition (Harmonic Sums (H-Sums)) For c i Z and n N we define S c1,...,c k (n) = n i 1 i 2 i k 1 sign(c 1 ) i 1 i c 1 1 sign(c k) i k i c k k k is called the depth and w = k i=1 c i is called the weight of the harmonic sum S c1,...,c k (n).

3 Definition (Harmonic Sums (H-Sums)) For c i Z and n N we define S c1,...,c k (n) = n i 1 i 2 i k 1 sign(c 1 ) i 1 i c 1 1 sign(c k) i k i c k k k is called the depth and w = k i=1 c i is called the weight of the harmonic sum S c1,...,c k (n). S 1 (n) = n i=1 1 i

4 Definition (Harmonic Sums (H-Sums)) For c i Z and n N we define S c1,...,c k (n) = n i 1 i 2 i k 1 sign(c 1 ) i 1 i c 1 1 sign(c k) i k i c k k k is called the depth and w = k i=1 c i is called the weight of the harmonic sum S c1,...,c k (n). S 1 (n) = S 2, 3 (n) = n 1 i i=1 n i j=1 i=1 ( 1) j j 3 i 2

5 Definition (Harmonic Sums (H-Sums)) For c i Z and n N we define S c1,...,c k (n) = n i 1 i 2 i k 1 sign(c 1 ) i 1 i c 1 1 sign(c k) i k i c k k k is called the depth and w = k i=1 c i is called the weight of the harmonic sum S c1,...,c k (n). S 1 (n) = S 2, 3 (n) = n 1 i i=1 n i j=1 i=1 Blümlein, Hoffman, Moch, Vermaseren,... ( 1) j j 3 i 2

6 Definition (S-Sums) For c i N and x i R we define S c1,...,c k (x 1,..., x k ; n) = n i 1 i 2 i k 1 x i 1 1 i c 1 1 x ik k i c. k k k is called the depth and w = k i=1 c i is called the weight of the S-sum S c1,...,c k (x 1,..., x k ; n).

7 Definition (S-Sums) For c i N and x i R we define S c1,...,c k (x 1,..., x k ; n) = n i 1 i 2 i k 1 x i 1 1 i c 1 1 x ik k i c. k k k is called the depth and w = k i=1 c i is called the weight of the S-sum S c1,...,c k (x 1,..., x k ; n). S 2,3 (4, 3; n) = n i=1 4 i i j=1 3j j 3 i 2

8 Definition (S-Sums) For c i N and x i R we define S c1,...,c k (x 1,..., x k ; n) = n i 1 i 2 i k 1 x i 1 1 i c 1 1 x ik k i c. k k k is called the depth and w = k i=1 c i is called the weight of the S-sum S c1,...,c k (x 1,..., x k ; n). S 2,3 (4, 3; n) = S 2,3 (1, 1; n) = n i=1 n i=1 4 i i j=1 3j j 3 i 2 1 i i ( 1) j j=1 j 3 i 2 = S 2, 3 (n)

9 Definition (Cyclotomic Harmonic Sums (C-Sums)) Let a i, k N, b i, n N and c i Z we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (n) = sign(c 1 ) i 1 sign(c 2 ) i 2 = (a 1 i 1 + b 1 ) c 1 (a 2 i 2 + b 2 ) c 2 sign(c k ) ik (a k i k + b k ) c k n i 1 i k 1 k is called the depth and w = k i=1 c i is called the weight of the cyclotomic harmonic sum S (a1,b 1,c 1 ),...,(a k,b k,c k ) (n).

10 Definition (Cyclotomic Harmonic Sums (C-Sums)) Let a i, k N, b i, n N and c i Z we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (n) = sign(c 1 ) i 1 sign(c 2 ) i 2 = (a 1 i 1 + b 1 ) c 1 (a 2 i 2 + b 2 ) c 2 sign(c k ) ik (a k i k + b k ) c k n i 1 i k 1 k is called the depth and w = k i=1 c i is called the weight of the cyclotomic harmonic sum S (a1,b 1,c 1 ),...,(a k,b k,c k ) (n). S (2,1,2),(3,2,1) (n) = n i=1 i j=1 1 3j+2 (2i + 1) 2

11 Definition (Cyclotomic Harmonic Sums (C-Sums)) Let a i, k N, b i, n N and c i Z we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (n) = sign(c 1 ) i 1 sign(c 2 ) i 2 = (a 1 i 1 + b 1 ) c 1 (a 2 i 2 + b 2 ) c 2 sign(c k ) ik (a k i k + b k ) c k n i 1 i k 1 k is called the depth and w = k i=1 c i is called the weight of the cyclotomic harmonic sum S (a1,b 1,c 1 ),...,(a k,b k,c k ) (n). S (2,1,2),(3,2,1) (n) = S (1,0,2),(1,0, 3) (n) = n i j=1 1 3j+2 (2i + 1) 2 i=1 n i j=1 i=1 ( 1) j (1j+0) 3 (1i + 0) 2 = S 2, 3 (n)

12 Definition (Cyclotomic S-Sums (CS-Sums)) Let a i, c i, n, k N, b i N and x i R we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (x 1, x 2,..., x k, n) = = n i 1 i k 1 x i 1 1 (a 1 i 1 + b 1 ) c 1 x i 2 2 (a 2 i 2 + b 2 ) c x ik k 2 (a k i k + b k ) c k k is called the depth and w = k i=1 c i is called the weight of the Cyclotomic S-Sums S (a1,b 1,c 1 ),...,(a k,b k,c k ) (x 1,..., x k ; n).

