Symbolic summation for Feynman integrals

Transcription

1 Symbolic summation for Feynman integrals Flavia Stan Research Institute for Symbolic Computation (RISC) Johannes Kepler University Linz, Austria INRIA Rocquencourt, Project Algorithms Université Paris 13, Seminar CALIN 22. March 2010

2 A Feynman parameter integral I = where N N and ɛ > 0. (1 w) 1 ɛ/2 z ɛ/2 (1 z) ɛ/2 (1 wz) 1 ɛ (1 w) N+1 dw dz,

3 Symbolic summation methods Feynman integrals as multiple sums Feynman integrals as Mellin-Barnes integrals

4 Part 1: Symbolic summation methods

5 Hypergeometric series A simple example 3F 2 ( a, b, c d, e ) ; x = n=0 (a) n (b) n (c) n (d) n (e) n x n n! = 1 Γ(d)Γ(e) Γ(b)Γ(c)Γ(d b)γ(e c) t b 1 (1 t) d b 1 u c 1 (1 u) e c 1 (1 utx) a dt du

6 3F 2 ( a, b, c d, e ) ; x = n=0 t n {}}{ (a) n (b) n (c) n x n (d) n (e) n n! where the Pochhammer symbol / rising factorial is (a) n := a(a + 1)... (a + n 1) = if a C and a + n 0, 1, 2,... Γ(a + n) Γ(n) t n+1 t n = (a + n)(b + n)(c + n) (d + n)(e + n) x n + 1

7 An illustrative application for summation methods 1 For n k 1 prove the following identity: 1+( 1) ( )( ) k 1 n n+k ( 1) m n = 2 k m k 1 m m 0 m k ( ) k n m 1 question asked by P. van der Kamp

8 An illustrative application for summation methods 1 For n k 1 prove the following identity: 1+( 1) n+k k 1 Initial values: ( )( ) k 1 n ( 1) m n n = 2 k m k 1 m m=0 if n = 1 then k = 1 if n = 2 then k {1, 2} m=k ( ) k n m 1 question asked by P. van der Kamp

9 Zeilberger s Algorithm 2,3 Given the summation problem B SUM[n] := f (m, n) m=b Zeilberger s Algorithm delivers certificate recurrences of the form c i (n)f (m, n + i) = m [g(m, n)] i S where g(m, n) = rat(m, n)f (m, n) and m [g(m, n)] := g(m + 1, n) g(m, n). 2 D. Zeilberger A fast algorithm for proving terminating hypergeometric identities. Discrete Mathematics, P. Paule, M. Schorn A Mathematica version of Zeilberger s algorithm for proving binomial coefficient identities. J. Symbolic Comput

10 Summing over the certificate recurrences c i (n)f (m, n + i) = m [g(m, n)] i S we obtain recurrences for the sum SUM[n] := B m=b f (m, n) c i (n)sum[n+i] = g(b + 1, n) g(b, n). i S Remark: Zeilberger s algorithm also deals with definite summation problems the bounds b, B can lineary depend on n.

11 Back to the illustrative example 1+( 1) k+n k 1 ( )( ) k 1 n ( 1) m n n ( ) k = 2 k m k 1 m n m m=0 m=k }{{} T (n) We extend the sum T (n) to a summation problem with standard boundary conditions: T (n) := n m=k ( ) k ( ) k (m k + ɛ)! = lim n m ɛ 0 n m (m k)! m= }{{} t(n,ɛ)

12 Zeilberger s algorithm delivers a recurrence for t(n, ɛ) (ɛ+n +1)t(n, ɛ)+(ɛ k +1)t(n + 1, ɛ)+(k n 2)t(n + 2, ɛ) = 0 Taking here ɛ 0, we come to the conclusion that T (n) satisfies (n + 1)T (n) + (1 k)t (n + 1) + (k n 2)T (n + 2) = 0 while for the LHS of the identity we computed the same recurrence (n + 1)SUM[n] + (1 k)sum[n+1] + (k n 2)SUM[n+2] = 0.

13 Zeilberger s algorithm delivers a recurrence for t(n, ɛ) (ɛ+n +1)t(n, ɛ)+(ɛ k +1)t(n + 1, ɛ)+(k n 2)t(n + 2, ɛ) = 0 Taking here ɛ 0, we come to the conclusion that T (n) satisfies (n + 1)T (n) + (1 k)t (n + 1) + (k n 2)T (n + 2) = 0 while for the LHS of the identity we computed the same recurrence (n + 1)SUM[n] + (1 k)sum[n+1] + (k n 2)SUM[n+2] = 0.

