Derivatives and special values of higher-order Tornheim zeta - PowerPoint PPT Presentation

2. Some special functions For each s ∈ C , the polylogarithm of order s is defined by ∞ z n � Li s ( z ) := ( | z | < 1 ) . n s n = 1 Special cases: z Li 0 ( z ) = 1 − z , Li 1 ( z ) = − log ( 1 − z ) , Li s ( 1 ) = ζ ( s ) ( Re ( s ) > 1 ) . Lemma For s ∈ C \ N , and for | log z | < 2 π , ∞ ζ ( s − m ) log m z � + Γ( 1 − s )( − log z ) s − 1 . Li s ( z ) = m ! m = 0 Karl Dilcher Tornheim zeta function

• This is a well-known representation; Karl Dilcher Tornheim zeta function

• This is a well-known representation; • There is a (more complicated) variant that holds also for s ∈ N . Karl Dilcher Tornheim zeta function

• This is a well-known representation; • There is a (more complicated) variant that holds also for s ∈ N . Connection with Tornheim zeta function: Lemma For t > 0 and r , s > 1 , � ∞ x t − 1 Li r ( e − x ) Li s ( e − x ) d x . Γ( t ) W ( r , s , t ) = 0 Karl Dilcher Tornheim zeta function

• This is a well-known representation; • There is a (more complicated) variant that holds also for s ∈ N . Connection with Tornheim zeta function: Lemma For t > 0 and r , s > 1 , � ∞ x t − 1 Li r ( e − x ) Li s ( e − x ) d x . Γ( t ) W ( r , s , t ) = 0 Proof: Use Euler’s integral for Γ( s ) with an easy substitution: � ∞ Γ( s ) = n s e − nx x s − 1 d x . 0 Replace s by t and n by n + m : Karl Dilcher Tornheim zeta function

� ∞ 1 1 x t − 1 e − ( n + m ) x d x ( n + m ) t = ( Re ( t ) > 0 ) . Γ( t ) 0 Karl Dilcher Tornheim zeta function

� ∞ 1 1 x t − 1 e − ( n + m ) x d x ( n + m ) t = ( Re ( t ) > 0 ) . Γ( t ) 0 Plug into definition of W ( r , s , t ) and change order of summation and integration (legitimate): � ∞ � � ∞ � ∞ � e − nx e − mx 1 � � x t − 1 W ( r , s , t ) = d x n r m s Γ( t ) 0 n = 1 m = 1 � ∞ 1 x t − 1 Li r ( e − x ) Li s ( e − x ) d x . = Γ( t ) 0 QED Karl Dilcher Tornheim zeta function

3. Crandall’s free parameter formula The main tool for our results is a remarkable identity due to Richard Crandall (1947–2012). Karl Dilcher Tornheim zeta function

3. Crandall’s free parameter formula The main tool for our results is a remarkable identity due to Richard Crandall (1947–2012). Karl Dilcher Tornheim zeta function

It uses a free parameter and is convenient for both • theoretical results, and • computations. Karl Dilcher Tornheim zeta function

It uses a free parameter and is convenient for both • theoretical results, and • computations. We first need another special function: The (upper) incomplete Gamma function is defined by � ∞ y a − 1 e − y d y . Γ( a , z ) := z Karl Dilcher Tornheim zeta function

It uses a free parameter and is convenient for both • theoretical results, and • computations. We first need another special function: The (upper) incomplete Gamma function is defined by � ∞ y a − 1 e − y d y . Γ( a , z ) := z Obviously, Γ( a , 0 ) = Γ( a ) . Karl Dilcher Tornheim zeta function

