PresentationCopyPhiladelphiaAreaNumberTheorySeminarTalk12716

Outline
Introduction
Main Theorem
Explicit Formula for the Mean Square of a
Dirichlet L-Function of Prime Power Modulus
Frank Romascavage, III
Thesis under the direction of Professor Djordje Mili´cevi´c
Bryn Mawr College Graduate School of Arts and Sciences
Philadelphia Area Number Theory Seminar
Bryn Mawr College
December 7, 2016
Frank Romascavage, III Mean Square of a Dirichlet L-function of Modulus pβ

Outline
Introduction
Main Theorem
Introduction
Abstract
Review
Motivation
Atkinson’s Formula for the Mean Square
Motohashi’s Formula
Analogy
Results
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
How We Get A2(g), A3(g), and A4(g)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Abstract
We seek to derive an explicit formula for the mean square of a
Dirichlet L-function of prime power modulus.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Abstract
Motohashi derived an explicit formula for the mean square (second
power moment) and also the fourth power moment of the Riemann
Zeta Function.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Abstract
Zeta Function.
There are many estimates on the power moments for a Dirichlet
L-function, but this work will not include an error term.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Abstract
Zeta Function.
The values of the L-functions at the central value often have deep
arithmetic signiﬁcance, but are diﬃcult to study individually.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Abstract
Zeta Function.
The values of the L-functions at the central value often have deep
arithmetic signiﬁcance, but are diﬃcult to study individually.
In several important problems in analytic number theory, including
how primes are distributed, estimates such as power moments can
give important information about these central values in families.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Overview of Review
In this section, we will brieﬂy discuss such topics as the Divisor
Function, Gamma Function, Fourier Transform, Mellin Transform,
Dirichlet Characters, and Dirichlet L-series. We will quickly go over
key concepts of each of these respective areas that will be utilized
throughout the rest of the work.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Divisor Function
Let d be the divisor function deﬁned by
d(n) =
w|n
1.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Divisor Function
Let d be the divisor function deﬁned by
d(n) =
w|n
1.
Let ∆ be the Delta function given by
∆(x) =
n≤x
d(n) − x(log x + 2γ − 1) −
1
4
,
where in this sum the last term is to be divided by 2 if x ∈ Z.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gamma Function
Let Γ : C(−Z ∪ {0}) → C be deﬁned by
Γ(x) =
∞
0
tx−1
e−t
dt, (Re(x) > 0)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gamma Function
Γ(x) =
∞
0
tx−1
e−t
dt, (Re(x) > 0)
Functional equation: Γ(s + 1) = sΓ(s)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gamma Function
Γ(x) =
∞
0
tx−1
e−t
dt, (Re(x) > 0)
Euler’s reﬂection formula: Γ(z)Γ(1 − z) = π
sin(πz)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gamma Function
Γ(x) =
∞
0
tx−1
e−t
dt, (Re(x) > 0)
Euler’s reﬂection formula: Γ(z)Γ(1 − z) = π
sin(πz)
Stirling’s formula:
|Γ(σ + it)| = (2π)1/2|t|σ−1
2 e−π|t|/2 1 + O |t|−1

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Introduction to Dirichlet Characters
Let q ∈ N. Let (Z/qZ)∗
= {[a]q| gcd(a, q) = 1} and
C∗ = {z ∈ C|z = 0}. A homomorphism χ : (Z/qZ)∗
→ C∗ is
called a Dirichlet character of modulus q.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Introduction to Dirichlet Characters
Let q ∈ N. Let (Z/qZ)∗
= {[a]q| gcd(a, q) = 1} and
C∗ = {z ∈ C|z = 0}. A homomorphism χ : (Z/qZ)∗
→ C∗ is
called a Dirichlet character of modulus q.
χ(m) =
χ(m(mod q)) if gcd(m, q) = 1
0 if gcd(m, q) > 1