14 Zeilberger s algorithm delivers a recurrence for t(n, ɛ) (ɛ+n +1)t(n, ɛ)+(ɛ k +1)t(n + 1, ɛ)+(k n 2)t(n + 2, ɛ) = 0 Taking here ɛ 0, we come to the conclusion that T (n) satisfies (n + 1)T (n) + (1 k)t (n + 1) + (k n 2)T (n + 2) = 0 while for the LHS of the identity we computed the same recurrence (n + 1)SUM[n] + (1 k)sum[n+1] + (k n 2)SUM[n+2] = 0.

15 A multisum example 5 For r k 0, s 0 prove that: r k i=0 j=0 r k = s ( )( r j r + k i i=0 j=0 i j )( k + i s j s ( )( r j + 2 r + k i i j )( ) r s + j = r k i )( k + i s j )( ) r s + j 2 r k i MultiSum 4 delivers the same recurrence for both sides: (2k + r + 2)(2r s + 2)SUM[s,r] ( 2r 2 + 4kr + 8r + 2k 3ks + 8 ) SUM[s+1,r+1] (k r 2)(s + 2)SUM[s+2,r+2] = 0 4 K. Wegschaider, Computer generated proofs of binomial multi-sum identities, Diploma thesis, RISC, Johannes Kepler University, question asked by H. Markwig

16 A multisum example 5 For r k 0, s 0 prove that: r k i=0 j=0 r k = s ( )( r j r + k i i=0 j=0 i j )( k + i s j s ( )( r j + 2 r + k i i j )( ) r s + j = r k i )( k + i s j )( ) r s + j 2 r k i MultiSum 4 delivers the same recurrence for both sides: (2k + r + 2)(2r s + 2)SUM[s,r] ( 2r 2 + 4kr + 8r + 2k 3ks + 8 ) SUM[s+1,r+1] (k r 2)(s + 2)SUM[s+2,r+2] = 0 4 K. Wegschaider, Computer generated proofs of binomial multi-sum identities, Diploma thesis, RISC, Johannes Kepler University, question asked by H. Markwig

17 More general: WZ-summation methods WZ-summation methods deliver recurrences in elements of µ = (µ 1,..., µ l ) for the multiple sum: Sum(µ) := κ 1 κ r F (µ, κ 1,..., κ r ) where F (µ, κ) is a hypergeometric term in the variables µ = (µ 1,..., µ l ) and κ = (κ 1,..., κ r ). Remark: for r = 1 and l = 1 we have Zeilberger s Algorithm

18 Given a κ-free recurrence 6 satisfied by the summand a u,v (µ, α) F(µ + u, κ + v) = 0 (u,v) S one computes a certificate recurrence 7 where c m (µ) F(µ + m, κ) = m S r κi p j,k (µ, κ, ) F(µ + j, κ + k), i=1 (m,k) S i κi F (µ, κ) := F (µ, κ 1,..., κ i + 1,..., κ r ) F (µ, κ). 6 M.C. Fasenmyer, Some generalized hypergeometric polynomials, PhD thesis, Univ. of Michigan, H.S. Wilf and D. Zeilberger, An algorithmic proof theory for hypergeometric (ordinary and q) multisum/integral identities, Inventiones Mathematicae, 1992

19 Summary of Part 1 For the summation problem Sum(µ) := κ 1 κ r F (µ, κ 1,..., κ r ) WZ-methods give a certificate recurrence of the form m S ( c m (µ) F(µ + m, κ) = r κi i=1 (j,k) S i p j,k (µ, κ) F(µ + j, κ + k) ) By summing over the certificate recurrence we obtain a recurrence for the sum c m (µ) Sum (µ + m) = RHS Remark: m S κi F (µ, κ) := F (µ, κ 1,..., κ i + 1,..., κ r ) F (µ, κ)

20 Part 2: Feynman integrals as multiple sums 8 8 joint work with Carsten Schneider (RISC) and Johannes Blümlein (DESY)

21 Back to the Feynman integral example I = where N N and ɛ > 0. (1 w) 1 ɛ/2 z ɛ/2 (1 z) ɛ/2 (1 wz) 1 ɛ (1 w) N+1 dw dz,

22 .. seen as a hypergeometric series 1 I = (1 w) N ɛ/2 z ɛ/2 (1 z) ɛ/2 (1 wz) 1 ɛ dw dz = Γ ( 1 ɛ ) ( ) 2 Γ 1 + ɛ ( ) 2 1, 1 ɛ, ( ɛ N + 1 ɛ ) 2 3F 2 2, N + 2 ɛ ; }{{ ( ) } (1 ɛ) σ 1 + ɛ 2 ( ) σ (2) σ 0 σ N + 2 ɛ 2 σ }{{} SUM[N]

