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

Derivatives and special values of higher-order Tornheim zeta functions Karl Dilcher Dalhousie University Number Theory Seminar June 6, 2018 Karl Dilcher Tornheim zeta function Joint work with Hayley Tomkins (University of Ottawa; formerly


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Derivatives and special values

  • f higher-order Tornheim zeta functions

Karl Dilcher

Dalhousie University

Number Theory Seminar June 6, 2018

Karl Dilcher Tornheim zeta function

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Joint work with Hayley Tomkins (University of Ottawa; formerly Dalhousie)

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Partly based on earlier work with Jon Borwein (1951–2016)

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  • 1. Introduction

One of the best-known multiple zeta functions: W(r, s, t) :=

  • m,n≥1

1 mr 1 ns 1 (m + n)t .

Karl Dilcher Tornheim zeta function

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  • 1. Introduction

One of the best-known multiple zeta functions: W(r, s, t) :=

  • m,n≥1

1 mr 1 ns 1 (m + n)t . Converges for r, s, t ∈ C with Re(r + t) > 1, Re(s + t) > 1, Re(r + s + t) > 2.

Karl Dilcher Tornheim zeta function

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  • 1. Introduction

One of the best-known multiple zeta functions: W(r, s, t) :=

  • m,n≥1

1 mr 1 ns 1 (m + n)t . Converges for r, s, t ∈ C with Re(r + t) > 1, Re(s + t) > 1, Re(r + s + t) > 2. First investigated for r, s, t ∈ N by Tornheim (1950), independently by Mordell (1958) for the special case r = s = t.

Karl Dilcher Tornheim zeta function

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  • 1. Introduction

One of the best-known multiple zeta functions: W(r, s, t) :=

  • m,n≥1

1 mr 1 ns 1 (m + n)t . Converges for r, s, t ∈ C with Re(r + t) > 1, Re(s + t) > 1, Re(r + s + t) > 2. First investigated for r, s, t ∈ N by Tornheim (1950), independently by Mordell (1958) for the special case r = s = t. Therefore often called a Tornheim (double) sum or Mordell-Tornheim (double) sum or series.

Karl Dilcher Tornheim zeta function

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  • 1. Introduction

One of the best-known multiple zeta functions: W(r, s, t) :=

  • m,n≥1

1 mr 1 ns 1 (m + n)t . Converges for r, s, t ∈ C with Re(r + t) > 1, Re(s + t) > 1, Re(r + s + t) > 2. First investigated for r, s, t ∈ N by Tornheim (1950), independently by Mordell (1958) for the special case r = s = t. Therefore often called a Tornheim (double) sum or Mordell-Tornheim (double) sum or series. Witten (1991) studied a wider class; Zagier (1993) called them Witten zeta functions. Also often called Mordell-Tornheim-Witten sums.

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W(r, s, t) can be analytically continued, separately in each of the three variables, to all of C3; poles at r + s + t = 2, r + t = 1 − ℓ, s + t = 1 − ℓ (ℓ ∈ N ∪ {0}). (Akiyama, Egami, Matsumoto, 1999, independently).

Karl Dilcher Tornheim zeta function

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W(r, s, t) can be analytically continued, separately in each of the three variables, to all of C3; poles at r + s + t = 2, r + t = 1 − ℓ, s + t = 1 − ℓ (ℓ ∈ N ∪ {0}). (Akiyama, Egami, Matsumoto, 1999, independently). Romik (preprint, 2015) studied the analytic properties of W(r, s, t) and its analytic continutation in greater detail; introduced the function ω3(s) := W(s, s, s).