13 Definition (Cyclotomic S-Sums (CS-Sums)) Let a i, c i, n, k N, b i N and x i R we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (x 1, x 2,..., x k, n) = = n i 1 i k 1 x i 1 1 (a 1 i 1 + b 1 ) c 1 x i 2 2 (a 2 i 2 + b 2 ) c x ik k 2 (a k i k + b k ) c k k is called the depth and w = k i=1 c i is called the weight of the Cyclotomic S-Sums S (a1,b 1,c 1 ),...,(a k,b k,c k ) (x 1,..., x k ; n). S (2,1,2),(3,2,1) (4, 5, n) = n i=1 i 4 i j=1 j 5 3j+2 (2i + 1) 2

14 Definition (Cyclotomic S-Sums (CS-Sums)) Let a i, c i, n, k N, b i N and x i R we define S (a1,b 1,c 1 ),(a 2,b 2,c 2 ),...,(a k,b k,c k ) (x 1, x 2,..., x k, n) = = n i 1 i k 1 x i 1 1 (a 1 i 1 + b 1 ) c 1 x i 2 2 (a 2 i 2 + b 2 ) c x ik k 2 (a k i k + b k ) c k k is called the depth and w = k i=1 c i is called the weight of the Cyclotomic S-Sums S (a1,b 1,c 1 ),...,(a k,b k,c k ) (x 1,..., x k ; n). S (2,1,2),(3,2,1) (4, 5, n) = S (1,0,2),(1,0,3) (1, 1, n) = n i 4 i j=1 i=1 n i j=1 i=1 j 5 3j+2 (2i + 1) 2 ( 1) j (1j+0) 3 (1i + 0) 2 = S 2, 3 (n)

15 Connection between these nested sums Harmonic Sums

16 Connection between these nested sums S-Sums Harmonic Sums

17 Connection between these nested sums S-Sums Harmonic Sums Cyclotomic Sums

18 Connection between these nested sums Cyclotomic S-Sums S-Sums Harmonic Sums Cyclotomic Sums

19 Example: All N-Results for 3 Loop Ladder Graphs Joint work with J. Blümlein (DESY), C. Schneider (RISC) A. Hasselhuhn (DESY), S. Klein (RWTH) Consider, e.g., the diagram (containing three massive fermion propagators) = Around 1000 sums have to be calculated

20 A typical sum N 2 j+1 j=0 s=1 N+s j 2 r=0 2( 1) s+r ( j+1 s )( j+n+s 2 (N r)(r+1)(r+2)( j+n+σ+1)( j+n+σ+2)( j+n+s+σ)! σ=0 r )(N j)!(s 1)!σ!S 1 (r + 2)

21 A typical sum N 2 j+1 j=0 s=1 N+s j 2 r=0 2( 1) s+r ( j+1 s )( j+n+s 2 (N r)(r+1)(r+2)( j+n+σ+1)( j+n+σ+2)( j+n+s+σ)! σ=0 r )(N j)!(s 1)!σ!S 1 (r + 2) ( 2N 2 + 6N + 5 ) S 2 (N) 2 = + S 2, 1,2 (N) + S 2,1, 2 (N) 2(N + 1)(N + 2) mit, z.b., + S 2,1,1,1 ( 1, 2, 1 2, 1; N) + S 2,1,1,1(1, 1, 1, 2; N) S 2,1,1,1 (1, 1 2, 1, 2; N) = N i=1 i j=1 145 S sums occur k 2 l ( ) 1 j j l l=1 2 k k=1 j i 2

22 Algebraic Relations Harmonic Sums, S-sums, cyclotomic harmonic sums und cyclotomic S-sums form a quasi-shuffle algebra. There are relations of the form: S 1,1,2 (n) = 1 2 ( 2S1 (n) (S 3 (n) S 2,1 (n)) 2S 3,1 (n) + 2S 2,1,1 (n) + S 2 (n)s 1 (n) 2 + S 4 (n) ) S a1,a 1,a 2 (x 1, x 2, x 3 ; n) = S a2 (x 3 ; n) S a1,a 1 (x 1, x 2 ; n) + S a1,a 1+a 2 (x 1, x 2 x 3 ; n) S a1 (x 1 ; n) S a2,a 1 (x 3, x 2 ; n) S a2,2a 1 (x 3, x 1 x 2 ; n) + S a2,a 1,a 1 (x 3, x 2, x 1 ; n) S (2,1,2),(3,2,1),(2,1,2) (n) = 6S (2,1,1),(2,1,2) (n) 6S (2,1,2),(2,1,1) (n) + S (3,2,1) (n)s (2,1,2),(2,1,2) (n) +4S (2,1,2),(2,1,2) (n) + 9S (2,1,2),(3,2,1) (n) + 9S (3,2,1),(2,1,2) (n) S (2,1,2),(2,1,2),(3,2,1) (n) S (3,2,1),(2,1,2),(2,1,2) (n), S (3,2,1),(2,1,2),(2,1,2) (x 2, x 1, x 1 ; n) = 6S (2,1,1),(2,1,2) (x 1 x 2, x 1 ; n) 6S (2,1,2),(2,1,1) (x 1, x 1 x 2 ; n) +S (3,2,1) (x 2 ; n) S (2,1,2),(2,1,2) (x 1, x 1 ; n) + 2S (2,1,2),(2,1,2) (x 1, x 1 x 2 ; n) +2S (2,1,2),(2,1,2) (x 1 x 2, x 1 ; n) + 9S (2,1,2),(3,2,1) (x 1, x 1 x 2 ; n) + 9S (3,2,1),(2,1,2) (x 1 x 2, x 1 ; n) S (2,1,2),(2,1,2),(3,2,1) (x 1, x 1, x 2 ; n) S (2,1,2),(3,2,1),(2,1,2) (x 1, x 2, x 1 ; n)

23 Differential Relations integral representation ( S 1,2,1 2, 1 ) 1 2, 1; n 1 x1 1 x2 = 0 x x x3 x4 n 1 x x 4 1 dx 4dx 3 dx 2 dx 1

24 Differential Relations integral representation ( S 1,2,1 2, 1 ) 1 2, 1; n 1 x1 1 x2 = 0 x x 2 0 differentiation 1 x3 x4 n 1 x x 4 1 dx 4dx 3 dx 2 dx 1 ( S 1,2,1 2, 1 2, 1; n) n = x 1 1 x1 1 x2 1 x x3 x4 n log(x 4) x x 4 1 dx 4dx 3 dx 2 dx 1

25 Differential Relations integral representation ( S 1,2,1 2, 1 ) 1 2, 1; n 1 x1 1 x2 = 0 x x 2 0 differentiation 1 x3 x4 n 1 x x 4 1 dx 4dx 3 dx 2 dx 1 ( S 1,2,1 2, 1 2, 1; n) n = x 1 1 x1 1 x2 1 x x3 x4 n log(x 4) x x 4 1 dx 4dx 3 dx 2 dx 1 integration of right hand side ( S 1,2,1 2, 1 2, 1; n) n = 2H 0,0,1,2,1 (1) 4H 0,0,2,1,1 (1) H 0,1,0,2,1 (1) H 0,1,2,0,1 (1) 2H 0,2,0,1,1 (1) H 0,2,1,0,1 (1) + (H 2 (1)(H 0,1,2 (1) + H 0,2,1 (1)) + 2H 0,0,1,2 (1) + 2H 0,0,2,1 (1) ( ( 3(H0,1,2 (1) + H 0,2,1 (1)) +H 0,1,0,2 (1))S 1 (2; n) + S 2 ( ) + S 1,2 2, 1 )) 2 2 ; n ( H 2 (1)S 1,2,1 2, 1 ) ( 2, 1; n S 1 ( ) S 1,2,1 2, 1 ) ( 2, 1; n S 1,2,2 2, 1 ) 2, 1; n 2S 1,3,1 ( 2, 1 2, 1; n ) S 2,2,1 ( 2, 1 2, 1; n )

