Iterated Binomial Sums and their Associated Iterated Integrals - PowerPoint PPT Presentation

Connection between these structures 11 integral representation (inv. Mellin transform) C-Logs H-Logs M-Logs S-Sums H-Sums C-Sums � 1 � S (2 , 1 , − 1) ( n ) H (4 , 1) , (0 , 0) ( x ) S 1 , 2 2 , 1; n S − 1 , 2 ( n ) H − 1 , 1 ( x ) H 2 , 3 ( x ) x → c ∈ R Mellin transform n → ∞ x → 1 x → 1 � � 1 S − 1 , 2 ( ∞ ) S (2 , 1 , − 1) ( ∞ ) H (4 , 1) , (0 , 0) (1) H − 1 , 1 (1) H 2 , 3 ( c ) S 1 , 2 2 , 1; ∞ power series expansion

Iterated Integrals 12 � x � y 1 � y 2 � y k − 1 f 1 ( y 1 ) f 2 ( y 2 ) f 3 ( y 3 ) · · · f k ( y k ) dy k · · · dy 3 dy 2 dy 1 0 0 0 0 1 ◮ f i ( y ) = y − a i , a i ∈ {− 1 , 0 , 1 } harmonic polylogarithms 1 ◮ f i ( y ) = y − a i , a i ∈ R multiple polylogarithms y ji ◮ f i ( y ) = Φ ai ( y ) , j i ∈ N cyclotomic polylogarithms

Iterated Integrals 13 sign (1 − a − 0) f a ( x ) := , x − a f a 1 ( x ) 1 / 2 . . . f a k ( x ) 1 / 2 f { a 1 ,...,a k } ( x ) := k ≥ 2 , f a 0 ( x ) f a 1 ( x ) 1 / 2 . . . f a k ( x ) 1 / 2 f ( a 0 , { a 1 ,...,a k } ) ( x ) := k ≥ 1 , x j f ( a 1 ,...,a k ) ( x ) f ( { a 1 ,...,a k } ,j ) ( x ) := j ∈ { 1 , . . . , k − 2 } . Restricting to at most two root-singularities we are left with the following cases: sign (1 − a − 0) f a ( x ) := , x − a � f ( a, { b } ) ( x ) := f a ( x ) f b ( x ) , � � f { a,b } ( x ) := f a ( x ) f b ( x ) , � � f ( a, { b,c } ) ( x ) := f a ( x ) f b ( x ) f c ( x ) .

Iterated Integrals 14 Examples � x 1 H (2) ( x ) = 2 − xdx 0

Iterated Integrals 14 Examples � x 1 H (2) ( x ) = 2 − xdx 0 � x 1 H (2 , {− 4 } ) ( x ) = (2 − x ) √ x + 4 dx 0

Iterated Integrals 14 Examples � x 1 H (2) ( x ) = 2 − xdx 0 � x 1 H (2 , {− 4 } ) ( x ) = (2 − x ) √ x + 4 dx 0 � x 1 H (2 , {− 4 , 4 } ) ( x ) = (2 − x ) √ x + 4 √ 4 − xdx 0

Iterated Integrals 14 Examples � x 1 H (2) ( x ) = 2 − xdx 0 � x 1 H (2 , {− 4 } ) ( x ) = (2 − x ) √ x + 4 dx 0 � x 1 H (2 , {− 4 , 4 } ) ( x ) = (2 − x ) √ x + 4 √ 4 − xdx 0 � x � y 1 1 H (2 , {− 4 } ) , (4 , {− 1 } ) ( x ) = (2 − x ) √ x + 4 (4 − y ) √ 1 + ydydx 0 0

Iterated Integrals 14 Examples � x 1 H (2) ( x ) = 2 − xdx 0 � x 1 H (2 , {− 4 } ) ( x ) = (2 − x ) √ x + 4 dx 0 � x 1 H (2 , {− 4 , 4 } ) ( x ) = (2 − x ) √ x + 4 √ 4 − xdx 0 � x � y 1 1 H (2 , {− 4 } ) , (4 , {− 1 } ) ( x ) = (2 − x ) √ x + 4 (4 − y ) √ 1 + ydydx 0 0 � x � y � y 1 1 1 H (2) , ( {− 1 } ) , ( {− 1 } ) ( x ) = √ 1 + y √ 1 + z dzdydx (2 − x ) 0 0 0

Shuffle Algebra 15 � H p ( x ) H q ( x ) = H r ( x ) r = p ∃ q here p q represent all merges of p and q in which the relative ∃ orders of the elements of p and q are preserved.