Theorem (Crandall) Let r , s , t be complex variables with r �∈ N and s �∈ N . Then for any real θ > 0 we have Γ( t , ( m + n ) θ ) � Γ( t ) W ( r , s , t ) = m r n s ( m + n ) t m , n ≥ 1 ( − 1 ) u + v ζ ( r − u ) ζ ( s − v ) θ u + v + t � + u ! v !( u + v + t ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ r + q + t − 1 � + Γ( 1 − r ) q !( r + q + t − 1 ) q ≥ 0 ( − 1 ) q ζ ( r − q ) θ s + q + t − 1 � + Γ( 1 − s ) q !( s + q + t − 1 ) q ≥ 0 θ r + s + t − 2 + Γ( 1 − r )Γ( 1 − s ) r + s + t − 2 . Karl Dilcher Tornheim zeta function

Remarks: 1. Identity looks complicated at first, but is remarkably useful. Karl Dilcher Tornheim zeta function

Remarks: 1. Identity looks complicated at first, but is remarkably useful. 2. r ∈ N and s ∈ N must be exluded since this would lead to ζ ( 1 ) and Γ( z ) at negative integers. Karl Dilcher Tornheim zeta function

Remarks: 1. Identity looks complicated at first, but is remarkably useful. 2. r ∈ N and s ∈ N must be exluded since this would lead to ζ ( 1 ) and Γ( z ) at negative integers. 3. However, these singularities cancel, and a careful analysis gives a (more complicated) identity valid for all r , s , t ∈ C . Karl Dilcher Tornheim zeta function

Remarks: 1. Identity looks complicated at first, but is remarkably useful. 2. r ∈ N and s ∈ N must be exluded since this would lead to ζ ( 1 ) and Γ( z ) at negative integers. 3. However, these singularities cancel, and a careful analysis gives a (more complicated) identity valid for all r , s , t ∈ C . 4. This, and the above theorem, gives another analytic continuation to all of C 3 , with the exception of the known singularities. Karl Dilcher Tornheim zeta function

Sketch of proof: Use defining integral of Γ( a , z ) . A simple substitution gives � ∞ x t − 1 e − ( m + n ) x d x = Γ( t , ( m + n ) θ ) . ( m + n ) t θ Karl Dilcher Tornheim zeta function

Sketch of proof: Use defining integral of Γ( a , z ) . A simple substitution gives � ∞ x t − 1 e − ( m + n ) x d x = Γ( t , ( m + n ) θ ) . ( m + n ) t θ Use same argument as in the 2nd Lemma; break up integral: �� θ � ∞ 1 � � x t − 1 e − ( m + n ) x d x Γ( t ) W ( r , s , t ) = + m r n s 0 θ m , n ≥ 1 � θ Γ( t , ( m + n ) θ ) � x t − 1 Li r ( e − x ) Li s ( e − x ) d x . = m r n s ( m + n ) t + 0 m , n ≥ 1 Karl Dilcher Tornheim zeta function

Sketch of proof: Use defining integral of Γ( a , z ) . A simple substitution gives � ∞ x t − 1 e − ( m + n ) x d x = Γ( t , ( m + n ) θ ) . ( m + n ) t θ Use same argument as in the 2nd Lemma; break up integral: �� θ � ∞ 1 � � x t − 1 e − ( m + n ) x d x Γ( t ) W ( r , s , t ) = + m r n s 0 θ m , n ≥ 1 � θ Γ( t , ( m + n ) θ ) � x t − 1 Li r ( e − x ) Li s ( e − x ) d x . = m r n s ( m + n ) t + 0 m , n ≥ 1 Use the first Lemma, namely ζ ( s − m ) log m z ∞ � + Γ( 1 − s )( − log z ) s − 1 . Li s ( z ) = m ! m = 0 Karl Dilcher Tornheim zeta function

