On the method of brackets Armin Straub Tulane University, New - PowerPoint PPT Presentation

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluating bracket series Rule φ n f ( n ) � an + b � = 1 � | a | f ( n ∗ )Γ( − n ∗ ) , n where n ∗ is the solution of the equation an + b = 0 . Example � ∞ � x s − 1 e − x d x = φ n � n + s � = Γ( − n ∗ ) = Γ( s ) . 0 n Example � ∞ � ∞ e − ax 2 d x = � � φ n a n x 2 n d x = φ n a n � 2 n + 1 � � �� � 0 0 n n n ∗ = − 1 / 2 � π = 1 2 a n ∗ Γ( − n ∗ ) = 1 2 a Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ e − αx J 0 ( βx ) d x ◮ 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k ◮ Choosing k as a free variable: m ∗ = − 2 k − 1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k ◮ Choosing k as a free variable: m ∗ = − 2 k − 1 � 2 k Γ( − m ∗ ) � 2 k Γ(2 k + 1) φ k α m ∗ � β � β � � φ k α − 2 k − 1 = 2 k ! 2 k ! k k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k ◮ Choosing k as a free variable: m ∗ = − 2 k − 1 � 2 k Γ( − m ∗ ) � 2 k Γ(2 k + 1) φ k α m ∗ � β � β � � φ k α − 2 k − 1 = 2 k ! 2 k ! k k � � β � 2 k � 2 k = 1 � ( − 1) k α k 2 α k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k ◮ Choosing k as a free variable: m ∗ = − 2 k − 1 � 2 k Γ( − m ∗ ) � 2 k Γ(2 k + 1) φ k α m ∗ � β � β � � φ k α − 2 k − 1 = 2 k ! 2 k ! k k � � β � 2 k � 2 k = 1 1 � ( − 1) k = � α 2 + β 2 α k 2 α k with the series converging for β < α . Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More interesting example ∞ ( − 1) k ( x/ 2) 2 k + ν � ◮ Bessel function: J ν ( x ) = k ! Γ( k + ν + 1) k =0 � ∞ � 2 k 1 � β � e − αx J 0 ( βx ) d x = φ m,k α m k ! � m + 2 k + 1 � ◮ 2 0 m,k ◮ Choosing k as a free variable: m ∗ = − 2 k − 1 � 2 k Γ( − m ∗ ) � 2 k Γ(2 k + 1) φ k α m ∗ � β � β � � φ k α − 2 k − 1 = 2 k ! 2 k ! k k � � β � 2 k � 2 k = 1 1 � ( − 1) k = � α 2 + β 2 α k 2 α k with the series converging for β < α . ◮ Similarly, for m free: � − m − 1 Γ(1 / 2 + m/ 2) � β 1 1 � φ m α m Γ(1 / 2 − m/ 2) = . . . = � α 2 + β 2 2 2 m Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluating higher bracket series Rule φ n f ( n ) � an + b � = 1 � where an ∗ + b = 0 . | a | f ( n ∗ )Γ( − n ∗ ) , n Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluating higher bracket series Rule φ n f ( n ) � an + b � = 1 � where an ∗ + b = 0 . | a | f ( n ∗ )Γ( − n ∗ ) , n Rule (Evaluation) � φ { n } f ( n 1 , . . . , n r ) � a 11 n 1 + · · · a 1 r n r + b 1 � · · · � a r 1 n 1 + · · · a rr n r + b r � { n } 1 | det( A ) | f ( n ∗ 1 , . . . , n ∗ r )Γ( − n ∗ 1 ) · · · Γ( − n ∗ = r ) , where A = ( a ij ) and ( n ∗ i ) such that the brackets vanish. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluating higher bracket series Rule φ n f ( n ) � an + b � = 1 � where an ∗ + b = 0 . | a | f ( n ∗ )Γ( − n ∗ ) , n Rule (Evaluation) � φ { n } f ( n 1 , . . . , n r ) � a 11 n 1 + · · · a 1 r n r + b 1 � · · · � a r 1 n 1 + · · · a rr n r + b r � { n } 1 | det( A ) | f ( n ∗ 1 , . . . , n ∗ r )Γ( − n ∗ 1 ) · · · Γ( − n ∗ = r ) , where A = ( a ij ) and ( n ∗ i ) such that the brackets vanish. Rule (Combining) If there are more summation indices than brackets, free variables are chosen. Each choice produces a series. Those converging in a common region are added. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Multinomial expansions Rule (Multinomial) 1 � s + m 1 + · · · + m r � � φ { m } a m 1 · · · a m r ( a 1 + a 2 + · · · + a r ) s = 1 r Γ( s ) m 1 ,...,m r where φ { m } := φ m 1 · · · φ m r . Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Multinomial expansions Rule (Multinomial) 1 � s + m 1 + · · · + m r � � φ { m } a m 1 · · · a m r ( a 1 + a 2 + · · · + a r ) s = 1 r Γ( s ) m 1 ,...,m r where φ { m } := φ m 1 · · · φ m r . ◮ Follows from the integral representation of Γ( s ) : � ∞ Γ( s ) x s − 1 e − ( a 1 + ... + a r ) x d x ( a 1 + . . . + a r ) s = 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Multinomial expansions Rule (Multinomial) 1 � s + m 1 + · · · + m r � � φ { m } a m 1 · · · a m r ( a 1 + a 2 + · · · + a r ) s = 1 r Γ( s ) m 1 ,...,m r where φ { m } := φ m 1 · · · φ m r . ◮ Follows from the integral representation of Γ( s ) : � ∞ Γ( s ) x s − 1 e − ( a 1 + ... + a r ) x d x ( a 1 + . . . + a r ) s = 0 � ∞ r � � φ m i ( a i x ) m i d x x s − 1 = 0 m i i =1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Multinomial expansions Rule (Multinomial) 1 � s + m 1 + · · · + m r � � φ { m } a m 1 · · · a m r ( a 1 + a 2 + · · · + a r ) s = 1 r Γ( s ) m 1 ,...,m r where φ { m } := φ m 1 · · · φ m r . ◮ Follows from the integral representation of Γ( s ) : � ∞ Γ( s ) x s − 1 e − ( a 1 + ... + a r ) x d x ( a 1 + . . . + a r ) s = 0 � ∞ r � � φ m i ( a i x ) m i d x x s − 1 = 0 m i i =1 � φ { m } a m 1 · · · a m r = � s + m 1 + · · · + m r � . 1 r { m } Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m 1 � = φ j,n,m Γ( − αj ) � n + m − αj � � n + s � � m + t � j,n,m Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m 1 � = φ j,n,m Γ( − αj ) � n + m − αj � � n + s � � m + t � j,n,m n ∗ = − s , m ∗ = − t , j ∗ = − s + t α , | det | = α Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m 1 � = φ j,n,m Γ( − αj ) � n + m − αj � � n + s � � m + t � j,n,m n ∗ = − s , m ∗ = − t , j ∗ = − s + t α , | det | = α = 1 1 Γ( − αj ∗ )Γ( − n ∗ )Γ( − m ∗ )Γ( − j ∗ ) α Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m 1 � = φ j,n,m Γ( − αj ) � n + m − αj � � n + s � � m + t � j,n,m n ∗ = − s , m ∗ = − t , j ∗ = − s + t α , | det | = α � s + t � = 1 Γ( − αj ∗ )Γ( − n ∗ )Γ( − m ∗ )Γ( − j ∗ ) = 1 1 Γ( s )Γ( t ) Γ( s + t ) Γ α α α Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A two-dimensional example � ∞ � ∞ x s − 1 y t − 1 exp ( − ( x + y ) α ) d x d y 0 0 � ∞ � ∞ � x s − 1 y t − 1 ( x + y ) αj d x d y = φ j 0 0 j � ∞ � ∞ φ n,m x n y m � n + m − αj � � x s − 1 y t − 1 � = φ j d x d y Γ( − αj ) 0 0 j n,m 1 � = φ j,n,m Γ( − αj ) � n + m − αj � � n + s � � m + t � j,n,m n ∗ = − s , m ∗ = − t , j ∗ = − s + t α , | det | = α � s + t � = 1 Γ( − αj ∗ )Γ( − n ∗ )Γ( − m ∗ )Γ( − j ∗ ) = 1 1 Γ( s )Γ( t ) Γ( s + t ) Γ α α α ◮ Mathematica 7 cannot evaluate this integral. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional More dimensions ◮ This generalizes to arbitrary dimensions: Theorem � ∞ � ∞ n � exp ( − ( x 1 + . . . + x n ) α ) x s i − 1 · · · d x i i 0 0 i =1 � n � s 1 + . . . + s n � = 1 i =1 Γ( s i ) Γ( s 1 + . . . + s n )Γ α α Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 � ∞ x 2 k + s − 1 � α � 2 k 1 � = φ k (1 + x 2 ) λ d x 2 Γ( k + 1) 0 k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 � ∞ x 2 k + s − 1 � α � 2 k 1 � = φ k (1 + x 2 ) λ d x 2 Γ( k + 1) 0 k 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 � ∞ x 2 k + s − 1 � α � 2 k 1 � = φ k (1 + x 2 ) λ d x 2 Γ( k + 1) 0 k 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m ◮ 3 indices, 2 brackets: 1 free variable Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — k free � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m ◮ k free: m ∗ = − k − s 2 and n ∗ = − λ + k + s 2 . | det | = 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — k free � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m ◮ k free: m ∗ = − k − s 2 and n ∗ = − λ + k + s 2 . | det | = 2 1 � α � 2 k 1 � Γ( k + 1)Γ( − n ∗ )Γ( − m ∗ ) φ k 2Γ( λ ) 2 k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — k free � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m ◮ k free: m ∗ = − k − s 2 and n ∗ = − λ + k + s 2 . | det | = 2 1 � α � 2 k 1 � Γ( k + 1)Γ( − n ∗ )Γ( − m ∗ ) φ k 2Γ( λ ) 2 k ( − 1) k 1 � α � 2 k � Γ( λ − k − s 2 )Γ( k + s = 2 ) ( k !) 2 2Γ( λ ) 2 k Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — k free � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m ◮ k free: m ∗ = − k − s 2 and n ∗ = − λ + k + s 2 . | det | = 2 1 � α � 2 k 1 � Γ( k + 1)Γ( − n ∗ )Γ( − m ∗ ) φ k 2Γ( λ ) 2 k ( − 1) k 1 � α � 2 k � Γ( λ − k − s 2 )Γ( k + s = 2 ) ( k !) 2 2Γ( λ ) 2 k � = Γ( s 2 )Γ( λ − s � � s α 2 2 ) � 2 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — the contributions � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m � � � ◮ k free: Γ( s 2 )Γ( λ − s s α 2 2 ) � 2 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — the contributions � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m � � � ◮ k free: Γ( s 2 )Γ( λ − s s α 2 2 ) � 2 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 2 � Γ( − λ + s � � α 2 2 ) � α � 2 λ − s λ � ◮ n free: 2 ) 1 F 2 � 2Γ( λ + 1 − s 1 + λ − s 2 , 1 + λ − s 2 4 � 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — the contributions � ∞ x s − 1 J 0 ( αx ) (1 + x 2 ) λ d x 0 1 � α � 2 k 1 � = φ k,n,m Γ( k + 1) � n + m + λ � � 2 m + 2 k + s � Γ( λ ) 2 k,n,m � � � ◮ k free: Γ( s 2 )Γ( λ − s s α 2 2 ) � 2 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 2 � Γ( − λ + s � � α 2 2 ) � α � 2 λ − s λ � ◮ n free: 2 ) 1 F 2 � 2Γ( λ + 1 − s 1 + λ − s 2 , 1 + λ − s 2 4 � 2 � − 2 m − s Γ( m + λ )Γ( m + s ( − 1) m � α 2 ) 1 � ◮ m free: Γ(1 − m − s 2Γ( λ ) m ! 