Shuffle Algebra 15 � H p ( x ) H q ( x ) = H r ( x ) r = p ∃ q here p q represent all merges of p and q in which the relative ∃ orders of the elements of p and q are preserved. H a 1 ,a 2 ( x ) H b 1 ,b 2 ( x ) = H a 1 ,a 2 ,b 1 ,b 2 ( x ) + H a 1 ,b 1 ,a 2 ,b 2 ( x ) + H a 1 ,b 1 ,b 2 ,a 2 ( x ) + H b 1 ,b 2 ,a 1 ,a 2 ( x ) + H b 1 ,a 1 ,b 2 ,a 2 ( x ) + H b 1 ,a 1 ,a 2 ,b 2 ( x ) There are no additional algebraic relations among the iterated integrals if the alphabet is chosen carefully.

16 Mellin transform

Mellin transform of D-finite functions 17 ◮ Let K be a field of characteristic 0.

Mellin transform of D-finite functions 17 ◮ Let K be a field of characteristic 0. ◮ A function f = f ( x ) is called D-finite if there exist p d ( x ) , p d − 1 ( x ) , . . . , p 0 ( x ) ∈ K [ x ] (not all p i = 0 ) such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 .

Mellin transform of D-finite functions 17 ◮ Let K be a field of characteristic 0. ◮ A function f = f ( x ) is called D-finite if there exist p d ( x ) , p d − 1 ( x ) , . . . , p 0 ( x ) ∈ K [ x ] (not all p i = 0 ) such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . ◮ A sequence ( f n ) n ≥ 0 ∈ K N is called P-finite if there exist p d ( n ) , p d − 1 ( n ) , . . . , p 0 ( n ) ∈ K [ n ] (not all p i = 0 ) such that p d ( n ) f n + d + · · · + p 1 ( n ) f n +1 + p 0 ( n ) f n = 0 .

Mellin transform of D-finite functions 17 ◮ Let K be a field of characteristic 0. ◮ A function f = f ( x ) is called D-finite if there exist p d ( x ) , p d − 1 ( x ) , . . . , p 0 ( x ) ∈ K [ x ] (not all p i = 0 ) such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . ◮ A sequence ( f n ) n ≥ 0 ∈ K N is called P-finite if there exist p d ( n ) , p d − 1 ( n ) , . . . , p 0 ( n ) ∈ K [ n ] (not all p i = 0 ) such that p d ( n ) f n + d + · · · + p 1 ( n ) f n +1 + p 0 ( n ) f n = 0 . ◮ If f ( x ) is D -finite, then the coefficients f n of the formal power series expansion ∞ � f n x n f ( x ) = n =0 form a P -finite sequence.

Mellin transform of D-finite functions 17 ◮ Let K be a field of characteristic 0. ◮ A function f = f ( x ) is called D-finite if there exist p d ( x ) , p d − 1 ( x ) , . . . , p 0 ( x ) ∈ K [ x ] (not all p i = 0 ) such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . ◮ A sequence ( f n ) n ≥ 0 ∈ K N is called P-finite if there exist p d ( n ) , p d − 1 ( n ) , . . . , p 0 ( n ) ∈ K [ n ] (not all p i = 0 ) such that p d ( n ) f n + d + · · · + p 1 ( n ) f n +1 + p 0 ( n ) f n = 0 . ◮ If f ( x ) is D -finite, then the coefficients f n of the formal power series expansion ∞ � f n x n f ( x ) = n =0 form a P -finite sequence. ◮ The generating function of a P -finite sequence ( f n ) n ≥ 0 is D -finite.

D-finite to P-finite 18 � f n x n is D-finite such that ◮ Assume that f ( x ) = n ≥ 0 p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . (1)

D-finite to P-finite 18 � f n x n is D-finite such that ◮ Assume that f ( x ) = n ≥ 0 p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . (1) ◮ It is easy to check that j � � x k f ( j ) ( x ) = ( n + i − k ) f n + j − k x n (2) n ≥ 0 i =1

D-finite to P-finite 18 � f n x n is D-finite such that ◮ Assume that f ( x ) = n ≥ 0 p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . (1) ◮ It is easy to check that j � � x k f ( j ) ( x ) = ( n + i − k ) f n + j − k x n (2) n ≥ 0 i =1 ◮ Transform (1) according to this relation.