26 Number of Basis Sums Example (Harmonic sums) Number of w N S N A N D N H N AD N AH N DH N ADH

27 Number of Basis Sums For w 2 we have N S (w) = 2 3 w 1 N A (w) = 1 ( w ) µ 3 d w d d w N D (w) = 4 3 w 2 N H (w) = 2 3 w 1 2 w 1 N AD (w) = 1 ( w ) µ 3 d 1 ( ) w 1 µ 3 d w d w 1 d d w d w 1 N AH (w) = 1 ( w ) µ [3 d 2 d] w d d w N DH (w) = 4 3 w 2 2 w 2 N ADH (w) = 1 ( w ) [ µ 3 d 2 d] 1 ( ) w 1 [ µ 3 d 2 d] w d w 1 d d w d w 1

28 Number of Basis Sums We consider S (a1,b 1,c 1 ),...,(a k,b k,c k ) (n) with a i {1, 2}, b i {0, 1}, a i > b i and c i Z. Example (Cyclotomic harmonic sums) w N S N A N D N H1 N H1H2 N H1M N H1H2M N AD N AH1H2M N DH1H2M N ADH1H2M

29 Number of Basis Sums N S (w) = 4 5 w 1 w>1 1 ( w ) N A (w) = µ 5 d w d d w N D (w) = N S (w) N S (w 1) w>1 = 16 5 w 2 N H1 (w) = N H2 (w) = N M (w) = N S (w) 2 w 1 = 4 5 w 1 2 w 1 N (w) H1H2 = N S(w) (2 2 w 1 1) = 4 5 w 1 (2 2 w 1 1) N H1M(w) = N H2M(w) = N S (w) 2 2 w 1 = 4 5 w w 1 N H1H2M(w) = N S (w) (3 2 w 1 1) = 4 5 w 1 (3 2 w 1 1) N AD (w) = N A (w) N A (w 1) w>2 1 ( w ) = µ 5 d 1 ( ) w 1 µ 5 d w d w 1 d d w d (w 1) w>1 1 ( w ) N AH1H2M(w) = µ 5 d 3 1 ( w ) µ 2 d 1 w d w d d w d w N DH1H2M(w) = N H1H2M(w) N H1H2M(w 1) w>1 = 16 5 w w 2 N ADH1H2M(w) = N AH1H2M(w) N AH1H2M(w 1) w>2 1 ( w ) = µ (5 d 3 2 d ) w d d w 1 ( ) w 1 µ (5 d 3 2 d ) w 1 d d (w 1)

30 Number of Basis Sums We consider S a1,...,a k (x 1,..., x k ; n) with x i {1, 1, 1/2, 1/2, 2, 2}. Example (S-Sums) Each of the indices {1/2, 1/2, 2, 2} is allowed to appear just once in each sum. Number of Weight N S N A N D N AD

31 Example: continued

32 Example: continued = C. Schneider s Sigma.m Around 1000 sums are calculated containing in total 533 S sums

33 Example: continued = C. Schneider s Sigma.m Around 1000 sums are calculated containing in total 533 S sums = HarmonicSums.m After elimination the following sums remain: S 4 (N), S 3 (N), S 2 (N), S 1 (N), S 2 (N), S 3 (N), S 4 (N), S 3,1 (N), S 2,1 (N), S 2, 2 (N), S 2,1 (N), S 3,1 (N), S 2,1,1 (N), S 2,1,1 (N)

34 Harmonic Polylogarithms (H-Logs) We define f : {0, 1, 1} (0, 1) R by f (0, x) = 1 x, f (1, x) = f ( 1, x) = 1 1 x, x.

35 Harmonic Polylogarithms (H-Logs) We define f : {0, 1, 1} (0, 1) R by f (0, x) = 1 x, f (1, x) = f ( 1, x) = 1 1 x, x. Definition (Harmonic Polylogarithms (H-Logs)) Let m i { 1, 0, 1} we define for x (0, 1) : H(x) = 1, 1 k! (log x)k, if (m 1,..., m k ) H m1,m 2,...,m k (x) = = 0 x 0 f m 1 (y)h m2,...,m k (y)dy, otherwise.

36 Multiple Polylogarithms (M-Logs) Let a R and q = We define f as follows: { a, if a > 0, otherwise f a : (0, q) R { 1 f a (x) = x, if a = 0 otherwise. 1 a sign(a) x,

37 Multiple Polylogarithms (M-Logs) Let a R and q = We define f as follows: { a, if a > 0, otherwise f a : (0, q) R { 1 f a (x) = x, if a = 0 otherwise. 1 a sign(a) x, Definition (Multiple Polylogarithms (M-Logs)) Let m i R and let q = min mi >0 m i we define for x (0, q) : H(x) = 1, 1 k! (log x)k, if (m 1,..., m k ) = 0 H m1,m 2,...,m k (x) = x 0 f m 1 (y)h m2,...,m k (y)dy, otherwise.

38 The first cyclotomic polynomials are given by Φ 1 (x) = x 1 Φ 2 (x) = x + 1 Φ 3 (x) = x 2 + x + 1 Φ 4 (x) = x Φ 5 (x) = x 4 + x 3 + x 2 + x + 1 Φ 6 (x) = x 2 x + 1 Φ 7 (x) = x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 8 (x) = x Φ 9 (x) = x 6 + x Φ 10 (x) = x 4 x 3 + x 2 x + 1 Φ 11 (x) = x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 + x + 1 Φ 12 (x) = x 4 x 2 + 1, etc.

39 Cyclotomic Harmonic Polylogarithms (C-Logs) We now define the alphabet { } { } 1 x l A := x Φ k (x) k N, 0 l < ϕ(k), where Φ k (x) denotes the kth cyclotomic polynomial and ϕ is Euler s totient function.