2 2 ) m This series diverges. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — harvesting Theorem � ∞ � � � (1 + x 2 ) λ d x = Γ( s 2 )Γ( λ − s x s − 1 s α 2 2 ) � 2 J 0 ( αx ) 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 0 2 � � � Γ( − λ + s α 2 � α � 2 λ − s 2 ) λ � + 2 ) 1 F 2 � 2Γ( λ + 1 − s 1 + λ − s 2 , 1 + λ − s 2 4 � 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — harvesting Theorem � ∞ � � � (1 + x 2 ) λ d x = Γ( s 2 )Γ( λ − s x s − 1 s α 2 2 ) � 2 J 0 ( αx ) 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 0 2 � � � Γ( − λ + s α 2 � α � 2 λ − s 2 ) λ � + 2 ) 1 F 2 � 2Γ( λ + 1 − s 1 + λ − s 2 , 1 + λ − s 2 4 � 2 Corollary ( s = 2 ) � ∞ � λ K λ ( α ) x � α J 0 ( αx ) (1 + x 2 ) λ +1 d x = 2 λ ! 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Bessel integral — harvesting Theorem � ∞ � � � (1 + x 2 ) λ d x = Γ( s 2 )Γ( λ − s x s − 1 s α 2 2 ) � 2 J 0 ( αx ) 1 F 2 � 1 , 1 − λ + s 2Γ( λ ) 4 � 0 2 � � � Γ( − λ + s α 2 � α � 2 λ − s 2 ) λ � + 2 ) 1 F 2 � 2Γ( λ + 1 − s 1 + λ − s 2 , 1 + λ − s 2 4 � 2 Corollary ( s = 2 ) � ∞ � λ K λ ( α ) x � α J 0 ( αx ) (1 + x 2 ) λ +1 d x = 2 λ ! 0 Corollary � ∞ x (1 + x 2 ) 3 / 2 d x = e − α J 0 ( αx ) 0 Armin Straub On the method of brackets

� � � � � � The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Feynman diagram P 1 � � � � a 1 � � a 3 P 2 � � � � � � a 2 � � � � P 3 ◮ Propagator associated to the index a 1 has mass m 2 = ( P 1 + P 3 ) 2 = s ◮ P 2 1 = P 2 3 = 0 and P 2 Armin Straub On the method of brackets

� � � � � � The method of brackets Examples A Feynman diagram Troubles RMT Distributional A Feynman diagram P 1 � � � � a 1 � � a 3 P 2 � � � � � � a 2 � � � � P 3 ◮ Propagator associated to the index a 1 has mass m 2 = ( P 1 + P 3 ) 2 = s ◮ P 2 1 = P 2 3 = 0 and P 2 ◮ D -dimensional representation in Minkowski space is given by � d D q 1 G = [( P 1 + q ) 2 − m 2 ] a 1 [( P 3 − q ) 2 ] a 2 [ q 2 ] a 3 . iπ D/ 2 E. E. Boos and A. I. Davydychev. “A method for evaluating massive Feynman integrals.” Jour. Phys. A , 41 , 1991. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional The associated Feynman integral ◮ Schwinger parametrization leads to G = ( − 1) − D/ 2 H with � 3 j =1 Γ( a j ) ∞ ∞ ∞ x 1 x 2 e x 1 m 2 e − x 1+ x 2+ x 3 s � � � x a 1 − 1 x a 2 − 1 x a 3 − 1 H := ( x 1 + x 2 + x 3 ) D/ 2 d x 1 d x 2 d x 3 . 1 2 3 0 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional The associated Feynman integral ◮ Schwinger parametrization leads to G = ( − 1) − D/ 2 H with � 3 j =1 Γ( a j ) ∞ ∞ ∞ x 1 x 2 e x 1 m 2 e − x 1+ x 2+ x 3 s � � � x a 1 − 1 x a 2 − 1 x a 3 − 1 H := ( x 1 + x 2 + x 3 ) D/ 2 d x 1 d x 2 d x 3 . 