D-finite to P-finite 18 � f n x n is D-finite such that ◮ Assume that f ( x ) = n ≥ 0 p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . (1) ◮ It is easy to check that j � � x k f ( j ) ( x ) = ( n + i − k ) f n + j − k x n (2) n ≥ 0 i =1 ◮ Transform (1) according to this relation. ◮ Equate coefficients of same powers of x on both sides.

D-finite to P-finite 18 � f n x n is D-finite such that ◮ Assume that f ( x ) = n ≥ 0 p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . (1) ◮ It is easy to check that j � � x k f ( j ) ( x ) = ( n + i − k ) f n + j − k x n (2) n ≥ 0 i =1 ◮ Transform (1) according to this relation. ◮ Equate coefficients of same powers of x on both sides. ◮ We get a linear recurrence equation with polynomial coefficients, satisfied by ( f n ) n ≥ 0 .

D-finite to P-finite 19 Consider � x � τ 1 1 � f n x n . f ( x ) = (1 + τ 1 ) (1 − τ 2 ) dτ 2 dτ 1 = 0 0 n ≥ 0

D-finite to P-finite 19 Consider � x � τ 1 1 � f n x n . f ( x ) = (1 + τ 1 ) (1 − τ 2 ) dτ 2 dτ 1 = 0 0 n ≥ 0 We can derive the differential equation: ( x + 1)( x − 1) f ′′′ ( x ) + (3 x − 1) f ′′ ( x ) + f ′ ( x ) = 0 .

D-finite to P-finite 19 Consider � x � τ 1 1 � f n x n . f ( x ) = (1 + τ 1 ) (1 − τ 2 ) dτ 2 dτ 1 = 0 0 n ≥ 0 We can derive the differential equation: ( x + 1)( x − 1) f ′′′ ( x ) + (3 x − 1) f ′′ ( x ) + f ′ ( x ) = 0 . Expanding leads to x 2 f ′′′ ( x ) − f ′′′ ( x ) + 3 xf ′′ ( x ) − f ′′ ( x ) + f ′ ( x ) = 0 .

D-finite to P-finite 19 Consider � x � τ 1 1 � f n x n . f ( x ) = (1 + τ 1 ) (1 − τ 2 ) dτ 2 dτ 1 = 0 0 n ≥ 0 We can derive the differential equation: ( x + 1)( x − 1) f ′′′ ( x ) + (3 x − 1) f ′′ ( x ) + f ′ ( x ) = 0 . Expanding leads to x 2 f ′′′ ( x ) − f ′′′ ( x ) + 3 xf ′′ ( x ) − f ′′ ( x ) + f ′ ( x ) = 0 . Using (2) results in: ∞ ∞ ∞ ( n + 1) f n +1 x n + 3 n ( n + 1) f n +1 x n + � � � ( n − 1) n ( n + 1) f n +1 x n n =0 n =0 n =0 ∞ ∞ ( n + 1)( n + 2)( n + 3) f n +3 x n = 0 � � ( n + 1)( n + 2) f n +2 x n − − n =0 n =0

D-finite to P-finite 19 Consider � x � τ 1 1 � f n x n . f ( x ) = (1 + τ 1 ) (1 − τ 2 ) dτ 2 dτ 1 = 0 0 n ≥ 0 We can derive the differential equation: ( x + 1)( x − 1) f ′′′ ( x ) + (3 x − 1) f ′′ ( x ) + f ′ ( x ) = 0 . Expanding leads to x 2 f ′′′ ( x ) − f ′′′ ( x ) + 3 xf ′′ ( x ) − f ′′ ( x ) + f ′ ( x ) = 0 . Using (2) results in: ∞ ∞ ∞ ( n + 1) f n +1 x n + 3 n ( n + 1) f n +1 x n + � � � ( n − 1) n ( n + 1) f n +1 x n n =0 n =0 n =0 ∞ ∞ ( n + 1)( n + 2)( n + 3) f n +3 x n = 0 � � ( n + 1)( n + 2) f n +2 x n − − n =0 n =0 Hence ( n + 1) 3 f n +1 − ( n + 2)( n + 1) f n +2 − ( n + 2)( n + 3)( n + 1) f n +3 = 0 holds for ( f n ) n ≥ 0 .

Mellin transform of D-finite functions 20 Let f ( x ) be a D -finite function such that � 1 x n f ( x ) dx M [ f ( x )]( n ) := 0 exists and let p i ( x ) ∈ K [ x ] such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 .