Sketch of proof: Use defining integral of Γ( a , z ) . A simple substitution gives � ∞ x t − 1 e − ( m + n ) x d x = Γ( t , ( m + n ) θ ) . ( m + n ) t θ Use same argument as in the 2nd Lemma; break up integral: �� θ � ∞ 1 � � x t − 1 e − ( m + n ) x d x Γ( t ) W ( r , s , t ) = + m r n s 0 θ m , n ≥ 1 � θ Γ( t , ( m + n ) θ ) � x t − 1 Li r ( e − x ) Li s ( e − x ) d x . = m r n s ( m + n ) t + 0 m , n ≥ 1 Use the first Lemma, namely ζ ( s − m ) log m z ∞ � + Γ( 1 − s )( − log z ) s − 1 . Li s ( z ) = m ! m = 0 Expand and then integrate. QED Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. LHS: s Γ( s ) → 1. Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. LHS: s Γ( s ) → 1. RHS: Almost all terms disappear, except Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. LHS: s Γ( s ) → 1. RHS: Almost all terms disappear, except – 2nd row for u = v = 0; get ζ ( 0 ) 2 = ( − 1 / 2 ) 2 = 1 / 4; Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. LHS: s Γ( s ) → 1. RHS: Almost all terms disappear, except – 2nd row for u = v = 0; get ζ ( 0 ) 2 = ( − 1 / 2 ) 2 = 1 / 4; – 3rd row for q = 1; get 2 Γ( 0 )( − 1 ) ζ ( − 1 ) / 2 = − ζ ( − 1 ) = 1 / 12. Karl Dilcher Tornheim zeta function

First application: Set r = s = t ; then Γ( s , ( m + n ) θ ) � Γ( s ) ω 3 ( s ) = ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 + Γ( 1 − s ) s θ 3 s − 2 3 s − 2 . Fix θ > 0, multiply both sides by s and let s → 0. LHS: s Γ( s ) → 1. RHS: Almost all terms disappear, except – 2nd row for u = v = 0; get ζ ( 0 ) 2 = ( − 1 / 2 ) 2 = 1 / 4; – 3rd row for q = 1; get 2 Γ( 0 )( − 1 ) ζ ( − 1 ) / 2 = − ζ ( − 1 ) = 1 / 12. Together: ω 3 ( 0 ) = 1 / 3. Karl Dilcher Tornheim zeta function

Remarks: 1. This last identity also immediately gives the singularities of ω 3 ( s ) and their residues. Karl Dilcher Tornheim zeta function

Remarks: 1. This last identity also immediately gives the singularities of ω 3 ( s ) and their residues. 2. With only a small variation we can prove a more general result: Define ω 3 ( s ; τ ) := W ( s , s , τ s ) . Karl Dilcher Tornheim zeta function

Remarks: 1. This last identity also immediately gives the singularities of ω 3 ( s ) and their residues. 2. With only a small variation we can prove a more general result: Define ω 3 ( s ; τ ) := W ( s , s , τ s ) . Then for any τ > 0 we have τ + 1 ζ ( − 1 ) = 1 2 τ 5 τ + 3 ω 3 ( 0 ; τ ) = ζ ( 0 ) 2 − τ + 1 , 12 Karl Dilcher Tornheim zeta function

Remarks: 1. This last identity also immediately gives the singularities of ω 3 ( s ) and their residues. 2. With only a small variation we can prove a more general result: Define ω 3 ( s ; τ ) := W ( s , s , τ s ) . Then for any τ > 0 we have τ + 1 ζ ( − 1 ) = 1 2 τ 5 τ + 3 ω 3 ( 0 ; τ ) = ζ ( 0 ) 2 − τ + 1 , 12 and in particular, ω 3 ( 0 ) = ω 3 ( 0 ; 1 ) = 1 3 . Karl Dilcher Tornheim zeta function

4. Derivative at the origin Let’s return to Crandall’s identity; multiply both sides by s :  Γ( s , ( m + n ) θ )  � s Γ( s ) ω 3 ( s ) = s ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 +Γ( 1 − s ) s θ 3 s − 2 � . 3 s − 2 Karl Dilcher Tornheim zeta function