23 MultiSum 9 computes a certificate recurrence satisfied by its summand Out[0]= certrec = (N+1)(2N ɛ+2)f [N, σ] (N ɛ+1)(2n+ɛ+2)f [N+1, σ] = σ[( σ 1)(2N ɛ + 2)F [N, σ]] The sum representation of I, denoted by SUM[N], satisfies the recurrence Out[0]= rec = (N + 1)(2N ɛ + 2)SUM[N] (N ɛ + 1)(2N + ɛ + 2)SUM[N+1] == Γ ( 1 ɛ 2 ) Γ ( ɛ 2 + 1). 9 K. Wegschaider, Computer generated proofs of binomial multi-sum identities, Diploma thesis, RISC, Johannes Kepler University, 1997

24 Unfolding recurrences Given the first order inhomogeneous recurrence SUM[N+1] c(n)sum[n] = h(n) one can write SUM[N+1] =c(n)sum[n] + h(n) =c(n) (c(n 1)SUM[N-1] + h(n 1)) + h(n) =... ( n ) n 1 i SUM[N+1] = c(i) SUM[1] + h(n i) c(j) + h(n) i=1 i=1 j=0

25 ΠΣ extensions 10 n ih i = i=1 n i i=1 i j=1 1 j =? K(i, H i ) σ(h i ) = H i + 1 i+1 K(i) σ(i) = i + 1 K = Q Solve in the field K(i, H i ), σ the difference equation: (g) = σ(g) g = i H i Sigma delivers the solution: n ih i =g(n + 1) g(1) = i=1 g = n(n 1) 4 i(i 1) 4 (2H n 1) (2H i 1) 10 C. Schneider Symbolic Summation Assists Combinatorics, Sém. Lothar. Combin. 2007

26 Higher order recurrences SUM[n+2] α(n)sum[n+1] β(n)sum[n] = h(n) can be rewritten using operator notation ( S 2 n α(n)s n β(n)i ) SUM[n] = h(n) Using Hyper 11 we find a factorization of the form if it exists. (S n c(n)i) (S n d(n)i)sum[n] = h(n) 11 M. Petkovsek Hypergeometric solutions of linear recurrences with polynomial coefficients J. Symbolic Comp. 1992

27 Higher order recurrences SUM[n+2] α(n)sum[n+1] β(n)sum[n] = h(n) can be rewritten using operator notation ( S 2 n α(n)s n β(n)i ) SUM[n] = h(n) Using Hyper 11 we find a factorization of the form (S n c(n)i) (S n d(n)i)sum[n] = h(n) }{{} Ψ[n] 11 M. Petkovsek Hypergeometric solutions of linear recurrences with polynomial coefficients J. Symbolic Comp. 1992

28 A realistic example U (N, ɛ) := ( 1) N N 3 σ 0 =0 j 0 =0 N j 0 3 j 1 =0 j 0 +1 j 2 =0 ( j0 )( +1 N j0 ) 3 j 2 j 1 ( ɛ 2 +1) σ 0 ( ɛ) σ0 (j 1 +j 2 +3) σ0 (3 ɛ 2) j1 (j 1 +4) σ0 ( ɛ 2 +j 1+j 2 +4) σ0 (4 ɛ 2) j1 +j 2 Γ(j 1+j 2 +2)Γ(j 1 +j 2 +3)Γ(N j 0 1)Γ(N j 1 j 2 1) Γ(σ 0 +1)Γ(j 1 +4)Γ(N j 0 2) where N 3 is the Mellin moment and ɛ > 0 is the dimension regularization parameter.

29 Setting up the inhomogeneous recurrence E.g. SUM[N] = N j 0 σ=0 j 0=0 j 1=0 F[N, σ, j 0, j 1 ] and MultiSum delivers the certificate recurrence F[N + 1, σ, j 0, j 1 ] + (N + 2)F[N, σ, j 0, j 1 ] = = j0 [(j 1 j 0 + 1)F[N + 1, σ, j 0, j 1 ] + F[N, σ, j 0, j 1 ]] N j 0 σ=0 j 0=0 j 1=0

30 N j 0 σ=0 j 0=0 j 1=0 F(N + 1, σ, j 0, j 1 ) = SUM[N + 1] N+1 σ=0 j 1=0 F(N + 1, σ, N + 1, j 1 ) } {{ } shift compensation