40 Cyclotomic Harmonic Polylogarithms (C-Logs) We now define the alphabet { } { } 1 x l A := x Φ k (x) k N, 0 l < ϕ(k), where Φ k (x) denotes the kth cyclotomic polynomial and ϕ is Euler s totient function. Definition (Cyclotomic Harmonic Polylogarithms) Let m i A we define for x (0, 1) : H(x) = 1, 1 k! (log x)k, if (m 1,..., m k ) H m1,m 2,...,m k (x) = = ( 1 x,..., 1 x ) x 0 m 1H m2,...,m k (y)dy, otherwise. k is called the depth of H m (x).

41 Connection between these structures S-Sums H-Sums C-Sums ( ) S 12 1,2, 1; n S 1,2 (n) S (2,1, 1) (n)

42 Connection between these structures S-Sums H-Sums C-Sums C-Logs ( ) S 12 1,2, 1; n S 1,2 (n) S (2,1, 1) (n) H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x)

43 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x)

44 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x) Mellin transform

45 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x) n Mellin transform ( ) S 12 1,2, 1; S 1,2 ( ) S (2,1, 1) ( )

46 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x) n Mellin transform x 1 x 1 x c R ( ) S 12 1,2, 1; S 1,2 ( ) S (2,1, 1) ( ) H (4,1),(0,0) (1) H 1,1 (1) H 2,3 (c)

47 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x) n Mellin transform x 1 x 1 x c R ( ) S 12 1,2, 1; S 1,2 ( ) S (2,1, 1) ( ) H (4,1),(0,0) (1) H 1,1 (1) H 2,3 (c) power series expansion

48 Connection between these structures integral representation (inv. Mellin transform) S-Sums ( ) S 12 1,2, 1; n H-Sums S 1,2 (n) C-Sums S (2,1, 1) (n) C-Logs H (4,1),(0,0) (x) H-Logs H 1,1 (x) M-Logs H 2,3 (x) n Mellin transform x 1 x 1 x c R ( ) S 12 1,2, 1; S 1,2 ( ) S (2,1, 1) ( ) H (4,1),(0,0) (1) H 1,1 (1) H 2,3 (c) power series expansion

49 Asymptotic Expansion of Harmonic Sums We say that the function f : R R is expanded in an asymptotic series a k f (x) x k, x, k=0 where a k are constants, if for all K 0 R K (x) = f (x) K k=0 ( ) a k 1 x k = o x K, x.

50 Why do we need these expansions of harmonic sums? E.g., for limits of the form lim n n ( S 2 (n) ζ 2 S 2,2 (n) + 7 ζ for the approximation of the values of analytic continued harmonic sums at the complex plane S 2, 3 ( i) )

51 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

52 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

53 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

54 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

55 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

56 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

57 Basic Idea ϕ(x) ϕ(1 x) = {}}{ 1 S 1,3 (n) = ( 1) n x n H 1,0,0 (x) x dx + const }{{} a k+1 k! n(n + 1)... (n + k) = k=0 b k n k k=1 ( S 1,3 (n) ( 1) n 1 4n n 4 5 8n n 6 +( 1) n ( 1 2n 1 4n n 4 1 4n 6 a k x k k=0 b 1 = a 1 k 2 b k = ( 1) l Sk l l a k l(k l)! l=0 ) + 3 log(2) ζ 3 4 ) ζ 3 19 ζ2 2 40

58 Asymptotic Expansions this basic idea can be turned into an algorithm: S 1,3(n) ( 1) n ( 1 4n n 4 5 8n n 6 ) + ( 3 log(2) ζ3 1 + ( 1) n 4 2n 1 4n n 4 1 4n 6 ) ζ 3 19 ζ2 2 40

59 Asymptotic Expansions this basic idea can be turned into an algorithm: S 1,3(n) ( 1) n ( 1 4n n 4 5 8n n 6 extension to cyclotomic sums: ) + ( 3 log(2) ζ3 1 + ( 1) n 4 2n 1 4n n 4 1 4n 6 S (2,1,2),(1,0,1) (n) 4 log(2)s (2,1, 1) ( ) 2 8 log(2)s (2,1, 1) ( ) 4 log(2) + 1 9n 3 1 4n + 7ζ3 ( n n 2 1 ) (log(n) + γ) 4n ) ζ 3 19 ζ2 2 40

60 Asymptotic Expansions this basic idea can be turned into an algorithm: S 1,3(n) ( 1) n ( 1 4n n 4 5 8n n 6 extension to cyclotomic sums: ) + ( 3 log(2) ζ3 1 + ( 1) n 4 2n 1 4n n 4 1 4n 6 S (2,1,2),(1,0,1) (n) 4 log(2)s (2,1, 1) ( ) 2 8 log(2)s (2,1, 1) ( ) 4 log(2) + 1 9n 3 1 4n + 7ζ3 ( n n 2 1 ) (log(n) + γ) 4n extension to a subset of the S-Sums: S 2,1(1, 1 3 )(n) S1,2 ( 1 3, 1; ) + 3 n ( 3 ( + (3 n 3 2n n 2 + 3n 2n 3 3 4n n ( 1 6n S 3 ( 1 3 ; ) + H 3(1)3 n ( 3 2n 3 1 2n 2 ) H 0,3(1) + 3 n ( 3 2n 2 1 n ) + 3 n 4n 3 ) ζ 3 19 ζ n n 2 1 ) ( ) 13 S 2 2n ; )) ) ( ) 13 + ζ 2 S 1 ;

61 Asymptotic Expansions this basic idea can be turned into an algorithm: S 1,3(n) ( 1) n ( 1 4n n 4 5 8n n 6 extension to cyclotomic sums: ) + ( 3 log(2) ζ3 1 + ( 1) n 4 2n 1 4n n 4 1 4n 6 S (2,1,2),(1,0,1) (n) 4 log(2)s (2,1, 1) ( ) 2 8 log(2)s (2,1, 1) ( ) 4 log(2) + 1 9n 3 1 4n + 7ζ3 ( n n 2 1 ) (log(n) + γ) 4n extension to a subset of the S-Sums: S 2,1(1, 1 3 )(n) S1,2 ( 1 3, 1; ) + 3 n ( 3 ( + (3 n 3 2n n 2 + 3n 2n 3 3 4n n ( 1 6n S 3 ( 1 3 ; ) + H 3(1)3 n ( 3 2n 3 1 2n 2 extension to a subset of the cyclotomic S-Sums: S (2,1,2) ( 1 2 ; n ) ) H 0,3(1) + 3 n ( 3 2n 2 1 n ) + 3 n 4n 3 ) ζ 3 19 ζ n n 2 1 ) ( ) 13 S 2 2n ; )) ) ( ) 13 + ζ 2 S 1 ; ( ( ) ) π 2 6 8H 1 0,1 2 + log 2 (2) ( 24 2 n n n 4 5 4n ) 4n 2 + 2n