4. Derivative at the origin Let’s return to Crandall’s identity; multiply both sides by s :  Γ( s , ( m + n ) θ )  � s Γ( s ) ω 3 ( s ) = s ( mn ( m + n )) s m , n ≥ 1 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 � + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 0 +Γ( 1 − s ) s θ 3 s − 2 � . 3 s − 2 Now isolate the singularies in s in the large brackets on the RHS; bring them to the left: Karl Dilcher Tornheim zeta function

s Γ( s ) ω 3 ( s ) − ζ ( s ) 2 θ 2 + Γ( 1 − s ) ζ ( s − 1 ) θ 2 s  Γ( s , ( m + n ) θ )  � = s ( mn ( m + n )) s m , n ≥ 1 + 2 Γ( 1 − s ) ζ ( s ) θ 2 s − 1 2 s − 1 + Γ( 1 − s ) 2 θ 3 s − 2 3 s − 2 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( u , v ) � =( 0 , 0 )  ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 �  . + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 2 Karl Dilcher Tornheim zeta function

s Γ( s ) ω 3 ( s ) − ζ ( s ) 2 θ 2 + Γ( 1 − s ) ζ ( s − 1 ) θ 2 s  Γ( s , ( m + n ) θ )  � = s ( mn ( m + n )) s m , n ≥ 1 + 2 Γ( 1 − s ) ζ ( s ) θ 2 s − 1 2 s − 1 + Γ( 1 − s ) 2 θ 3 s − 2 3 s − 2 ( − 1 ) u + v ζ ( s − u ) ζ ( s − v ) θ u + v + s � + u ! v !( u + v + s ) u , v ≥ 0 ( u , v ) � =( 0 , 0 )  ( − 1 ) q ζ ( s − q ) θ 2 s + q − 1 �  . + 2 Γ( 1 − s ) q !( 2 s + q − 1 ) q ≥ 2 Derivative at s = 0 of LHS becomes ω ′ 3 ( 0 ) − 5 12 γ − 1 2 log 2 π − 5 12 log θ + ζ ′ ( − 1 ) . Karl Dilcher Tornheim zeta function

Derivative at s = 0 of RHS amounts to evaluating [ . . . ] at s = 0. Karl Dilcher Tornheim zeta function

Derivative at s = 0 of RHS amounts to evaluating [ . . . ] at s = 0. Since θ > 0 is a free variable, we consider θ → 0. Karl Dilcher Tornheim zeta function

Derivative at s = 0 of RHS amounts to evaluating [ . . . ] at s = 0. Since θ > 0 is a free variable, we consider θ → 0. There are singularities at θ = 0; however, they cancel. Karl Dilcher Tornheim zeta function

Derivative at s = 0 of RHS amounts to evaluating [ . . . ] at s = 0. Since θ > 0 is a free variable, we consider θ → 0. There are singularities at θ = 0; however, they cancel. Some key ingredients: � ∞ du � Γ( 0 , ( m + n ) θ ) = ( e θ u − 1 ) 2 u . 1 m , n ≥ 1 (Easy manipulation using definition of Γ( a , z ) ). Karl Dilcher Tornheim zeta function

Derivative at s = 0 of RHS amounts to evaluating [ . . . ] at s = 0. Since θ > 0 is a free variable, we consider θ → 0. There are singularities at θ = 0; however, they cancel. Some key ingredients: � ∞ du � Γ( 0 , ( m + n ) θ ) = ( e θ u − 1 ) 2 u . 1 m , n ≥ 1 (Easy manipulation using definition of Γ( a , z ) ). � ∞ t α − 1 ( e t − 1 ) 2 dt = Γ( α )( ζ ( α − 1 ) − ζ ( α )) ( Re ( α ) > 2 ) . 0 (An integral in Gradshteyn & Ryzhik). Karl Dilcher Tornheim zeta function

Everything put together, we get (after some work) � Γ( 0 , ( m + n ) θ ) m , n ≥ 1 = 1 2 log ( 2 π ) − 5 γ 12 + ζ ′ ( − 1 ) − 1 2 θ 2 − 5 1 θ + 12 log θ + O ( θ ) . Karl Dilcher Tornheim zeta function