31 j 0 σ=0 j 1=0 j0 [(j 1 j 0 + 1)F(N + 1, σ, j 0, j 1 )] j 0 =N+1 k=0 = j 0 j 0 =N+1 = (j 1 j 0 + 1)F(N + 1, σ, j 0, j 1 ) σ=0 j 1=0 j 0 =0 }{{} -boundary sums N+1 F(N + 1, σ, j 0, j 0 ) σ=0 j 0=1 } {{ } telescoping compensation

32 The inhomogeneous recurrence + + SUM[N + 1] + (N + 2)SUM[N] = N+1 σ=0 j 1=0 j 0 (j 1 j 0 + 1)F[N + 1, σ, j 0, j 1 ] σ=0 j 1=0 j 0 σ=0 j 1=0 F[N, σ, j 0, j 1 ] j 0=N+1 j 0=0 N+1 σ=0 j 0=1 F[N + 1, σ, N + 1, j 1 ] j 0=N+1 j 0=0 N+1 σ=0 j 0=1 F[N + 1, σ, j 0, j 0 ] F[N, σ, j 0, j 0 ] where the RHS of the recurrence contains shift and telescoping compensating sums as well as delta boundary sums.

33 SUM[N] := N j 0 σ=0 j 0=0 j 1=0 F[N, σ, j 0, j 1 ]

34 N j 0 σ=0 j 0=0 j 1=0 F[N, σ, j 0 + 1, j 1 ] = σ=0 N+1 j 0=1 j 0 1 j 1=0 F[N, σ, j 0, j 1 ]

35 N j 0 σ=0 j 0=0 j 1=0 F[N + 1, σ, j 0 + 1, j 1 ] = σ=0 N+1 j 0=1 j 0 1 j 1=0 F[N + 1, σ, j 0, j 1 ]

36 SUM[N] = + σ=0 j 1=0 N 1 j 0 σ=0 j 0=0 j 1=0 N F[N, σ, N, j 1 ] F[N, σ, j 0, j 1 ]

37 Part 3: Feynman integrals as Mellin-Barnes integrals

38 The Mellin transform The Mellin transform of a locally integrable function f : (0, ) C is f (s) = 0 x s 1 f (x)dx =: M[f ; s] defined usually on an infinite strip a < Re(s) < b. The inversion formula f (x) = 1 2πi c+i c i x s f (s)ds has a contour of integration placed in the strip of analyticity a < c < b.

39 For example 1 (1 x) a = 1 Γ ( s) Γ (a + s) ( x) s ds 2πi B Γ (a) where the contour B separates the set of ascending poles {0, 1, 2, 3,... } from the set of descending poles of the integrand { a, a 1, a 2,... }

40 Let s ask the question Γ(a + s)γ( s) ( z) s ds =? B Γ(a) }{{} Ψ z,a(s) B B is a Barnes contour of integration puting the poles of Γ (a + s) to left and those of Γ ( s) to the right

41 S n Answer: B Γ(a + s)γ( s) ( z) s ds =? Γ(a) }{{} Ψ z,a(s) B n n C n := B n S n for all n 0 and lim Ψ z,a (s)ds = 0 n S n B then Ψ z,a (s)ds = lim n C n Ψ z,a (s)ds }{{} ( 2πi) n Res(Ψ(t),k) k=0

42 Back to our Feynman integral I = where N N and ɛ > 0. (1 w) 1 ɛ/2 z ɛ/2 (1 z) ɛ/2 (1 wz) 1 ɛ (1 w) N+1 dw dz,

43 1 I = 0 dw 1 0 dz (1 w) N ɛ/2 z ɛ/2 (1 z) ɛ/2 1 Γ ( s) Γ (1 ɛ + s) ( wz) s ds 2πi C s Γ (1 ɛ) }{{} 1/(1 wz) 1 ɛ = 1 Γ ( s) Γ (1 ɛ + s) ( 1) s 2πi C s Γ (1 ɛ) 1 0 z ɛ/2+s (1 z) ɛ/2 dz } {{ } B(ɛ/2+s+1, ɛ/2+1) 1 0 w s (1 w) N ɛ/2 dw ds } {{ } B(s+1,N ɛ/2+1)