Karl Dilcher Tornheim zeta function

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W(r, s, t) can be analytically continued, separately in each of the three variables, to all of C3; poles at r + s + t = 2, r + t = 1 − ℓ, s + t = 1 − ℓ (ℓ ∈ N ∪ {0}). (Akiyama, Egami, Matsumoto, 1999, independently). Romik (preprint, 2015) studied the analytic properties of W(r, s, t) and its analytic continutation in greater detail; introduced the function ω3(s) := W(s, s, s). Already known to Mordell: ω3(2n) = cnπ6n, cn ∈ Q (n ≥ 1), where cn can be given explicitly,

Karl Dilcher Tornheim zeta function

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W(r, s, t) can be analytically continued, separately in each of the three variables, to all of C3; poles at r + s + t = 2, r + t = 1 − ℓ, s + t = 1 − ℓ (ℓ ∈ N ∪ {0}). (Akiyama, Egami, Matsumoto, 1999, independently). Romik (preprint, 2015) studied the analytic properties of W(r, s, t) and its analytic continutation in greater detail; introduced the function ω3(s) := W(s, s, s). Already known to Mordell: ω3(2n) = cnπ6n, cn ∈ Q (n ≥ 1), where cn can be given explicitly, e.g., ω3(4) = 19 273648375π12.

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Further properties: ω3(−n) = 0 (n = 1, 2, 3, . . .),

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Further properties: ω3(−n) = 0 (n = 1, 2, 3, . . .), and for n ≥ 0, ω3(2n + 1) = −4

n

  • k=0

4n − 2k + 1 2n

  • ζ(2k)ζ(6n − 2k + 3)

(Zagier, 1994; Huard, Williams, and Nan-Yue, 1996);

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Further properties: ω3(−n) = 0 (n = 1, 2, 3, . . .), and for n ≥ 0, ω3(2n + 1) = −4

n

  • k=0

4n − 2k + 1 2n

  • ζ(2k)ζ(6n − 2k + 3)

(Zagier, 1994; Huard, Williams, and Nan-Yue, 1996); e.g., ω3(5) = −2 9π4ζ(11) − 70 3 π2ζ(13) + 252 ζ(15).

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Further properties: ω3(−n) = 0 (n = 1, 2, 3, . . .), and for n ≥ 0, ω3(2n + 1) = −4

n

  • k=0

4n − 2k + 1 2n

  • ζ(2k)ζ(6n − 2k + 3)

(Zagier, 1994; Huard, Williams, and Nan-Yue, 1996); e.g., ω3(5) = −2 9π4ζ(11) − 70 3 π2ζ(13) + 252 ζ(15). Also, ω3(s) has simple poles at s = 2 3 and s = 1 2 − k, k = 0, 1, 2, . . . and no other singularities. (Romik, 2015, who also determined the residues).

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),
  • ω′

3(s),

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),
  • ω′

3(s),

  • and some generalizations.

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),
  • ω′

3(s),

  • and some generalizations.

Important: As we shall see, care must be taken in how we approach (r, s, t) = (0, 0, 0) in W(r, s, t).

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),
  • ω′

3(s),

  • and some generalizations.

Important: As we shall see, care must be taken in how we approach (r, s, t) = (0, 0, 0) in W(r, s, t).

  • 1. Romik (2015) showed:

ω3(0) = 1 3.

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Main purpose of this talk: Study the behaviour at the origin of

  • ω3(s),
  • ω′

3(s),

  • and some generalizations.

Important: As we shall see, care must be taken in how we approach (r, s, t) = (0, 0, 0) in W(r, s, t).

  • 1. Romik (2015) showed:

ω3(0) = 1 3. Here, we’ll give a different proof.

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Dan Romik (UC Davis)

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  • 2. In the same paper, Romik showed:

ω′

3(0) = 1

12(1 + γ) + 3 4 log(2π) − 2ζ′(−1) + 1 2 ∞

−∞

ζ( 3

2 + it)ζ(− 3 2 − it)

( 3

2 + it) cosh(πt)

dt = 1.83787706640934548356 . . .

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  • 2. In the same paper, Romik showed:

ω′

3(0) = 1

12(1 + γ) + 3 4 log(2π) − 2ζ′(−1) + 1 2 ∞

−∞

ζ( 3

2 + it)ζ(− 3 2 − it)

( 3

2 + it) cosh(πt)

dt = 1.83787706640934548356 . . . Jon Borwein used high-precision evaluations and the integer relation algorithm PSLQ to conjecture the stunningly simple expression

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  • 2. In the same paper, Romik showed:

ω′

3(0) = 1

12(1 + γ) + 3 4 log(2π) − 2ζ′(−1) + 1 2 ∞

−∞

ζ( 3

2 + it)ζ(− 3 2 − it)

( 3

2 + it) cosh(πt)

dt = 1.83787706640934548356 . . . Jon Borwein used high-precision evaluations and the integer relation algorithm PSLQ to conjecture the stunningly simple expression ω′

3(0) = log(2π).