Everything put together, we get (after some work) � Γ( 0 , ( m + n ) θ ) m , n ≥ 1 = 1 2 log ( 2 π ) − 5 γ 12 + ζ ′ ( − 1 ) − 1 2 θ 2 − 5 1 θ + 12 log θ + O ( θ ) . Finally: Theorem ω ′ 3 ( 0 ) = log ( 2 π ) . Karl Dilcher Tornheim zeta function

Everything put together, we get (after some work) � Γ( 0 , ( m + n ) θ ) m , n ≥ 1 = 1 2 log ( 2 π ) − 5 γ 12 + ζ ′ ( − 1 ) − 1 2 θ 2 − 5 1 θ + 12 log θ + O ( θ ) . Finally: Theorem ω ′ 3 ( 0 ) = log ( 2 π ) . As before, with small modifications we get more generally 3 ( 0 ; τ ) = τ + 1 log ( 2 π ) + ( τ − 1 ) τ ω ′ ζ ′ ( − 1 ) . 2 τ + 1 (Recall: ω 3 ( s ; τ ) := W ( s , s , τ s ) .) Karl Dilcher Tornheim zeta function

5. Extensions 1. A multi-dimensional analogue: For n ≥ 2 define 1 � W ( r 1 , . . . , r n , t ) := m r 1 1 . . . m r n n ( m 1 + . . . m n ) t m 1 ,..., m n ≥ 1 Studied by Matsumoto (2000), Bailey & Borwein (2015), and others. Karl Dilcher Tornheim zeta function

5. Extensions 1. A multi-dimensional analogue: For n ≥ 2 define 1 � W ( r 1 , . . . , r n , t ) := m r 1 1 . . . m r n n ( m 1 + . . . m n ) t m 1 ,..., m n ≥ 1 Studied by Matsumoto (2000), Bailey & Borwein (2015), and others. In analogy to before, define ω n + 1 ( s ) := W ( s , . . . , s , s ) . Karl Dilcher Tornheim zeta function

5. Extensions 1. A multi-dimensional analogue: For n ≥ 2 define 1 � W ( r 1 , . . . , r n , t ) := m r 1 1 . . . m r n n ( m 1 + . . . m n ) t m 1 ,..., m n ≥ 1 Studied by Matsumoto (2000), Bailey & Borwein (2015), and others. In analogy to before, define ω n + 1 ( s ) := W ( s , . . . , s , s ) . Hayley Tomkins (honours thesis, 2016) showed that ω n + 1 ( 0 ) = ( − 1 ) n n + 1 holds for n ≤ 7, and conjectured that it is true for all n . Karl Dilcher Tornheim zeta function

Meanwhile proved, using higher-order convolution identities for Bernoulli numbers and polynomials. (Joint with Armin Straub, 2016, unpublished). Karl Dilcher Tornheim zeta function

Meanwhile proved, using higher-order convolution identities for Bernoulli numbers and polynomials. (Joint with Armin Straub, 2016, unpublished). A more complicated convolution identity shows that ω n + 1 ( − k ) = 0 for all integers n ≥ 2 and k ≥ 1. Karl Dilcher Tornheim zeta function

Meanwhile proved, using higher-order convolution identities for Bernoulli numbers and polynomials. (Joint with Armin Straub, 2016, unpublished). A more complicated convolution identity shows that ω n + 1 ( − k ) = 0 for all integers n ≥ 2 and k ≥ 1. This is analogous to the zeta function identity ζ ( − k ) = 0 for k = 2 , 4 , 6 , . . . . Karl Dilcher Tornheim zeta function

Also proved by Tomkins: ω ′ 4 ( 0 ) = − log ( 2 π ) + ζ ′ ( − 2 ) = − log ( 2 π ) − ζ ( 3 ) 4 π 2 . Karl Dilcher Tornheim zeta function

Also proved by Tomkins: ω ′ 4 ( 0 ) = − log ( 2 π ) + ζ ′ ( − 2 ) = − log ( 2 π ) − ζ ( 3 ) 4 π 2 . How about ω ′ n + 1 ( 0 ) for n ≥ 4? Karl Dilcher Tornheim zeta function