62 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations

63 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations compute the inverse Mellin transform of harmonic sums and their generalizations, this leads to harmonic polylogarithms and their generalizations

64 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations compute the inverse Mellin transform of harmonic sums and their generalizations, this leads to harmonic polylogarithms and their generalizations find algebraic and structural relations of harmonic polylogarithms and their generalizations

65 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations compute the inverse Mellin transform of harmonic sums and their generalizations, this leads to harmonic polylogarithms and their generalizations find algebraic and structural relations of harmonic polylogarithms and their generalizations apply algebraic and structural relations to harmonic sums

66 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations compute the inverse Mellin transform of harmonic sums and their generalizations, this leads to harmonic polylogarithms and their generalizations find algebraic and structural relations of harmonic polylogarithms and their generalizations apply algebraic and structural relations to harmonic sums calculate the asymptotic expansion of harmonic sums and their generalizations

67 The Package HarmonicSums The package HarmonicSums offers functions to find algebraic and structural relations of harmonic sums and their generalizations compute the inverse Mellin transform of harmonic sums and their generalizations, this leads to harmonic polylogarithms and their generalizations find algebraic and structural relations of harmonic polylogarithms and their generalizations apply algebraic and structural relations to harmonic sums calculate the asymptotic expansion of harmonic sums and their generalizations perform several other tasks not mentioned in this talk (see, e.g., my PhD thesis, April 2012)

68 Almkvist Zeilberger Algorithm Two point Feynman integrals in D-dimensional Minkowski space with one time- and (D 1) Euclidean space dimensions, ε = D 4 and ε R with ε 1 of the structure d D p 1 (2π) D... d D p k N (p 1,... p k ; p; m 1... m k ;, N) (2π) D ( p1 2 + m2 1 )l 1... ( p 2 k + mk 2)l k are of relevance for many physical processes at high energy colliders, such as the Large Hadron Collider, LHC, and others. V δ V

69 Almkvist Zeilberger Algorithm Two point Feynman integrals in D-dimensional Minkowski space with one time- and (D 1) Euclidean space dimensions, ε = D 4 and ε R with ε 1 of the structure d D p 1 (2π) D... d D p k N (p 1,... p k ; p; m 1... m k ;, N) (2π) D ( p1 2 + m2 1 )l 1... ( p 2 k + mk 2)l k are of relevance for many physical processes at high energy colliders, such as the Large Hadron Collider, LHC, and others. These integrals can be transformed to integrals of the form V δ V od u d... o1 u 1 F (N; x 1,..., x d, ε)dx 1... dx d

70 Recurrence for the Integrand Apagodu and Zeilberger 2006 where a(x 1,...,x d ) F (n; x 1,..., x d ) = q(n; x 1,..., x d ) e b(x 1,...,x d ) P ( ) S p (x 1,..., x d ) αp s(x1,..., x d ) n, t(x 1,..., x d ) a(x 1,..., x d ), b(x 1,..., x d ), s(x 1,..., x d ), t(x 1,..., x d ), q(n; x 1,..., x d ) K[x 1,..., x d ] and S p (x 1,..., x d ) K[x 1,..., x d ], α p K for 1 p P. With e.g., K = Q(ε, n). p=1

71 Recurrence for the Integrand Apagodu and Zeilberger 2006 where a(x 1,...,x d ) F (n; x 1,..., x d ) = q(n; x 1,..., x d ) e b(x 1,...,x d ) P ( ) S p (x 1,..., x d ) αp s(x1,..., x d ) n, t(x 1,..., x d ) a(x 1,..., x d ), b(x 1,..., x d ), s(x 1,..., x d ), t(x 1,..., x d ), q(n; x 1,..., x d ) K[x 1,..., x d ] and S p (x 1,..., x d ) K[x 1,..., x d ], α p K for 1 p P. With e.g., K = Q(ε, n). p=1 Then there exist L N, e 0 (n), e 1 (n),..., e L (n) K[n], not all zero, and R i (n; x 1,..., x d ) K(n, x 1,..., x d ) such that G i (n; x 1,..., x d ) := R i (n; x 1,..., x d )F (n; x 1,..., x d ) satisfy L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ). i=0 i=1

72 Recurrence for the Integrand Apagodu and Zeilberger 2006 where a(x 1,...,x d ) F (n; x 1,..., x d ) = q(n; x 1,..., x d ) e b(x 1,...,x d ) P ( ) S p (x 1,..., x d ) αp s(x1,..., x d ) n, t(x 1,..., x d ) a(x 1,..., x d ), b(x 1,..., x d ), s(x 1,..., x d ), t(x 1,..., x d ), q(n; x 1,..., x d ) K[x 1,..., x d ] and S p (x 1,..., x d ) K[x 1,..., x d ], α p K for 1 p P. With e.g., K = Q(ε, n). p=1 Then there exist L N, e 0 (n), e 1 (n),..., e L (n) K[n], not all zero, and R i (n; x 1,..., x d ) K(n, x 1,..., x d ) such that G i (n; x 1,..., x d ) := R i (n; x 1,..., x d )F (n; x 1,..., x d ) satisfy L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ). i=0 i=1

73 Recurrence for the Integral 1 We now consider the integral od o1 a(n) :=... F (n; x 1,..., x d )dx 1... dx d, u d u 1 where for F (n; x 1,..., x d ) we have L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ). i=0 i=1

74 Recurrence for the Integral 1 We now consider the integral od o1 a(n) :=... F (n; x 1,..., x d )dx 1... dx d, u d u 1 where for F (n; x 1,..., x d ) we have L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ). i=0 i=1 If F (n;..., x i 1, u i, x i+1,... ) = F (n;..., x i 1, o i, x i+1,... ) = 0, we also have G i (n;..., x i 1, u i, x i+1,... ) = G i (n;..., x i 1, o i, x i+1,... ) = 0