44 I = Γ ( 1 ɛ ) ( 2 Γ N ɛ 2 + 1) 4πiΓ(1 ɛ) ( ) Γ(s + 1 ɛ)γ s ɛ 2 C s (s + 1)Γ ( N + s ɛ 2 + 2) Γ( s)( 1)s ds MultiSum computes a certificate recurrence satisfied by its integrand Out[0]= certrec = (N +1)(2N ɛ+2)f [N, s] (N ɛ+1)(2n +ɛ+2)f [N +1, s] = s[( s 1)(2N ɛ + 2)F [N, s]]

45 The contour C s separates the ascending chain of poles of Γ( s) from the descending chains coming from Γ(s+1 ɛ)γ(s+1+ ɛ 2) (s+1). 1 ɛ Λ m C s On the RHS we have an improper integral s [( s 1)(2N ɛ + 2)F[N,s]] ds C s = (2N ɛ + 2) lim (s + 1)F[N,s] ds = Γ ( 1 ɛ ) ( m 2 Γ ɛ 2 + 1) Λ m where F[N,s] denotes the integrand.

46 For the Feynman parameter integral I = (1 w) 1 ɛ/2 z ɛ/2 (1 z) ɛ/2 (1 wz) 1 ɛ (1 w) N+1 dw dz, the sum representation satisfies the recurrence Out[0]= rec = (N + 1)(2N ɛ + 2)SUM[N] (N ɛ + 1)(2N + ɛ + 2)SUM[N + 1] == Γ ( 1 ɛ 2 ) Γ ( ɛ 2 + 1) and the Mellin-Barnes integral representation satisfies the same In[1]:= rec/.sum INT Out[1]= (N + 1)(2N ɛ + 2)INT[N] (N ɛ + 1)(2N + ɛ + 2)INT[N + 1] == Γ ( 1 ɛ 2 ) Γ ( ɛ 2 + 1).

47 Recurrences for Mellin-Barnes integrals For the integration problem Int(µ) := B κ1... B κr F (µ, κ 1,..., κ r ) dκ WZ-methods give a certificate recurrence of the form m S ( c m (µ) F(µ + m, κ) = r κi i=1 (j,k) S i p j,k (µ, κ) F(µ + j, κ + k) ) By integrating over the certificate recurrence we obtain a recurrence for the integral c m (µ) Int (µ + m) = RHS m S

48 One more example Identity from the Table x ρ 1 (1 x) β γ n 2F 1 ( n, β γ ) ; x dx = Γ(γ)Γ(ρ)Γ(β γ + 1)Γ(γ ρ + n) Γ(γ + n)γ(γ ρ)γ(β γ + ρ + 1), where n N, Re ρ > 0 and Re(β γ) > n I.S. Gradshteyn, I.M. Ryzhik, Table of Integrals, Series, and Products

49 Recall the Mellin transform slide.. The Mellin transform of a locally integrable function f : (0, ) C is f (s) = 0 x s 1 f (x)dx =: M[f ; s] defined usually on an infinite strip a < Re(s) < b. Remark: For a polynomial function f (x) = (1 x) n we have a = 0 and b = n = no strip of analyticity.

50 A generalized definition of the Mellin transform 13 Decomposing f (x) such that f (x) = f 1 (x) + f 2 (x), e.g., f 1 (x) = { f (x), x [0, 1) 0, x [1, ), f 2 (x) = { 0, x [0, 1) f (x), x [1, ) we have M[f ; z] = M[f 1 ; z] + M[f 2 ; z]. In our case, f (x) = (1 x) n and its Mellin transform is [ ] Γ(z) Γ( n z) M[f ; z] = Γ(n + 1) + ( 1)n. Γ(n + z + 1) Γ(1 z) 13 N. Bleistein, R.A. Handelsman Asymptotic Expansions of Integrals, 1967

51 ..we arrive at the Mellin-Barnes integral representation: 2F 1 ( n, β γ ) ; x = = Γ(γ)Γ(n + 1) Γ(β) 1 2πi [ δ+i δ i +( 1) n η+i Γ(s) Γ(β s) Γ(n + s + 1) Γ(γ s) x s ds η i ] Γ( n s) Γ(β s) Γ(1 s) Γ(γ s) x s ds where Re β > δ > 0 and η < n.

52 Using the Mellin transform and WZ-methods we can now prove identities of the form 14 0 ( )( ) α + m β + n e x x α+β L α m(x)l β n(x)dx = ( 1) m+n (α+β)! n m where L α m(x) are the Laguerre polynomials. 14 Karen Kohl, FS. An algorithmic approach to the Mellin transform method

53 Conclusions real-world applications of symbolic summation methods hypergeometric summation with non-standard bounds computing recurrences for nested Mellin-Barnes integrals