Recall: ω ′ 3 ( 0 ) = log ( 2 π ) , ω ′ 4 ( 0 ) = − log ( 2 π ) + ζ ′ ( − 2 ) . Karl Dilcher Tornheim zeta function

Recall: ω ′ 3 ( 0 ) = log ( 2 π ) , ω ′ 4 ( 0 ) = − log ( 2 π ) + ζ ′ ( − 2 ) . Bailey & Borwein found experimentally: ω ′ 5 ( 0 ) = log ( 2 π ) − 2 ζ ′ ( − 2 ) ω ′ 6 ( 0 ) = − log ( 2 π ) + 35 12 ζ ′ ( − 2 ) + 1 12 ζ ′ ( − 4 ) ω ′ 7 ( 0 ) = log ( 2 π ) − 15 4 ζ ′ ( − 2 ) − 1 4 ζ ′ ( − 4 ) . . . ω ′ 33633600 ζ ′ ( − 2 ) − . . . − 1162377216000 ζ ′ ( − 16 ) . 19 ( 0 ) = log ( 2 π ) − 344499373 1 Karl Dilcher Tornheim zeta function

Recall: ω ′ 3 ( 0 ) = log ( 2 π ) , ω ′ 4 ( 0 ) = − log ( 2 π ) + ζ ′ ( − 2 ) . Bailey & Borwein found experimentally: ω ′ 5 ( 0 ) = log ( 2 π ) − 2 ζ ′ ( − 2 ) ω ′ 6 ( 0 ) = − log ( 2 π ) + 35 12 ζ ′ ( − 2 ) + 1 12 ζ ′ ( − 4 ) ω ′ 7 ( 0 ) = log ( 2 π ) − 15 4 ζ ′ ( − 2 ) − 1 4 ζ ′ ( − 4 ) . . . ω ′ 33633600 ζ ′ ( − 2 ) − . . . − 1162377216000 ζ ′ ( − 16 ) . 19 ( 0 ) = log ( 2 π ) − 344499373 1 What are the coefficients in these expressions? Karl Dilcher Tornheim zeta function

Theorem For any n ≥ 2 we have ⌊ n − 1 2 ⌋ n + 1 ( 0 ) = ( − 1 ) n log ( 2 π ) + ω ′ � s ( n , 2 j + 1 ) ζ ′ ( − 2 j ) , 2 ( n − 1 )! j = 1 where s ( n , k ) are the Stirling numbers of the first kind. Karl Dilcher Tornheim zeta function

Theorem For any n ≥ 2 we have ⌊ n − 1 2 ⌋ n + 1 ( 0 ) = ( − 1 ) n log ( 2 π ) + ω ′ � s ( n , 2 j + 1 ) ζ ′ ( − 2 j ) , 2 ( n − 1 )! j = 1 where s ( n , k ) are the Stirling numbers of the first kind. (Note: ζ ′ ( 0 ) = − 1 2 log ( 2 π ) ). Karl Dilcher Tornheim zeta function

Theorem For any n ≥ 2 we have ⌊ n − 1 2 ⌋ n + 1 ( 0 ) = ( − 1 ) n log ( 2 π ) + ω ′ � s ( n , 2 j + 1 ) ζ ′ ( − 2 j ) , 2 ( n − 1 )! j = 1 where s ( n , k ) are the Stirling numbers of the first kind. (Note: ζ ′ ( 0 ) = − 1 2 log ( 2 π ) ). Proof uses similar methods as that of the original case n = 2. Karl Dilcher Tornheim zeta function

Theorem For any n ≥ 2 we have ⌊ n − 1 2 ⌋ n + 1 ( 0 ) = ( − 1 ) n log ( 2 π ) + ω ′ � s ( n , 2 j + 1 ) ζ ′ ( − 2 j ) , 2 ( n − 1 )! j = 1 where s ( n , k ) are the Stirling numbers of the first kind. (Note: ζ ′ ( 0 ) = − 1 2 log ( 2 π ) ). Proof uses similar methods as that of the original case n = 2. Recall: n s ( n , k ) x k = x ( x − 1 ) . . . ( x − n + 1 ); � k = 0 Karl Dilcher Tornheim zeta function