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  • 2. In the same paper, Romik showed:

ω′

3(0) = 1

12(1 + γ) + 3 4 log(2π) − 2ζ′(−1) + 1 2 ∞

−∞

ζ( 3

2 + it)ζ(− 3 2 − it)

( 3

2 + it) cosh(πt)

dt = 1.83787706640934548356 . . . Jon Borwein used high-precision evaluations and the integer relation algorithm PSLQ to conjecture the stunningly simple expression ω′

3(0) = log(2π).

The main part of this talk concerns proving this and a generalization.

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  • 2. Some special functions

For each s ∈ C, the polylogarithm of order s is defined by Lis(z) :=

  • n=1

zn ns (|z| < 1).

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  • 2. Some special functions

For each s ∈ C, the polylogarithm of order s is defined by Lis(z) :=

  • n=1

zn ns (|z| < 1). Special cases: Li0(z) = z 1 − z , Li1(z) = − log(1 − z), Lis(1) = ζ(s) (Re(s) > 1).

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  • 2. Some special functions

For each s ∈ C, the polylogarithm of order s is defined by Lis(z) :=

  • n=1

zn ns (|z| < 1). Special cases: Li0(z) = z 1 − z , Li1(z) = − log(1 − z), Lis(1) = ζ(s) (Re(s) > 1). Lemma For s ∈ C \ N, and for | log z| < 2π, Lis(z) =

  • m=0

ζ(s − m)logm z m! + Γ(1 − s)(− log z)s−1.

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  • This is a well-known representation;

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  • This is a well-known representation;
  • There is a (more complicated) variant that holds also for

s ∈ N.

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  • 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, Γ(t) W(r, s, t) = ∞ xt−1Lir(e−x)Lis(e−x)dx.

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  • 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, Γ(t) W(r, s, t) = ∞ xt−1Lir(e−x)Lis(e−x)dx. Proof: Use Euler’s integral for Γ(s) with an easy substitution: Γ(s) = ns ∞ e−nxxs−1dx. Replace s by t and n by n + m:

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1 (n + m)t = 1 Γ(t) ∞ xt−1e−(n+m)xdx (Re(t) > 0).

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1 (n + m)t = 1 Γ(t) ∞ xt−1e−(n+m)xdx (Re(t) > 0). Plug into definition of W(r, s, t) and change order of summation and integration (legitimate): W(r, s, t) = 1 Γ(t) ∞ xt−1 ∞

  • n=1

e−nx nr ∞

  • m=1

e−mx ms

  • dx

= 1 Γ(t) ∞ xt−1Lir(e−x)Lis(e−x)dx. QED

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  • 3. Crandall’s free parameter formula

The main tool for our results is a remarkable identity due to Richard Crandall (1947–2012).

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  • 3. Crandall’s free parameter formula

The main tool for our results is a remarkable identity due to Richard Crandall (1947–2012).

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It uses a free parameter and is convenient for both

  • theoretical results, and
  • computations.

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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 Γ(a, z) := ∞

z

ya−1e−ydy.

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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 Γ(a, z) := ∞

z

ya−1e−ydy. Obviously, Γ(a, 0) = Γ(a).

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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)W(r, s, t) =

  • m,n≥1

Γ(t, (m + n)θ) mrns(m + n)t +

  • u,v≥0

(−1)u+v ζ(r − u)ζ(s − v)θu+v+t u!v!(u + v + t) + Γ(1 − r)

  • q≥0

(−1)q ζ(s − q)θr+q+t−1 q!(r + q + t − 1) + Γ(1 − s)

  • q≥0

(−1)q ζ(r − q)θs+q+t−1 q!(s + q + t − 1) + Γ(1 − r)Γ(1 − s) θr+s+t−2 r + s + t − 2.