Mellin transform of D-finite functions 20 Let f ( x ) be a D -finite function such that � 1 x n f ( x ) dx M [ f ( x )]( n ) := 0 exists and let p i ( x ) ∈ K [ x ] such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . Since we have ( − 1) p ( n + m )! M [ x m f ( p ) ( x )]( n ) = ( n + m − p )! M [ f ( x )]( n + m − p ) p − 1 ( − 1) i ( n + m )! � ( n + m − i )! f ( p − 1 − i ) (1) . + i =0

Mellin transform of D-finite functions 20 Let f ( x ) be a D -finite function such that � 1 x n f ( x ) dx M [ f ( x )]( n ) := 0 exists and let p i ( x ) ∈ K [ x ] such that p d ( x ) f ( d ) ( x ) + · · · + p 1 ( x ) f ′ ( x ) + p 0 ( x ) f ( x ) = 0 . Since we have ( − 1) p ( n + m )! M [ x m f ( p ) ( x )]( n ) = ( n + m − p )! M [ f ( x )]( n + m − p ) p − 1 ( − 1) i ( n + m )! � ( n + m − i )! f ( p − 1 − i ) (1) . + i =0 We can conclude: Proposition If the Mellin transform of a D -finite function is defined i.e., the � 1 0 x n f ( x ) dx exist, then it is P -finite. integral

Mellin transform of D-finite functions 21 Given a D -finite function f ( x ) . Find an expression F ( n ) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have � 1 x n f ( x ) dx = F ( n ) . M [ f ( x )]( n ) := 0

Mellin transform of D-finite functions 21 Given a D -finite function f ( x ) . Find an expression F ( n ) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have � 1 x n f ( x ) dx = F ( n ) . M [ f ( x )]( n ) := 0 Method: 1. Compute a D -finite differential equation for f ( x ) .

Mellin transform of D-finite functions 21 Given a D -finite function f ( x ) . Find an expression F ( n ) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have � 1 x n f ( x ) dx = F ( n ) . M [ f ( x )]( n ) := 0 Method: 1. Compute a D -finite differential equation for f ( x ) . 2. Use the proposition above to compute a P -finite recurrence for M [ f ( x )]( n ) .

Mellin transform of D-finite functions 21 Given a D -finite function f ( x ) . Find an expression F ( n ) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have � 1 x n f ( x ) dx = F ( n ) . M [ f ( x )]( n ) := 0 Method: 1. Compute a D -finite differential equation for f ( x ) . 2. Use the proposition above to compute a P -finite recurrence for M [ f ( x )]( n ) . 3. Compute initial values for the recurrence.

Mellin transform of D-finite functions 21 Given a D -finite function f ( x ) . Find an expression F ( n ) given as a linear combination of indefinite nested sums such that for all n ∈ N (from a certain point on) we have � 1 x n f ( x ) dx = F ( n ) . M [ f ( x )]( n ) := 0 Method: 1. Compute a D -finite differential equation for f ( x ) . 2. Use the proposition above to compute a P -finite recurrence for M [ f ( x )]( n ) . 3. Compute initial values for the recurrence. 4. Solve the recurrence to get a closed form representation for M [ f ( x )]( n ) .

Mellin transform of D-finite functions 22 We want to compute the Mellin transform of √ 1 − τ � x f ( x ) := dτ. 1 + τ 0

Mellin transform of D-finite functions 22 We want to compute the Mellin transform of √ 1 − τ � x f ( x ) := dτ. 1 + τ 0 We find that ( − 3 + x ) f ( x ) ′ + 2( − 1 + x )(1 + x ) f ( x ) ′′ = 0

Mellin transform of D-finite functions 22 We want to compute the Mellin transform of √ 1 − τ � x f ( x ) := dτ. 1 + τ 0 We find that ( − 3 + x ) f ( x ) ′ + 2( − 1 + x )(1 + x ) f ( x ) ′′ = 0 which leads to the recurrence √ 1 − τ � 1 6 dτ = − 2( n − 1) n M [ f ( x )]( n − 2) + 3 n M [ f ( x )]( n − 1) 1 + τ 0 +( n + 1)(2 n + 3) M [ f ( x )]( n ) .