75 Recurrence for the Integral 1 We now consider the integral od o1 a(n) :=... F (n; x 1,..., x d )dx 1... dx d, u d u 1 where for F (n; x 1,..., x d ) we have L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ). i=0 i=1 If F (n;..., x i 1, u i, x i+1,... ) = F (n;..., x i 1, o i, x i+1,... ) = 0, we also have G i (n;..., x i 1, u i, x i+1,... ) = G i (n;..., x i 1, o i, x i+1,... ) = 0 and hence a(n) satisfies the homogenous linear recurrence L e i (n)a(n + i) = 0. i=0

76 Recurrence for the Integral 2 We again consider the integral a(n) := od o1... F (n; x 1,..., x d )dx 1... dx d, u d u 1 suppose F (n; x 1,..., x d ) does not vanish at the bounds

77 Recurrence for the Integral 2 We again consider the integral a(n) := od u d... o1 u 1 F (n; x 1,..., x d )dx 1... dx d, suppose F (n; x 1,..., x d ) does not vanish at the bounds G i (n; x 1,..., x d ) does not have to vanish at the bounds

78 Recurrence for the Integral 2 We again consider the integral a(n) := od u d... o1 u 1 F (n; x 1,..., x d )dx 1... dx d, suppose F (n; x 1,..., x d ) does not vanish at the bounds G i (n; x 1,..., x d ) does not have to vanish at the bounds then force the G i to vanish at the integration bounds by modifying the ansatz, and look for G i of the form G i (n; x 1,..., x d ) = G i (n; x 1,..., x d )(x i u i )(x i o i ),

79 Recurrence for the Integral 2 We again consider the integral a(n) := od u d... o1 u 1 F (n; x 1,..., x d )dx 1... dx d, suppose F (n; x 1,..., x d ) does not vanish at the bounds G i (n; x 1,..., x d ) does not have to vanish at the bounds then force the G i to vanish at the integration bounds by modifying the ansatz, and look for G i of the form G i (n; x 1,..., x d ) = G i (n; x 1,..., x d )(x i u i )(x i o i ), hence a(n) satisfies again a homogenous linear recurrence of the form L e i (n)a(n + i) = 0. i=0

80 Example 1 1 I (ε, n) = 0 0 (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n):= dx 1 dx 2. Note that the integrand does not vanish at the integration bounds.

81 Example 1 1 I (ε, n) = 0 0 (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n):= dx 1 dx 2. Note that the integrand does not vanish at the integration bounds. Using our ansatz we find 2(n + 1)(ε n 2)F (n) (n + 2)(5ε 5n 13)F (n + 1) +(n + 3)(4ε 4n 13)F (n + 2) (n + 4)(ε n 4)F (n + 3) = D x1 F (n)(x 1 1) x 1 (x 1 + 1) x 2 ((n + 3)x 1 x 2 + 2) +D x2 F (n)(x 2 1) x 2 ( x 2 2 x 3 1 ( ε + n + 4) (ε 3)x 2 x (n + 2)x 2 x ).

82 Example 1 1 I (ε, n) = 0 0 (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n):= dx 1 dx 2. Note that the integrand does not vanish at the integration bounds. Using our ansatz we find 2(n + 1)(ε n 2)F (n) (n + 2)(5ε 5n 13)F (n + 1) +(n + 3)(4ε 4n 13)F (n + 2) (n + 4)(ε n 4)F (n + 3) = D x1 F (n)(x 1 1) x 1 (x 1 + 1) x 2 ((n + 3)x 1 x 2 + 2) +D x2 F (n)(x 2 1) x 2 ( x 2 2 x 3 1 ( ε + n + 4) (ε 3)x 2 x (n + 2)x 2 x ). Integration of this reccurence yields 2(n + 1)(ε n 2)I (ε, n) (n + 2)(5ε 5n 13)I (ε, n + 1) +(n + 3)(4ε 4n 13)I (ε, n + 2) (n + 4)(ε n 4)I (ε, n + 3) = 0.

83 Example 1 1 I (ε, n) = 0 0 (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n):= dx 1 dx 2. Note that the integrand does not vanish at the integration bounds. Using our ansatz we find 2(n + 1)(ε n 2)F (n) (n + 2)(5ε 5n 13)F (n + 1) +(n + 3)(4ε 4n 13)F (n + 2) (n + 4)(ε n 4)F (n + 3) = D x1 F (n)(x 1 1) x 1 (x 1 + 1) x 2 ((n + 3)x 1 x 2 + 2) +D x2 F (n)(x 2 1) x 2 ( x 2 2 x 3 1 ( ε + n + 4) (ε 3)x 2 x (n + 2)x 2 x ). Integration of this reccurence yields 2(n + 1)(ε n 2)I (ε, n) (n + 2)(5ε 5n 13)I (ε, n + 1) +(n + 3)(4ε 4n 13)I (ε, n + 2) (n + 4)(ε n 4)I (ε, n + 3) = 0. Solving the recurrence leads to ( n I (ε, n) = 1 1 n + 1 i + ε 1 21 ε i=1 n i=1 ) 2 i i + ε ε (2 ε 2) ε 1

84 Recurrence for the Integral 3 We look again at the integral od o1 a(n) := F (n; x 1,..., x d )dx 1... dx d. u d u 1 Suppose that we found L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ) i=0 i=1

85 Recurrence for the Integral 3 We look again at the integral od o1 a(n) := F (n; x 1,..., x d )dx 1... dx d. u d u 1 Suppose that we found L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ) i=0 i=1 By integration with respect to x 1,..., x d we get L e i (n)a(n + i) = i=0 d i=1 od d u d od oi 1 oi+1 i=1 u i 1 u d u i 1 u 1 oi 1 oi+1 o1 u i+1 u i+1 o1 u 1 O i (n)dx 1... dx i 1 dx i+1... dx d U i (n)dx 1... dx i 1 dx i+1... dx d with U i (n) := G i (n; x 1,..., x i 1, o i, x i+1..., x d ) O i (n) := G i (n; x 1,..., x i 1, u i, x i+1..., x d ).

86 Recurrence for the Integral 3 We look again at the integral od o1 a(n) := F (n; x 1,..., x d )dx 1... dx d. u d u 1 Suppose that we found L d e i (n)f (n + i; x 1,..., x d ) = D xi G i (n; x 1,..., x d ) i=0 i=1 By integration with respect to x 1,..., x d we get L e i (n)a(n + i) = i=0 d i=1 od d u d od oi 1 oi+1 i=1 u i 1 u d u i 1 u 1 oi 1 oi+1 o1 u i+1 u i+1 o1 u 1 O i (n)dx 1... dx i 1 dx i+1... dx d U i (n)dx 1... dx i 1 dx i+1... dx d with U i (n) := G i (n; x 1,..., x i 1, o i, x i+1..., x d ) O i (n) := G i (n; x 1,..., x i 1, u i, x i+1..., x d ). Note that there are 2 d integrals of dimension d 1, to compute.