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Remarks:

  • 1. Identity looks complicated at first, but is remarkably useful.

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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.

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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.

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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 C3, with the exception of the known singularities.

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Sketch of proof: Use defining integral of Γ(a, z). A simple substitution gives ∞

θ

xt−1e−(m+n)xdx = Γ(t, (m + n)θ) (m + n)t .

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Sketch of proof: Use defining integral of Γ(a, z). A simple substitution gives ∞

θ

xt−1e−(m+n)xdx = Γ(t, (m + n)θ) (m + n)t . Use same argument as in the 2nd Lemma; break up integral: Γ(t)W(r, s, t) =

  • m,n≥1

1 mrns θ + ∞

θ

  • xt−1e−(m+n)xdx

=

  • m,n≥1

Γ(t, (m + n)θ) mrns(m + n)t + θ xt−1Lir(e−x)Lis(e−x)dx.

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Sketch of proof: Use defining integral of Γ(a, z). A simple substitution gives ∞

θ

xt−1e−(m+n)xdx = Γ(t, (m + n)θ) (m + n)t . Use same argument as in the 2nd Lemma; break up integral: Γ(t)W(r, s, t) =

  • m,n≥1

1 mrns θ + ∞

θ

  • xt−1e−(m+n)xdx

=

  • m,n≥1

Γ(t, (m + n)θ) mrns(m + n)t + θ xt−1Lir(e−x)Lis(e−x)dx. Use the first Lemma, namely Lis(z) =

  • m=0

ζ(s − m)logm z m! + Γ(1 − s)(− log z)s−1.

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Sketch of proof: Use defining integral of Γ(a, z). A simple substitution gives ∞

θ

xt−1e−(m+n)xdx = Γ(t, (m + n)θ) (m + n)t . Use same argument as in the 2nd Lemma; break up integral: Γ(t)W(r, s, t) =

  • m,n≥1

1 mrns θ + ∞

θ

  • xt−1e−(m+n)xdx

=

  • m,n≥1

Γ(t, (m + n)θ) mrns(m + n)t + θ xt−1Lir(e−x)Lis(e−x)dx. Use the first Lemma, namely Lis(z) =

  • m=0

ζ(s − m)logm z m! + Γ(1 − s)(− log z)s−1. Expand and then integrate. QED

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2.

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0.

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0. LHS: sΓ(s) → 1.

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0. LHS: sΓ(s) → 1. RHS: Almost all terms disappear, except

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0. LHS: sΓ(s) → 1. RHS: Almost all terms disappear, except – 2nd row for u = v = 0; get ζ(0)2 = (−1/2)2 = 1/4;

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0. LHS: sΓ(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 2Γ(0)(−1)ζ(−1)/2 = −ζ(−1) = 1/12.

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First application: Set r = s = t; then Γ(s)ω3(s) =

  • m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) + Γ(1 − s)s θ3s−2 3s − 2. Fix θ > 0, multiply both sides by s and let s → 0. LHS: sΓ(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 2Γ(0)(−1)ζ(−1)/2 = −ζ(−1) = 1/12. Together: ω3(0) = 1/3.

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Remarks:

  • 1. This last identity also immediately gives the singularities of

ω3(s) and their residues.

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Remarks:

  • 1. This last identity also immediately gives the singularities of

ω3(s) and their residues.

  • 2. With only a small variation we can prove a more general

result: Define ω3(s; τ) := W(s, s, τs).

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Remarks:

  • 1. This last identity also immediately gives the singularities of

ω3(s) and their residues.

  • 2. With only a small variation we can prove a more general

result: Define ω3(s; τ) := W(s, s, τs). Then for any τ > 0 we have ω3(0; τ) = ζ(0)2 − 2 τ τ + 1ζ(−1) = 1 12 5 τ + 3 τ + 1 ,

Karl Dilcher Tornheim zeta function

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Remarks:

  • 1. This last identity also immediately gives the singularities of

ω3(s) and their residues.

  • 2. With only a small variation we can prove a more general

result: Define ω3(s; τ) := W(s, s, τs). Then for any τ > 0 we have ω3(0; τ) = ζ(0)2 − 2 τ τ + 1ζ(−1) = 1 12 5 τ + 3 τ + 1 , and in particular, ω3(0) = ω3(0; 1) = 1 3.