Mellin transform of D-finite functions 22 We want to compute the Mellin transform of √ 1 − τ � x f ( x ) := dτ. 1 + τ 0 We find that ( − 3 + x ) f ( x ) ′ + 2( − 1 + x )(1 + x ) f ( x ) ′′ = 0 which leads to the recurrence √ 1 − τ � 1 6 dτ = − 2( n − 1) n M [ f ( x )]( n − 2) + 3 n M [ f ( x )]( n − 1) 1 + τ 0 +( n + 1)(2 n + 3) M [ f ( x )]( n ) . Initial values can be computed easily

Mellin transform of D-finite functions 22 We want to compute the Mellin transform of √ 1 − τ � x f ( x ) := dτ. 1 + τ 0 We find that ( − 3 + x ) f ( x ) ′ + 2( − 1 + x )(1 + x ) f ( x ) ′′ = 0 which leads to the recurrence √ 1 − τ � 1 6 dτ = − 2( n − 1) n M [ f ( x )]( n − 2) + 3 n M [ f ( x )]( n − 1) 1 + τ 0 +( n + 1)(2 n + 3) M [ f ( x )]( n ) . Initial values can be computed easily and solving the recurrence leads to √ 1 − τ � 1 � � 4 n +1 1+ τ dτ − 2 0 ( − 1) n M [ f ( x )]( n ) = � + � 2 n n + 1 (2 n + 1)(2 n + 3) n √ 1 − τ 4( − 1) n � n 4 i � 1 1+ τ dτ i =1 (2 i +1) ( 2 i i ) 0 − + . n + 1 n + 1

Mellin transform of D-finite functions 23 ◮ we have a general method to express the Mellin transform of nested integrals which are D -finite to indefinite nested sums and products.

Mellin transform of D-finite functions 23 ◮ we have a general method to express the Mellin transform of nested integrals which are D -finite to indefinite nested sums and products. ◮ we can exploit the structure of our nested integrals instead.

Mellin transform of D-finite functions 23 ◮ we have a general method to express the Mellin transform of nested integrals which are D -finite to indefinite nested sums and products. ◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform.

Mellin transform of D-finite functions 23 ◮ we have a general method to express the Mellin transform of nested integrals which are D -finite to indefinite nested sums and products. ◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform. � 1 − ε � 1 − ε � � 1 x N f ( x ) dx (1 − ε ) N +1 f (1 − ε ) − dxx N +1 f ′ ( x ) = N + 1 0 0

Mellin transform of D-finite functions 23 ◮ we have a general method to express the Mellin transform of nested integrals which are D -finite to indefinite nested sums and products. ◮ we can exploit the structure of our nested integrals instead. ◮ we get direct rewrite rules to compute the Mellin transform. � 1 − ε � 1 − ε � � 1 x N f ( x ) dx (1 − ε ) N +1 f (1 − ε ) − dxx N +1 f ′ ( x ) = N + 1 0 0 � 1 − ε � � 1 − ε x N f ( x ) (4 a ) N dx f ( x ) √ x − a dx = √ x − a � 2 N � (2 N + 1) 0 0 N N � 2 i � � √ � i 1 − a − ε (1 − ε ) i f (1 − ε ) +2 (4 a ) i i =1 � 1 − ε �� dxx i √ x − af ′ ( x ) . − 0

Mellin transform of D-finite functions 24 1 � � � � H ∗ xh 1 ( x )H ∗ M h 1 ,..., h k ( x ) ( N ) = N + 1 M h 2 ,..., h k ( x ) ( N ) � � H ∗ h 1 ,..., h k ( x ) � √ xh 1 ( x )H ∗ 1 � M √ x ( N ) = N + 1 / 2 M h 2 ,..., h k ( x ) ( N ) � � � � 1 H ∗ H ∗ h 1 ,..., h k ( x ) (4 a ) N h 1 ,..., h k ( x ) M √ x − a ( N ) = dx √ x − a + � 2 N � (2 N + 1) 0 N � N � 2 i � � √ � � i x − ah 1 ( x )H ∗ +2 (4 a ) i M h 2 ,..., h k ( x ) ( i ) i =1 � � �� � 1 H ∗ � N � 2 N H ∗ h 1 ,..., h k ( x ) h 1 ,..., h k ( x ) � a M ( N ) = dx + � � 4 N x ( x − a ) x ( x − a ) 0 �� � � N (4 /a ) i x − a � h 1 ( x )H ∗ + � M h 2 ,..., h k ( x ) ( i ) � 2 i x i i =1 i

Some Properties of the Mellin transform 25 ◮ It inherits the linearity from the integral.