87 Example 1 1 I (ε, n) = 0 0 Applying the algorithm leads to (1 + x 1 x 2) n (1 + x 1) ε } {{ } F (n;x1,x2):= dx 1dx 2, (n + 1)F (n; x 1, x 2) + (n + 2)F (n + 1; x 1, x 2) = D x1 0 + D x2 x 2(x 1 x 2 + 1) n+1 (1 + x 1) ε

88 Example 1 1 I (ε, n) = 0 0 Applying the algorithm leads to (1 + x 1 x 2) n (1 + x 1) ε } {{ } F (n;x1,x2):= dx 1dx 2, (n + 1)F (n; x 1, x 2) + (n + 2)F (n + 1; x 1, x 2) = D x1 0 + D x2 x 2(x 1 x 2 + 1) n+1 and hence it follows by integration (1 + x 1) ε 1 1 (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = (x 1 + 1) n+1 ε dx 1 0dx 1. 0 } {{ } 0 I1(n) In the next step apply the algorithm to I 1(n); we find I 1(ε, n) = 4 2n 2 ε 2 ε (n + 2 ε).

89 Example 1 1 I (ε, n) = 0 0 Applying the algorithm leads to (1 + x 1 x 2) n (1 + x 1) ε } {{ } F (n;x1,x2):= dx 1dx 2, (n + 1)F (n; x 1, x 2) + (n + 2)F (n + 1; x 1, x 2) = D x1 0 + D x2 x 2(x 1 x 2 + 1) n+1 and hence it follows by integration (1 + x 1) ε 1 1 (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = (x 1 + 1) n+1 ε dx 1 0dx 1. 0 } {{ } 0 I1(n) In the next step apply the algorithm to I 1(n); we find Plugging in yields I 1(ε, n) = 4 2n 2 ε 2 ε (n + 2 ε). (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = 4 2n 2 ε 2 ε (n + 2 ε).

90 Example 1 1 I (ε, n) = 0 0 Applying the algorithm leads to (1 + x 1 x 2) n (1 + x 1) ε } {{ } F (n;x1,x2):= dx 1dx 2, (n + 1)F (n; x 1, x 2) + (n + 2)F (n + 1; x 1, x 2) = D x1 0 + D x2 x 2(x 1 x 2 + 1) n+1 and hence it follows by integration (1 + x 1) ε 1 1 (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = (x 1 + 1) n+1 ε dx 1 0dx 1. 0 } {{ } 0 I1(n) In the next step apply the algorithm to I 1(n); we find Plugging in yields I 1(ε, n) = 4 2n 2 ε 2 ε (n + 2 ε). (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = 4 2n 2 ε 2 ε (n + 2 ε). Solving this recurrence yields ( n ) I (ε, n) = 1 1 n n + 1 i + ε 1 2 i 21 ε i + ε ε (2 ε 2). ε 1 i=1 i=1

91 Laurent Series Expansion of the Integral we look again at the integral I(ε, n) := od o1 F (n; x 1,..., x d )dx 1... dx d. u d u 1 assume that we can write it in the form I(ε, n) = ε l I l (n). l= L

92 Laurent Series Expansion of the Integral we look again at the integral I(ε, n) := od u d o1 assume that we can write it in the form I(ε, n) = u 1 F (n; x 1,..., x d )dx 1... dx d. ε l I l (n). l= L find I L (n), I L+1 (n),..., I u (n) in terms of indefinite nested product-sum expressions.

93 Laurent Series Expansion of the Integral we look again at the integral I(ε, n) := od u d o1 assume that we can write it in the form I(ε, n) = u 1 F (n; x 1,..., x d )dx 1... dx d. ε l I l (n). l= L find I L (n), I L+1 (n),..., I u (n) in terms of indefinite nested product-sum expressions. compute a recurrence for I(ε, n) in the form a 0 (ε, n)t (ε, n)+ +a d (ε, n)t (ε, n+d) = h 0 (n)+ +h u (n)ε u +O(ε u+1 ); use one of the methods presented above

94 Laurent Series Expansion of the Integral we look again at the integral I(ε, n) := od u d o1 assume that we can write it in the form I(ε, n) = u 1 F (n; x 1,..., x d )dx 1... dx d. ε l I l (n). l= L find I L (n), I L+1 (n),..., I u (n) in terms of indefinite nested product-sum expressions. compute a recurrence for I(ε, n) in the form a 0 (ε, n)t (ε, n)+ +a d (ε, n)t (ε, n+d) = h 0 (n)+ +h u (n)ε u +O(ε u+1 ); use one of the methods presented above use algorithm FLSR implemented in Sigma.

95 Example I (ε, n) = (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n;x 1,x 2):= dx 1 dx 2,

96 Example I (ε, n) = (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n;x 1,x 2):= dx 1 dx 2, Applying the algorithm leads to (n + 1)F (n; x 1, x 2 ) + (n + 2)F (n + 1; x 1, x 2 ) = D x1 0 + D x2 x 2 (x 1 x 2 + 1) n+1 (1 + x 1 ) ε

97 Example I (ε, n) = (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n;x 1,x 2):= dx 1 dx 2, Applying the algorithm leads to (n + 1)F (n; x 1, x 2 ) + (n + 2)F (n + 1; x 1, x 2 ) = D x1 0 + D x2 x 2 (x 1 x 2 + 1) n+1 and hence it follows by integration 1 (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = (x 1 + 1) n+1 ε dx 1 0 } {{ } I 1(n) 1 0 (1 + x 1 ) ε 0dx 1.