1 2 3 0 0 0 ◮ First x n 1 + n 2 x n 2 x 1+ x 2+ x 3 s = x 1 x 2 � e x 1 m 2 e − φ { n } ( − 1) n 1 m 2 n 1 s n 2 1 2 ( x 1 + x 2 + x 3 ) n 2 n 1 ,n 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional The associated Feynman integral ◮ Schwinger parametrization leads to G = ( − 1) − D/ 2 H with � 3 j =1 Γ( a j ) ∞ ∞ ∞ x 1 x 2 e x 1 m 2 e − x 1+ x 2+ x 3 s � � � x a 1 − 1 x a 2 − 1 x a 3 − 1 H := ( x 1 + x 2 + x 3 ) D/ 2 d x 1 d x 2 d x 3 . 1 2 3 0 0 0 ◮ First x n 1 + n 2 x n 2 x 1+ x 2+ x 3 s = x 1 x 2 � e x 1 m 2 e − φ { n } ( − 1) n 1 m 2 n 1 s n 2 1 2 ( x 1 + x 2 + x 3 ) n 2 n 1 ,n 2 ◮ Then expand � D � 1 2 + n 2 + n 3 + n 4 + n 5 � φ { n } x n 3 1 x n 4 2 x n 5 ( x 1 + x 2 + x 3 ) D/ 2+ n 2 = 3 Γ( D 2 + n 2 ) n 3 ,n 4 ,n 5 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation ◮ The resulting bracket series is � D � 2 + n 2 + n 3 + n 4 + n 5 � φ { n } ( − m 2 ) n 1 s n 2 H = Γ( D 2 + n 2 ) { n } × � a 1 + n 1 + n 2 + n 3 � � a 2 + n 2 + n 4 � � a 3 + n 5 � . Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation ◮ The resulting bracket series is � D � 2 + n 2 + n 3 + n 4 + n 5 � φ { n } ( − m 2 ) n 1 s n 2 H = Γ( D 2 + n 2 ) { n } × � a 1 + n 1 + n 2 + n 3 � � a 2 + n 2 + n 4 � � a 3 + n 5 � . ◮ Possible choices for free variables are n 1 , n 2 , and n 4 . Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation ◮ The resulting bracket series is � D � 2 + n 2 + n 3 + n 4 + n 5 � φ { n } ( − m 2 ) n 1 s n 2 H = Γ( D 2 + n 2 ) { n } × � a 1 + n 1 + n 2 + n 3 � � a 2 + n 2 + n 4 � � a 3 + n 5 � . ◮ Possible choices for free variables are n 1 , n 2 , and n 4 . ◮ The series associated to n 2 converges for | s m 2 | < 1 : Theorem � � � a 1 + a 2 + a 3 − D 2 , a 2 s � H = η 2 · 2 F 1 � D m 2 � 2 with η 2 defined by � � � � D a 1 + a 2 + a 3 − D 2 − a 1 − a 2 − a 3 Γ( a 2 )Γ( a 3 )Γ 2 − a 2 − a 3 Γ − m 2 � D 2 � η 2 = . � � D Γ 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation II ◮ Similarly, the series associated to n 1 , n 4 converges for | m 2 s | < 1 : Theorem � � � a 1 + a 2 + a 3 − D m 2 2 , 1 + a 1 + a 2 + a 3 − D � H = η 1 · 2 F 1 � 1 + a 1 + a 3 − D s � 2 � � � 1 + a 2 − D m 2 2 , a 2 � + η 4 · 2 F 1 � 1 − a 1 − a 3 + D s � 2 with η 1 , η 4 defined by � � � � � � a 1 + a 2 + a 3 − D D D 2 − a 1 − a 2 − a 3 Γ( a 3 )Γ Γ 2 − a 1 − a 3 Γ 2 − a 2 − a 3 2 D η 1 = s , Γ ( D − a 1 − a 2 − a 3 ) � � � � a 1 + a 3 − D D 2 − a 1 − a 3 Γ( a 2 )Γ( a 3 )Γ Γ 2 − a 2 − a 3 − m 2 � D 2 η 4 = s − a 2 � . � � D Γ 2 − a 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation III ◮ Specialize to a 1 = a 2 = a 3 = 1 so that ∞ ∞ ∞ x 1 x 2 e x 1 m 2 e − x 1+ x 2+ x 3 s � � � H = ( x 1 + x 2 + x 3 ) D/ 2 d x 1 d x 2 d x 3 . 0 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Evaluation III ◮ Specialize to a 1 = a 2 = a 3 = 1 so that ∞ ∞ ∞ x 1 x 2 e x 1 m 2 e − x 1+ x 2+ x 3 s � � � H = ( x 1 + x 2 + x 3 ) D/ 2 d x 1 d x 2 d x 3 . 