Some Properties of the Mellin transform 25 ◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x , i.e., M [ f ( x )]( N + k ) = M [ x k f ( x )]( N ) .

Some Properties of the Mellin transform 25 ◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x , i.e., M [ f ( x )]( N + k ) = M [ x k f ( x )]( N ) . ◮ As a consequence we have the following summation formula � � � � N x x � c i M [ f ( x )]( i ) = c N M f ( x ) ( N ) − M f ( x ) (0) . x − 1 x − 1 c c i =1

Some Properties of the Mellin transform 25 ◮ It inherits the linearity from the integral. ◮ Shifts in N correspond to multiplication by powers of x , i.e., M [ f ( x )]( N + k ) = M [ x k f ( x )]( N ) . ◮ As a consequence we have the following summation formula � � � � N x x � c i M [ f ( x )]( i ) = c N M f ( x ) ( N ) − M f ( x ) (0) . x − 1 x − 1 c c i =1 ◮ Furthermore, the following properties are immediate, where a > 0 : d m M [ln( x ) m f ( x )] ( N ) = dN m M [ f ( x )]( N ) , 1 M [ f ( ax )]( N ) = a N +1 M [ f ( x ) θ ( a − x )]( N ) , a ≤ 1 , � N + 1 − a � 1 M [ f ( x a )]( N ) = a M [ f ( x )] . a

Some Properties of the Mellin transform 26 ◮ The Mellin-convolution of two real functions is defined by � 1 � 1 f ( x ) ∗ g ( x ) = dx 1 dx 2 δ ( x − x 1 x 2 ) f ( x 1 ) g ( x 2 ) . 0 0 The Mellin transform obeys the relation M [ f ( x ) ∗ g ( x )]( N ) = M [ f ( x )]( N ) · M [ g ( x )]( N ) .

Some Properties of the Mellin transform 26 ◮ The Mellin-convolution of two real functions is defined by � 1 � 1 f ( x ) ∗ g ( x ) = dx 1 dx 2 δ ( x − x 1 x 2 ) f ( x 1 ) g ( x 2 ) . 0 0 The Mellin transform obeys the relation M [ f ( x ) ∗ g ( x )]( N ) = M [ f ( x )]( N ) · M [ g ( x )]( N ) . ◮ The Mellin transformation for functions with + -prescription � 1 dx ( x N − 1) f ( x ) . M [[ f ( x )] + ]( N ) = 0 has similar properties.

27 Inverse Mellin transform

Inverse Mellin transform 28 Aim: represent our nested sums in terms of Mellin transforms in the form k � c N c 0 + j M [ f j ( x )]( N ) , (3) j =1 where the constants c j and functions f j ( x ) do not depend on N .

Inverse Mellin transform 28 Aim: represent our nested sums in terms of Mellin transforms in the form k � c N c 0 + j M [ f j ( x )]( N ) , (3) j =1 where the constants c j and functions f j ( x ) do not depend on N . ◮ achieved by virtue of the properties of the Mellin transform.

Inverse Mellin transform 28 Aim: represent our nested sums in terms of Mellin transforms in the form k � c N c 0 + j M [ f j ( x )]( N ) , (3) j =1 where the constants c j and functions f j ( x ) do not depend on N . ◮ achieved by virtue of the properties of the Mellin transform. ◮ as starting point we only need the following basic integral representations: � 1 � 1 = M ( N ) N x � � � 2 N � 4 N 1 = π M ( N ) � N x (1 − x ) � � 1 1 1 x √ 1 − x = 4 N M ( N ) . � 2 N � N N

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows:

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards.

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum i j i k − 1 N � � � a j ( i j ) a j +1 ( i j +1 ) · · · a k ( i k ) (4) i j =1 i j +1 =1 i k =1 we first set up an integral rep. for a j ( N ) of the form (3).

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum i j i k − 1 N � � � a j ( i j ) a j +1 ( i j +1 ) · · · a k ( i k ) (4) i j =1 i j +1 =1 i k =1 we first set up an integral rep. for a j ( N ) of the form (3). ◮ This may require the computation of Mellin convolutions.

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum i j i k − 1 N � � � a j ( i j ) a j +1 ( i j +1 ) · · · a k ( i k ) (4) i j =1 i j +1 =1 i k =1 we first set up an integral rep. for a j ( N ) of the form (3). ◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of i k − 1 N � � a j ( N ) a j +1 ( i j +1 ) · · · a k ( i k ) (5) i j +1 =1 i k =1 by Mellin convolution with the result for the inner sums.