Theorem For any n ≥ 2 we have ⌊ n − 1 2 ⌋ n + 1 ( 0 ) = ( − 1 ) n log ( 2 π ) + ω ′ � s ( n , 2 j + 1 ) ζ ′ ( − 2 j ) , 2 ( n − 1 )! j = 1 where s ( n , k ) are the Stirling numbers of the first kind. (Note: ζ ′ ( 0 ) = − 1 2 log ( 2 π ) ). Proof uses similar methods as that of the original case n = 2. Recall: n s ( n , k ) x k = x ( x − 1 ) . . . ( x − n + 1 ); � k = 0 s ( n , k ) = s ( n − 1 , k − 1 ) − ( n − 1 ) s ( n − 1 , k ) . Karl Dilcher Tornheim zeta function

2. Character analogues Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Karl Dilcher Tornheim zeta function

2. Character analogues Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by ( − 1 ) m ( − 1 ) n 1 � A ( r , s , t ) := ( m + n ) t . m r n s m , n ≥ 1 Karl Dilcher Tornheim zeta function

2. Character analogues Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by ( − 1 ) m ( − 1 ) n 1 � A ( r , s , t ) := ( m + n ) t . m r n s m , n ≥ 1 In analogy to ω 3 ( s ) , consider α 3 ( s ) := A ( s , s , s ) . Karl Dilcher Tornheim zeta function

2. Character analogues Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by ( − 1 ) m ( − 1 ) n 1 � A ( r , s , t ) := ( m + n ) t . m r n s m , n ≥ 1 In analogy to ω 3 ( s ) , consider α 3 ( s ) := A ( s , s , s ) . Analogue to Crandall’s formula is much simpler: ( − 1 ) m ( − 1 ) n Γ( s , ( m + n ) θ ) � Γ( s ) α 3 ( s ) = m r n s ( m + n ) s m , n ≥ 1 ( − 1 ) u + v η ( s − u ) η ( s − v ) θ u + v + s � + . u ! v !( u + v + s ) u , v ≥ 0 Karl Dilcher Tornheim zeta function

Here ∞ ( − 1 ) n + 1 � = ( 1 − 2 1 − s ) ζ ( s ) η ( s ) := n s n = 1 is the alternating zeta function. Karl Dilcher Tornheim zeta function

Here ∞ ( − 1 ) n + 1 � = ( 1 − 2 1 − s ) ζ ( s ) η ( s ) := n s n = 1 is the alternating zeta function. Some special values: 2 log π η ′ ( 0 ) = 1 η ( 1 ) = − log 2 , 2 . Karl Dilcher Tornheim zeta function

Here ∞ ( − 1 ) n + 1 � = ( 1 − 2 1 − s ) ζ ( s ) η ( s ) := n s n = 1 is the alternating zeta function. Some special values: 2 log π η ′ ( 0 ) = 1 η ( 1 ) = − log 2 , 2 . Following same procedure as before, we find α 3 ( 0 ) = η ( 0 ) 2 = 1 4 . Karl Dilcher Tornheim zeta function

Here ∞ ( − 1 ) n + 1 � = ( 1 − 2 1 − s ) ζ ( s ) η ( s ) := n s n = 1 is the alternating zeta function. Some special values: 2 log π η ′ ( 0 ) = 1 η ( 1 ) = − log 2 , 2 . Following same procedure as before, we find α 3 ( 0 ) = η ( 0 ) 2 = 1 4 . Furthermore, using methods of before: α ′ 3 ( 0 ) = 2 η ′ ( 0 ) − η ′ ( − 1 ) − 1 4 γ 4 γ + 3 ζ ′ ( − 1 ) . = log ( 2 π ) − 5 3 log 2 − 1 Karl Dilcher Tornheim zeta function