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gauss Sum
Let χ be a Dirichlet character of modulus q. Then, we have
that the Gauss sum of χ is given by
τ(χ) =
n(mod q)
χ(n)e2πin/q
.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Gauss Sum
Let χ be a Dirichlet character of modulus q. Then, we have
that the Gauss sum of χ is given by
τ(χ) =
n(mod q)
χ(n)e2πin/q
.
Note that we have the estimate |τ(χ)| ≤
√
q.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Orthogonality Relations
Let χ be a Dirichlet character of modulus q and χ0 be the trivial
Dirichlet character of modulus q. Then, we have that
a(mod q)
χ(a) =
ϕ(q) if χ = χ0
0 if χ = χ0.
and
χ(mod q)
χ(m) =
ϕ(q) if m ≡ 1(mod q)
0 otherwise.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Dirichlet L-series
A Dirichlet L-series is a function of the form
L(s, χ) =
∞
n=1
χ(n)
ns
.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Dirichlet L-series
A Dirichlet L-series is a function of the form
L(s, χ) =
∞
n=1
χ(n)
ns
.
In addition, when Re(s) > 1, we have that L(s, χ) can be
written as an Euler product:
p prime
1 −
χ(p)
ps
−1
.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Functional Equations for L(s, χ) and ζ(s)
Let χ be a primitive character of modulus q. Let
ξ(s, χ) =
q
π
s/2
Γ
s + aχ
2
L(s, χ),
where aχ = 0 when χ(−1) = 1 and aχ = 1 when
χ(−1) = −1. Then, we have that the following functional
equation holds
L(s, χ) =
τ(χ)
iaχ π1/2
q
π
−s
Γ
1 − s + aχ
2
Γ
s + aχ
2
−1
L(1−s, χ).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Functional Equations for L(s, χ) and ζ(s)
Let χ be a primitive character of modulus q. Let
ξ(s, χ) =
q
π
s/2
Γ
s + aχ
2
L(s, χ),
where aχ = 0 when χ(−1) = 1 and aχ = 1 when
χ(−1) = −1. Then, we have that the following functional
equation holds
L(s, χ) =
τ(χ)
iaχ π1/2
q
π
−s
Γ
1 − s + aχ
2
Γ
s + aχ
2
−1
L(1−s, χ).
In addition, recall the functional equation for the Riemann
zeta function: ζ(s) = 2sπs−1 sin 1
2sπ Γ(1 − s)ζ(1 − s).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Dealing with Non-Primitive Characters
Now, we will consider the case of χ not being a primitive character
of modulus q. If χ is a non-primitive character of conductor q |q,
and if χq is the primitive character equivalent to χ then
L(s, χ) = L(s, χq )
p|q,p q
1 −
χq (p)
ps
.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Fourier Transform
For suitable f : R → C, the Fourier Transform of f is Ff : R → C
deﬁned by
(Ff )(ξ) =
1
2π R
f (t)e−iξt
dt.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Fourier Transform for Finite Abelian Groups
If G is a ﬁnite abelian group, then for f : G → C, the Fourier
Transform (FT) of f , f : G → C (where G is the dual group of G),
is deﬁned by
f (χ) =
g∈G
f (g)χ(g).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Inverse Fourier Transform
For suitable g : R → C, the Inverse Fourier Transform of g is:
F−1g : R → C deﬁned by
(F−1
g)(t) =
R
g(ξ)eiξt
dξ.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Fourier Inversion Formula
Let f : G → C. The Fourier Inversion Formula (FIF) is given by
f (x) =
1
|G|
χ∈G
f (χ) χ(x),
where |G| is the order of G.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Mellin Transform
For suitable f : R+ → C, the Mellin Transform of f is:
Mf : {Re(s) > σ0} → C (where σ0 > 0) deﬁned by
(Mf )(s) =
R+
f (t)ts dt
t
.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Mellin Transform for Finite Abelian Groups
If G is a ﬁnite abelian group, then for f : G → C∗, the Mellin
Transform (MT) of f , denoted by Mf , is given by:
(Mf )(χ) =
y∈G
f (y)χ(y).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Inverse Mellin Transform
The Inverse Mellin Transform of g is: M−1g : R+ → C deﬁned by
(M−1
g)(t) =
1
2πi Re(s)=σ
g(s)t−s
ds
for any suitable σ (where σ > 0).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Mellin Inversion Formula
If G is a ﬁnite abelian group and f : G → C∗, then the Mellin
Inversion Formula is given by
f (g) =
1
|G|
χ∈G
(Mf )(χ)χ(g).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motivation Part 1
Why do we look at moments of L-functions?

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motivation Part 1
Lindel¨of hypothesis for the Riemann zeta function: For any
> 0, ζ 1
2 + it = O(t )

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motivation Part 1
Lindel¨of hypothesis for the Riemann zeta function: For any
> 0, ζ 1
2 + it = O(t )
The Lindel¨of hypothesis is true if and only if for all n ∈ N,
1
T
T
0 ζ 1
2 + it
2n
dt = O(T ).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motivation Part 2
Generalized Lindel¨of hypothesis (GLH): Suppose L(s, f ) is an
L-function which has a meromorphic continuation to the
entire s-plane and a functional equation relating L(s, f ) and
L(1 − s, f ). Then, for any > 0, L 1
2 + it, f = O(t ).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motivation Part 2
Generalized Lindel¨of hypothesis (GLH): Suppose L(s, f ) is an
L-function which has a meromorphic continuation to the
entire s-plane and a functional equation relating L(s, f ) and
L(1 − s, f ). Then, for any > 0, L 1
2 + it, f = O(t ).
We study moments of the form 1
T
T
0 L 1
2 + it, f
2n
dt, for
every n ∈ N to gain a better understanding of the GLH.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Investigate the asymptotics of
I(T) =
T
0 ζ(1
2 + it)
2
dt = T log T
2π + (2γ − 1)T + E(T)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
I(T) =
T
0 ζ(1
2 + it)
2
dt = T log T
2π + (2γ − 1)T + E(T)
Let N T and N = T
2π + N
2 − N2
4 + NT
2π
1/2
. Then,
E(T) = 1(T) + 2(T) + O(log2
T), where