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum i j i k − 1 N � � � a j ( i j ) a j +1 ( i j +1 ) · · · a k ( i k ) (4) i j =1 i j +1 =1 i k =1 we first set up an integral rep. for a j ( N ) of the form (3). ◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of i k − 1 N � � a j ( N ) a j +1 ( i j +1 ) · · · a k ( i k ) (5) i j +1 =1 i k =1 by Mellin convolution with the result for the inner sums. ◮ By the summation property we obtain an integral representation for (4).

Inverse Mellin transform 29 We obtain integral rep. for nested sums as follows: ◮ Starting from the innermost sum we move outwards. ◮ For each intermediate sum i j i k − 1 N � � � a j ( i j ) a j +1 ( i j +1 ) · · · a k ( i k ) (4) i j =1 i j +1 =1 i k =1 we first set up an integral rep. for a j ( N ) of the form (3). ◮ This may require the computation of Mellin convolutions. ◮ We obtain an integral representation of the same form of i k − 1 N � � a j ( N ) a j +1 ( i j +1 ) · · · a k ( i k ) (5) i j +1 =1 i k =1 by Mellin convolution with the result for the inner sums. ◮ By the summation property we obtain an integral representation for (4). ◮ Repeat until the outermost sum has been processed.

Inverse Mellin transform 30 � 1 N dx ( x 4 ) N − 1 1 1 � = √ 1 − x � � x − 4 2 i 0 i =1 i i � 1 � � N dx ( − 4 x ) N − 1 1 2 i 1 � ( − 1) i H ∗ = w 1 ( x ) x + 1 i i π 0 4 i =1 � 1 N i � � ln( x ) + H ∗ dx ( − 2 x ) N − 1 1 2 j � w 28 ( x ) � � � ( − 2) j = − √ x + 1 � � j 6 2 2 i 0 2 i =1 i 2 j =1 i � 1 dx ( x 4 ) N − 1 − 2 H ∗ w 3 ( x ) . 3 x − 4 0

31 Asymptotic Expansion of Nested Sums

Asymptotic Expansion of Nested Sums 32 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 � 1 K � a k � R K ( x ) = f ( x ) − x k = o , x → ∞ . x K k =0

Asymptotic Expansion of Nested Sums 33 Why do we need these expansions of nested sums? E.g., ◮ for limits of the form � � S 2 ( n ) − ζ 2 − S 2 , 2 ( n ) + 7 ζ 2 2 n →∞ n lim 10 ◮ for the approximation of the values of analytic continued nested sums at the complex plane S 2 , − 3 ( − 2 . 5 + 2 i )

Asymptotic Expansion of Nested Sums 34 Basic Idea � �� � � 1 H 1 , 0 , 0 ( x ) ( − 1) n x n S − 1 , 3 ( n ) = dx + const 1 + x 0 � �� �

Asymptotic Expansion of Nested Sums 34 Basic Idea ϕ ( x ) � �� � � 1 H 1 , 0 , 0 ( x ) ( − 1) n x n S − 1 , 3 ( n ) = dx + const 1 + x 0 � �� �

Asymptotic Expansion of Nested Sums 34 Basic Idea ∞ � a k x k ϕ ( x ) − → ϕ (1 − x ) = k =0 � �� � � 1 H 1 , 0 , 0 ( x ) ( − 1) n x n S − 1 , 3 ( n ) = dx + const 1 + x 0 � �� �

Asymptotic Expansion of Nested Sums 34 Basic Idea ∞ � a k x k ϕ ( x ) − → ϕ (1 − x ) = k =0 � �� � � 1 H 1 , 0 , 0 ( x ) ( − 1) n x n S − 1 , 3 ( n ) = dx + const 1 + x 0 � �� � ∞ a k +1 k ! � n ( n + 1) . . . ( n + k ) k =0

Asymptotic Expansion of Nested Sums 34 Basic Idea ∞ � a k x k ϕ ( x ) − → ϕ (1 − x ) = k =0 � �� � � 1 H 1 , 0 , 0 ( x ) ( − 1) n x n S − 1 , 3 ( n ) = dx + const 1 + x 0 � �� � ∞ ∞ a k +1 k ! b k � � n ( n + 1) . . . ( n + k ) = n k k =0 k =1