0 0 0 ◮ Then with D = 4 − 2 ǫ : Corollary For | s m 2 | < 1 , � � 1 + ǫ, 1 � s � H = ( − m 2 ) − 1 − ǫ Γ( ǫ − 1) 2 F 1 . � m 2 2 − ǫ � Corollary For | s m 2 | > 1 , � ǫ, 1 � � � − 2 ǫ � H = s − 1 − ǫ Γ( − ǫ ) 2 Γ(1 + ǫ ) 1 − m 2 m 2 − m − 2 ǫ Γ( ǫ ) � ǫs 2 F 1 . � Γ(1 − 2 ǫ ) s 1 − ǫ s � Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Thoughts ◮ The method of brackets produces evaluations for the different regions of the kinematic variables. ◮ Alternative to introducing Mellin-Barnes representations ◮ Most aspects of this process are automatable. ◮ Karen Kohl is working on an implementation in SAGE. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals ◮ Studied by Bailey, Borwein, Crandall: � ∞ � ∞ C n,k = 4 1 d u 1 · · · d u n · · · �� n � k +1 n ! u 1 u n 0 0 j =1 ( u j + 1 /u j ) Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals ◮ Studied by Bailey, Borwein, Crandall: � ∞ � ∞ C n,k = 4 1 d u 1 · · · d u n · · · �� n � k +1 n ! u 1 u n 0 0 j =1 ( u j + 1 /u j ) � ∞ = 2 n − k +1 t k K n 0 ( t ) d t n ! k ! 0 D. H. Bailey, J. M. Borwein and R. E. Crandall. “Integrals of the Ising class.” Jour. Phys. A , 39 , 2006. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals ◮ Studied by Bailey, Borwein, Crandall: � ∞ � ∞ C n,k = 4 1 d u 1 · · · d u n · · · �� n � k +1 n ! u 1 u n 0 0 j =1 ( u j + 1 /u j ) � ∞ = 2 n − k +1 t k K n 0 ( t ) d t n ! k ! 0 ◮ C 1 , 1 = 2 , C 2 , 1 = 1 , C 3 , 1 = L − 3 (2) , C 4 , 1 = 7 12 ζ (3) D. H. Bailey, J. M. Borwein and R. E. Crandall. “Integrals of the Ising class.” Jour. Phys. A , 39 , 2006. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals ◮ Studied by Bailey, Borwein, Crandall: � ∞ � ∞ C n,k = 4 1 d u 1 · · · d u n · · · �� n � k +1 n ! u 1 u n 0 0 j =1 ( u j + 1 /u j ) � ∞ = 2 n − k +1 t k K n 0 ( t ) d t n ! k ! 0 ◮ C 1 , 1 = 2 , C 2 , 1 = 1 , C 3 , 1 = L − 3 (2) , C 4 , 1 = 7 12 ζ (3) , C 5 , 1 =?? D. H. Bailey, J. M. Borwein and R. E. Crandall. “Integrals of the Ising class.” Jour. Phys. A , 39 , 2006. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 � � n 2 − n 4 � k ! { n } Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 � � n 2 − n 4 � k ! { n } ◮ 4 indices, 3 brackets: 1 free variable Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 � � n 2 − n 4 � k ! { n } ◮ 4 indices, 3 brackets: 1 free variable ◮ n 1 free: n ∗ 3 = n 1 and n ∗ 2 = n ∗ 4 = − n 1 − k +1 2 2 � φ n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 � � n 2 − n 4 � k ! { n } ◮ 4 indices, 3 brackets: 1 free variable ◮ n 1 free: n ∗ 3 = n 1 and n ∗ 2 = n ∗ 4 = − n 1 − k +1 2 2 � φ n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 ( − 1) n 1 = 2 � 2 ) 2 Γ( − n 1 ) Γ( n 1 + 1)Γ( n 1 + k +1 k ! n 1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — n = 2 � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 � � n 2 − n 4 � k ! { n } ◮ 4 indices, 3 brackets: 1 free variable ◮ n 1 free: n ∗ 3 = n 1 and n ∗ 2 = n ∗ 4 = − n 1 − k +1 2 2 � φ n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 ( − 1) n 1 = 2 � 2 ) 2 Γ( − n 1 ) Γ( n 1 + 1)Γ( n 1 + k +1 k ! n 1 ◮ No luck for all choices of free variables. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — perturbing ◮ C 2 ,k is the case ε → 0 of � ∞ � ∞ d x d y 2 x 1 − ε y ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 + ε � � n 2 − n 4 � k ! { n } Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — perturbing ◮ C 2 ,k is the case ε → 0 of � ∞ � ∞ d x d y 2 x 1 − ε y ( x + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 + ε � � n 2 − n 4 � k ! { n } ◮ n 1 free: n ∗ 3 = n 1 + ε and n ∗ 2 = n ∗ 4 = − n 1 − k +1+ ε 2 2 � φ n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 ( − 1) n 1 = 2 � ) 2 Γ( − n 1 − ε ) Γ( n 1 + 1)Γ( n 1 + k +1+ ε 2 k ! n 1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — perturbing ◮ C 2 ,k is the case ε → 0 , A → 1 of � ∞ � ∞ d x d y 2 x 1 − ε y ( Ax + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } A n 1 � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 + ε � � n 2 − n 4 � k ! { n } ◮ n 1 free: n ∗ 3 = n 1 + ε and n ∗ 2 = n ∗ 4 = − n 1 − k +1+ ε 2 2 � φ n 1 A n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 ( − 1) n 1 = 2 � Γ( n 1 + 1) A n 1 Γ( n 1 + k +1+ ε ) 2 Γ( − n 1 − ε ) 2 k ! n 1 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — perturbing ◮ C 2 ,k is the case ε → 0 , A → 1 of � ∞ � ∞ d x d y 2 x 1 − ε y ( Ax + 1 /x + y + 1 /y ) k +1 0 0 = 2 � φ { n } A n 1 � n 1 + n 2 + n 3 + n 4 + k + 1 � � n 1 − n 3 + ε � � n 2 − n 4 � k ! { n } ◮ n 1 free: n ∗ 3 = n 1 + ε and n ∗ 2 = n ∗ 4 = − n 1 − k +1+ ε 2 2 � φ n 1 A n 1 Γ( − n ∗ 2 )Γ( − n ∗ 3 )Γ( − n ∗ 4 ) k ! n 1 ( − 1) n 1 = 2 � Γ( n 1 + 1) A n 1 Γ( n 1 + k +1+ ε ) 2 Γ( − n 1 − ε ) 2 k ! n 1 ◮ Combined with n 3 : � � � 1 + ε � � 1 − ε � 2 , 1 + ε 2 , 1 − ε 2 + 2 � � 2 ) 22 F 1 k ! A − ε Γ( ε )Γ(1 − ε 2 ) 22 F 1 k !Γ( − ε )Γ(1 + ε 2 2 � A � A � � 1 + ε 1 − ε Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 � ∞ � ∞ ( xy ) k d x d y = 2 ( xy [ x + y ] + [ x + y ]) k +1 0 0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 � ∞ � ∞ ( xy ) k d x d y = 2 ( xy [ x + y ] + [ x + y ]) k +1 0 0 � � = 2 � φ { n } ( xy ) n 1 + k ( x + y ) n 1 + n 2 � n 1 + n 2 + k + 1 � d x d y k ! n 1 ,n 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 � ∞ � ∞ ( xy ) k d x d y = 2 ( xy [ x + y ] + [ x + y ]) k +1 0 0 � � = 2 � φ { n } ( xy ) n 1 + k ( x + y ) n 1 + n 2 � n 1 + n 2 + k + 1 � d x d y k ! n 1 ,n 2 = 2 φ { n } � n 1 + n 2 + k + 1 � � n 3 + n 4 − n 1 − n 2 � � k ! Γ( − n 1 − n 2 ) { n } × � n 1 + n 3 + k + 1 � � n 1 + n 4 + k + 1 � Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 � ∞ � ∞ ( xy ) k d x d y = 2 ( xy [ x + y ] + [ x + y ]) k +1 0 0 � � = 2 � φ { n } ( xy ) n 1 + k ( x + y ) n 1 + n 2 � n 1 + n 2 + k + 1 � d x d y k ! n 1 ,n 2 = 2 φ { n } � n 1 + n 2 + k + 1 � � n 3 + n 4 − n 1 − n 2 � � k ! Γ( − n 1 − n 2 ) { n } × � n 1 + n 3 + k + 1 � � n 1 + n 4 + k + 1 � = Γ ( − n ∗ 1 ) Γ ( − n ∗ 2 ) Γ ( − n ∗ 3 ) Γ ( − n ∗ 4 ) Γ( k + 1)Γ ( − n ∗ 1 − n ∗ 2 ) ◮ n ∗ 1 = n ∗ 2 = n ∗ 3 = n ∗ 4 = − k +1 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ising Integrals — minding form � ∞ � ∞ d x d y C 2 ,k = 2 xy ( x + 1 /x + y + 1 /y ) k +1 0 0 � ∞ � ∞ ( xy ) k d x d y = 2 ( xy [ x + y ] + [ x + y ]) k +1 0 0 � � = 2 � φ { n } ( xy ) n 1 + k ( x + y ) n 1 + n 2 � n 1 + n 2 + k + 1 � d x d y k ! n 1 ,n 2 = 2 φ { n } � n 1 + n 2 + k + 1 � � n 3 + n 4 − n 1 − n 2 � � k ! Γ( − n 1 − n 2 ) { n } × � n 1 + n 3 + k + 1 � � n 1 + n 4 + k + 1 � � k +1 � 4 = Γ ( − n ∗ 1 ) Γ ( − n ∗ 2 ) Γ ( − n ∗ 3 ) Γ ( − n ∗ = Γ 4 ) 2 Γ( k + 1) 2 . Γ( k + 1)Γ ( − n ∗ 1 − n ∗ 2 ) ◮ n ∗ 1 = n ∗ 2 = n ∗ 3 = n ∗ 4 = − k +1 2 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Challenges ◮ The form of the integrand makes a huge difference. How can it be automatically optimized? Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Challenges ◮ The form of the integrand makes a huge difference. How can it be automatically optimized? ◮ More complicated integrals/bracket series need to be perturbed. How to automatize the insertion of the necessary parameters? Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional The method of brackets � ∞ x s − 1 d x � s � := 0 Rule (Multinomial) 1 � s + m 1 + · · · + m r � � φ { m } a m 1 · · · a m r ( a 1 + a 2 + · · · + a r ) s = r 1 Γ( s ) m 1 ,...,m r Rule (Evaluation) � φ { n } f ( n 1 , . . . , n r ) � a 11 n 1 + · · · a 1 r n r + b 1 � · · · � a r 1 n 1 + · · · a rr n r + b r � 1 { n } | det | f ( n ∗ 1 , . . . , n ∗ r )Γ( − n ∗ 1 ) · · · Γ( − n ∗ = r ) , Rule (Combining) If there are more summation indices than brackets, free variables are chosen. Each choice produces a series. Those converging in a common region are added. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ramanujan’s Master Theorem Rule φ n λ ( n ) � an + b � = 1 � | a | λ ( n ∗ )Γ( − n ∗ ) , n where n ∗ is the solution of the equation an + b = 0 . Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ramanujan’s Master Theorem Rule φ n λ ( n ) � an + b � = 1 � | a | λ ( n ∗ )Γ( − n ∗ ) , n where n ∗ is the solution of the equation an + b = 0 . ◮ Therefore: � ∞ � x s − 1 f ( x ) d x = φ n λ ( n ) � n + s � 0 n ∞ ( − 1) n � λ ( n ) x n f ( x ) = n ! n =0 Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional Ramanujan’s Master Theorem Rule φ n λ ( n ) � an + b � = 1 � | a | λ ( n ∗ )Γ( − n ∗ ) , n where n ∗ is the solution of the equation an + b = 0 . ◮ Therefore: � ∞ � x s − 1 f ( x ) d x = φ n λ ( n ) � n + s � = λ ( − s )Γ( s ) 0 n ∞ ( − 1) n � λ ( n ) x n f ( x ) = n ! n =0 ◮ This is Ramanujan’s Master Theorem. Armin Straub On the method of brackets

The method of brackets Examples A Feynman diagram Troubles RMT Distributional A joke in the sense of Littlewood Theorem (Ramanujan’s Master Theorem) � ∞ � � 1! λ (1) + x 2 λ (0) − x x s − 1 2! λ (2) − · · · d x = Γ( s ) λ ( − s ) 0 ◮ Nearly discovered as early as 1847 by Glaisher and O’Kinealy. Armin Straub On the method of brackets