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  • 4. Derivative at the origin

Let’s return to Crandall’s identity; multiply both sides by s: sΓ(s)ω3(s) = s  

m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) +Γ(1 − s)s θ3s−2 3s − 2

  • .

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  • 4. Derivative at the origin

Let’s return to Crandall’s identity; multiply both sides by s: sΓ(s)ω3(s) = s  

m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +

  • u,v≥0

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) + 2Γ(1 − s)

  • q≥0

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1) +Γ(1 − s)s θ3s−2 3s − 2

  • .

Now isolate the singularies in s in the large brackets on the RHS; bring them to the left:

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sΓ(s)ω3(s) − ζ(s)2θ2 + Γ(1 − s)ζ(s − 1)θ2s = s  

m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +2Γ(1 − s)ζ(s) θ2s−1 2s − 1 + Γ(1 − s)2 θ3s−2 3s − 2 +

  • u,v≥0

(u,v)=(0,0)

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) +2Γ(1 − s)

  • q≥2

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1)   .

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sΓ(s)ω3(s) − ζ(s)2θ2 + Γ(1 − s)ζ(s − 1)θ2s = s  

m,n≥1

Γ(s, (m + n)θ) (mn(m + n))s +2Γ(1 − s)ζ(s) θ2s−1 2s − 1 + Γ(1 − s)2 θ3s−2 3s − 2 +

  • u,v≥0

(u,v)=(0,0)

(−1)u+v ζ(s − u)ζ(s − v)θu+v+s u!v!(u + v + s) +2Γ(1 − s)

  • q≥2

(−1)q ζ(s − q)θ2s+q−1 q!(2s + q − 1)   . Derivative at s = 0 of LHS becomes ω′

3(0) − 5 12γ − 1 2 log 2π − 5 12 log θ + ζ′(−1).

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Derivative at s = 0 of RHS amounts to evaluating [. . .] at s = 0.

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

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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.

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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:

  • m,n≥1

Γ(0, (m + n)θ) = ∞

1

du (eθu − 1)2u . (Easy manipulation using definition of Γ(a, z)).

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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:

  • m,n≥1

Γ(0, (m + n)θ) = ∞

1

du (eθu − 1)2u . (Easy manipulation using definition of Γ(a, z)). ∞ tα−1 (et − 1)2 dt = Γ(α)(ζ(α − 1) − ζ(α)) (Re(α) > 2). (An integral in Gradshteyn & Ryzhik).

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Everything put together, we get (after some work)

  • m,n≥1

Γ(0, (m + n)θ) = 1 2 log(2π) − 5γ 12 + ζ′(−1) − 1 θ + 1 2θ2 − 5 12 log θ + O(θ).

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Everything put together, we get (after some work)

  • m,n≥1

Γ(0, (m + n)θ) = 1 2 log(2π) − 5γ 12 + ζ′(−1) − 1 θ + 1 2θ2 − 5 12 log θ + O(θ). Finally: Theorem ω′

3(0) = log(2π).

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Everything put together, we get (after some work)

  • m,n≥1

Γ(0, (m + n)θ) = 1 2 log(2π) − 5γ 12 + ζ′(−1) − 1 θ + 1 2θ2 − 5 12 log θ + O(θ). Finally: Theorem ω′

3(0) = log(2π).

As before, with small modifications we get more generally ω′

3(0; τ) = τ + 1

2 log(2π) + (τ − 1)τ τ + 1 ζ′(−1). (Recall: ω3(s; τ) := W(s, s, τs).)

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  • 5. Extensions
  • 1. A multi-dimensional analogue:

For n ≥ 2 define W(r1, . . . , rn, t) :=

  • m1,...,mn≥1

1 mr1

1 . . . mrn n (m1 + . . . mn)t

Studied by Matsumoto (2000), Bailey & Borwein (2015), and

  • thers.