98 Example I (ε, n) = (1 + x 1 x 2 ) n (1 + x 1 ) ε } {{ } F (n;x 1,x 2):= dx 1 dx 2, Applying the algorithm leads to (n + 1)F (n; x 1, x 2 ) + (n + 2)F (n + 1; x 1, x 2 ) = D x1 0 + D x2 x 2 (x 1 x 2 + 1) n+1 and hence it follows by integration 1 (n + 1)I (ε, n) + (n + 2)I (ε, n + 1) = (x 1 + 1) n+1 ε dx 1 0 } {{ } I 1(n) In the next step apply the method to I 1 (n); we find I 1 (ε, n) = 2n+2 1 n ε ( 2 n+2 (H 1 (1)(n + 2) 1) 1 ) (n + 2) (1 + x 1 ) ε 0dx 1. + ε2 ( 2 n+1 ( H 1 (1) 2 (n + 2) 2 2H 1 (1)(n + 2) + 2 ) 1 ) (n + 2) 3 + O ( ε 3).

99 Plugging in and solving the resulting recurrence by means of FLSR and combining the solution with the initial values of the integral yields the result. I (ε, n) = S 1(2; n) n + 1 S 1(n) n (n + 1)3 + 4 ( n + 2 n 1) (n + 1) 2 + 2n(n + 1) 2 2(n + 1) ( 4 ( 2 n+2 (n + 1) 2 n+2 n(n + 1) +ε H 1 (1) 2(n + 1) 4 S ) 1(2; n) n + 1 ( 2(n + 1)n 2 + 4(n + 1)n + 2(n + 1) ) S 2 (2; n) + 2(n + 1) 4 S 2(n) n + 1 ) ( ( + 2n+2 (n + 1) 2(n + 1) 2(n + 1) 4 + ε 2 H 1 (1) 2 S1 (2; n) 2(n + 1) + 2n+1 (n 2 ) + 2n + 1) 2(n + 1) 4 ( 2 n+2 n 2 n+2 +H 1 (1) 2(n + 1) 4 S ) ) 2(2; n) + S 3(2; n) n + 1 n + 1 S 3(n) n n+2 2 2(n + 1) 4 +O ( ε 3).

100 The Package MultiIntegrate In[1]:= << Sigma.m In[2]:= << HarmonicSums.m In[3]:= << EvaluateMultiSums.m In[4]:= << MultiIntegrate.m Sigma - A summation package by Carsten Schneider -RISC Linz- V 1.0 (7/7/11) HarmonicSums by Jakob Ablinger -RISC Linz- Version 1.0 (1/3/12) EvaluateMultiSums by Carsten Schneider -RISC Linz- Version 1.0 (11/8/11) MultiIntegrate by Jakob Ablinger -RISC Linz- Version 1.0 (1/3/12) In[5]:= mazintegrate[(1 + x1 x2) n /(1 + x1) ε, {n}, {{x1, 0, 1}, {x2, 0, 1}}] Out[5]= n ι 1 =1 1 ι 1 ɛ+1 n 1 2 n ι 1 =1 2 ι 1 ι 1 +ɛ 1 (n + 1)2 ɛ + 2 ɛ 2 (n + 1)(ɛ 1)2 ɛ

101 The Package MultiIntegrate The following integral occurs in the direct computation of a 3-loop diagram of the ladder-type: 1 1 ( (s(x 1) + t(u 1) + 1) n 0 0 (w 1)(z 1)(sx s + tu t u + 1)(sx s + tu t x + 1) 1 (z( s + tu t + 1) + x((s 1)z + 1)) n + (z 1)(sx s + tu t x + 1) swx + sw + sxz sz tuw + tuz + tw tz + uw u w xz + x + z 1 (u((t 1)w + 1) w(s( x) + s + t 1)) n ) + dudx (w 1)(sx s + tu t u + 1) swx sw sxz + sz + tuw tuz tw + tz uw + u + w + xz x z

102 The Package MultiIntegrate The following integral occurs in the direct computation of a 3-loop diagram of the ladder-type: 1 1 ( (s(x 1) + t(u 1) + 1) n 0 0 (w 1)(z 1)(sx s + tu t u + 1)(sx s + tu t x + 1) 1 (z( s + tu t + 1) + x((s 1)z + 1)) n + (z 1)(sx s + tu t x + 1) swx + sw + sxz sz tuw + tuz + tw tz + uw u w xz + x + z 1 (u((t 1)w + 1) w(s( x) + s + t 1)) n ) + dudx (w 1)(sx s + tu t u + 1) swx sw sxz + sz + tuw tuz tw + tz uw + u + w + xz x z ( ) ( ) 1 (s + t 1)w(z 1) sw sz + z 1 (s + t 1)w(z 1) z(sw sz + z 1) = (S 1,1, ; n S 1,1, ; n (w 1)(z 1)(s + t 1) sw sz + z 1 z 1 sw sz + z 1 w(z 1) ( ) ( ) ( ) (s + t 1)w(z 1) (t 1)(sw sz + z 1) s + t 1 s + t 1 1)(t 1) S 1,1, ; n + S 1,1, (1 t)w; n + S 1,1, (s ; n sw sz + z 1 (s + t 1)(z 1) t 1 t 1 s + t 1 ( ) ( ) ( ) (s + t 1)w(z 1) (sw sz + z 1)(tz 1) s + t 1 s + t 1 +S 1,1, ; n S 1,1 sw sz + z 1 (s + t 1)w(z 1) t 1, 1 t; n 1)(t 1) + S 1,1, (s ; n s 1 s + t 1 ( ) ( ) ( ) (s + t 1)(w 1)z tw + w + tz 1 s + t 1 1)(sw 1) s + t 1 +S 1,1, ; n S 1,1, (t ; n + S 1,1, (1 s)z; n (t 1)w tz + 1 w 1 t 1 s + t 1 s 1 ( ) ( ) ( ) (s + t 1)(w 1)z (s 1)( tw + w + tz 1) s + t 1 s + t 1 S 1,1, ; n S 1,1 (t 1)w tz + 1 (s + t 1)(w 1) s 1, 1 s; n 1)(tz 1) S 1,1, (s ; n s 1 s + t 1 ( ) ( ) (s + t 1)(w 1)z 1)w tz + 1) (s + t 1)(w 1)z 1)((t 1)w tz + 1) S 1,1, w((t ; n + S 1,1, (sw ; n (t 1)w tz + 1 (w 1)z (t 1)w tz + 1 (s + t 1)(w 1)z S 1,1(1, (1 s)z; n) + S 1,1(1, z sz; n) + S 2((1 s)z; n) S 1,1(1, (1 t)w; n) + S 1,1(1, w tw; n) S 2(( s t + 1)w; n) ) S 2(( s t + 1)z; n) + 2S 2( s t + 1; n) S 2(1 s; n) + S 2((1 t)w; n) S 2(1 t; n)