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  • 5. Extensions
  • 1. A multi-dimensional analogue:

For n ≥ 2 define W(r1, . . . , rn, t) :=

  • m1,...,mn≥1

1 mr1

1 . . . mrn n (m1 + . . . mn)t

Studied by Matsumoto (2000), Bailey & Borwein (2015), and

  • thers. In analogy to before, define

ωn+1(s) := W(s, . . . , s, s).

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  • 5. Extensions
  • 1. A multi-dimensional analogue:

For n ≥ 2 define W(r1, . . . , rn, t) :=

  • m1,...,mn≥1

1 mr1

1 . . . mrn n (m1 + . . . mn)t

Studied by Matsumoto (2000), Bailey & Borwein (2015), and

  • thers. In analogy to before, define

ωn+1(s) := W(s, . . . , s, s). Hayley Tomkins (honours thesis, 2016) showed that ωn+1(0) = (−1)n n + 1 holds for n ≤ 7, and conjectured that it is true for all n.

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

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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 ωn+1(−k) = 0 for all integers n ≥ 2 and k ≥ 1.

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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 ωn+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, . . ..

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Also proved by Tomkins: ω′

4(0) = − log(2π) + ζ′(−2)

= − log(2π) − ζ(3) 4π2 .

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Also proved by Tomkins: ω′

4(0) = − log(2π) + ζ′(−2)

= − log(2π) − ζ(3) 4π2 . How about ω′

n+1(0) for n ≥ 4?

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Recall: ω′

3(0) = log(2π),

ω′

4(0) = − log(2π) + ζ′(−2).

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Recall: ω′

3(0) = log(2π),

ω′

4(0) = − log(2π) + ζ′(−2).

Bailey & Borwein found experimentally: ω′

5(0) = log(2π) − 2ζ′(−2)

ω′

6(0) = − log(2π) + 35 12ζ′(−2) + 1 12ζ′(−4)

ω′

7(0) = log(2π) − 15 4 ζ′(−2) − 1 4ζ′(−4)

. . . ω′

19(0) = log(2π) − 344499373 33633600 ζ′(−2) − . . . − 1 1162377216000ζ′(−16).

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Recall: ω′

3(0) = log(2π),

ω′

4(0) = − log(2π) + ζ′(−2).

Bailey & Borwein found experimentally: ω′

5(0) = log(2π) − 2ζ′(−2)

ω′

6(0) = − log(2π) + 35 12ζ′(−2) + 1 12ζ′(−4)

ω′

7(0) = log(2π) − 15 4 ζ′(−2) − 1 4ζ′(−4)

. . . ω′

19(0) = log(2π) − 344499373 33633600 ζ′(−2) − . . . − 1 1162377216000ζ′(−16).

What are the coefficients in these expressions?

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Theorem For any n ≥ 2 we have ω′

n+1(0) = (−1)n log(2π) + 2 (n−1)! ⌊ n−1

2 ⌋

  • j=1

s(n, 2j + 1)ζ′(−2j), where s(n, k) are the Stirling numbers of the first kind.

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Theorem For any n ≥ 2 we have ω′

n+1(0) = (−1)n log(2π) + 2 (n−1)! ⌊ n−1

2 ⌋

  • j=1

s(n, 2j + 1)ζ′(−2j), where s(n, k) are the Stirling numbers of the first kind. (Note: ζ′(0) = − 1

2 log(2π)).

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Theorem For any n ≥ 2 we have ω′

n+1(0) = (−1)n log(2π) + 2 (n−1)! ⌊ n−1

2 ⌋

  • j=1

s(n, 2j + 1)ζ′(−2j), where s(n, k) are the Stirling numbers of the first kind. (Note: ζ′(0) = − 1

2 log(2π)).

Proof uses similar methods as that of the original case n = 2.

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Theorem For any n ≥ 2 we have ω′

n+1(0) = (−1)n log(2π) + 2 (n−1)! ⌊ n−1

2 ⌋

  • j=1

s(n, 2j + 1)ζ′(−2j), where s(n, k) are the Stirling numbers of the first kind. (Note: ζ′(0) = − 1

2 log(2π)).

Proof uses similar methods as that of the original case n = 2. Recall:

n

  • k=0

s(n, k)xk = x(x − 1) . . . (x − n + 1);

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Theorem For any n ≥ 2 we have ω′

n+1(0) = (−1)n log(2π) + 2 (n−1)! ⌊ n−1

2 ⌋

  • j=1

s(n, 2j + 1)ζ′(−2j), where s(n, k) are the Stirling numbers of the first kind. (Note: ζ′(0) = − 1

2 log(2π)).

Proof uses similar methods as that of the original case n = 2. Recall:

n

  • k=0

s(n, k)xk = x(x − 1) . . . (x − n + 1); s(n, k) = s(n − 1, k − 1) − (n − 1)s(n − 1, k).

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  • 2. Character analogues

Introduced and studied by Bailey & Borwein (Math. Comp. 2016).

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  • 2. Character analogues

Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by A(r, s, t) :=

  • m,n≥1

(−1)m mr (−1)n ns 1 (m + n)t .

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  • 2. Character analogues

Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by A(r, s, t) :=

  • m,n≥1

(−1)m mr (−1)n ns 1 (m + n)t . In analogy to ω3(s), consider α3(s) := A(s, s, s).

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  • 2. Character analogues

Introduced and studied by Bailey & Borwein (Math. Comp. 2016). Easiest case: Alternating Tornheim zeta function, defined by A(r, s, t) :=

  • m,n≥1

(−1)m mr (−1)n ns 1 (m + n)t . In analogy to ω3(s), consider α3(s) := A(s, s, s). Analogue to Crandall’s formula is much simpler: Γ(s)α3(s) =

  • m,n≥1

(−1)m mr (−1)n ns Γ(s, (m + n)θ) (m + n)s +

  • u,v≥0

(−1)u+v η(s − u)η(s − v)θu+v+s u!v!(u + v + s) .

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Here η(s) :=

  • n=1

(−1)n+1 ns = (1 − 21−s)ζ(s) is the alternating zeta function.

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Here η(s) :=

  • n=1

(−1)n+1 ns = (1 − 21−s)ζ(s) is the alternating zeta function. Some special values: η(1) = − log 2, η′(0) = 1 2 log π 2.

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Here η(s) :=

  • n=1

(−1)n+1 ns = (1 − 21−s)ζ(s) is the alternating zeta function. Some special values: η(1) = − log 2, η′(0) = 1 2 log π 2. Following same procedure as before, we find α3(0) = η(0)2 = 1 4.

Karl Dilcher Tornheim zeta function

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Here η(s) :=

  • n=1

(−1)n+1 ns = (1 − 21−s)ζ(s) is the alternating zeta function. Some special values: η(1) = − log 2, η′(0) = 1 2 log π 2. Following same procedure as before, we find α3(0) = η(0)2 = 1 4. Furthermore, using methods of before: α′

3(0) = 2η′(0) − η′(−1) − 1 4γ

= log(2π) − 5

3 log 2 − 1 4γ + 3ζ′(−1).

Karl Dilcher Tornheim zeta function

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Character analogues in general:

  • On the one hand it will be easier because the Crandall-like

formula will always be simpler than in the principal case.

Karl Dilcher Tornheim zeta function

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Character analogues in general:

  • On the one hand it will be easier because the Crandall-like

formula will always be simpler than in the principal case.

  • On the other hand, obtaining explicit values and derivatives at

the origin will be difficult if not impossible in general.

Karl Dilcher Tornheim zeta function

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Character analogues in general:

  • On the one hand it will be easier because the Crandall-like

formula will always be simpler than in the principal case.

  • On the other hand, obtaining explicit values and derivatives at

the origin will be difficult if not impossible in general. General Remark: Many of the results in this talk were first obtained experimentally before they were proved.

Karl Dilcher Tornheim zeta function

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Character analogues in general:

  • On the one hand it will be easier because the Crandall-like

formula will always be simpler than in the principal case.

  • On the other hand, obtaining explicit values and derivatives at

the origin will be difficult if not impossible in general. General Remark: Many of the results in this talk were first obtained experimentally before they were proved. Knowing what to expect provides a great deal of guidance, as well as certainty when it’s done.

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Thank you

Karl Dilcher Tornheim zeta function