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
I(T) =
T
0 ζ(1
2 + it)
2
dt = T log T
2π + (2γ − 1)T + E(T)
Let N T and N = T
2π + N
2 − N2
4 + NT
2π
1/2
. Then,
E(T) = 1(T) + 2(T) + O(log2
T), where
1(T) =
√
2 T
2π
1/4
n≤N(−1)nd(n)n−3/4e(T, n)
cos(f (T, n)),

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
I(T) =
T
0 ζ(1
2 + it)
2
dt = T log T
2π + (2γ − 1)T + E(T)
Let N T and N = T
2π + N
2 − N2
4 + NT
2π
1/2
. Then,
E(T) = 1(T) + 2(T) + O(log2
T), where
1(T) =
√
2 T
2π
1/4
n≤N(−1)nd(n)n−3/4e(T, n)
cos(f (T, n)),
2(T) = −2 n≤N d(n)n−1/2 log T
2πn
−1
cos(g(T, n)),

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
I(T) =
T
0 ζ(1
2 + it)
2
dt = T log T
2π + (2γ − 1)T + E(T)
Let N T and N = T
2π + N
2 − N2
4 + NT
2π
1/2
. Then,
E(T) = 1(T) + 2(T) + O(log2
T), where
1(T) =
√
2 T
2π
1/4
n≤N(−1)nd(n)n−3/4e(T, n)
cos(f (T, n)),
2(T) = −2 n≤N d(n)n−1/2 log T
2πn
−1
cos(g(T, n)),
e(T, n) = 1 + O n
T , f (T, n) = 4π πn
2T − π
4 +
O n3/2T−1/2 , g(T, n) = T log T
2πn − T + π
4

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
What Motohashi Accomplished Part 1
Motohashi was able to obtain an exact formula for
L1(g) =
∞
−∞ ζ 1
2 + it
2
g(t) dt when g is a regular
function that satisﬁes g(r) = O (|r| + 1)−A
for |Im(r)| ≤ A
where A is a large positive constant.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motohashi’s formula:
L1(g) =
∞
−∞
Re
Γ
Γ
1
2
+ it + 2cE − log(2π) g(t) dt + 2π
Re g
1
2
i + 4
∞
n=1
d(n)
∞
0
(y(y + 1))−1/2
gc log 1 +
1
y
cos (2πny) dy,
where cE is Euler’s constant and
gc(x) =
∞
−∞
g(t) cos(xt) dt.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motohashi achieved this result in several steps.
First, examine J(u, v; g) =
∞
−∞ ζ(u + it)ζ(v − it)g(t) dt in
order to get a better understanding of L1(g)

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Motohashi achieved this result in several steps.
First, examine J(u, v; g) =
∞
−∞ ζ(u + it)ζ(v − it)g(t) dt in
order to get a better understanding of L1(g)
Second, divide J(u, v; g) into diagonal and oﬀ-diagonal sums
in the area of absolute convergence, i.e.
J(u, v; g) =
m=n
+
m<n
+
m>n
m−u
n−v
g∗
log
n
m
= ζ(u + v)g∗
(0) + J1(u, v; g) + J1(v, u; g),
where g∗ is the Fourier Transform of g.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Third, continue J1 meromorphically to the domain
|u|, |v| < B = cA where c is a small constant and A is a large
constant.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
constant.
This occurs by using the Mellin Transform, Stirling’s Formula,
the Inverse Mellin Transform, and then moving the path of
integration by contour integration.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
constant.
Next, the functional equation of the Riemann zeta function is
utilized.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
constant.
Next, the functional equation of the Riemann zeta function is
utilized.
Since the region |u|, |v| < B = cA is symmetric, we can take
u = v = α.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Fourth, the path of integration is moved again by contour
integration.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Fourth, the path of integration is moved again by contour
integration.
Fifth, α = 1
2 is examined. Singularities are canceled out as α
approaches 1
2, the remaining integral is transformed, and then
the desired result is obtained.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Analogy
We want to ﬁnd an analogous formula for
L1(g) =
∞
−∞
ζ
1
2
+ it
2
g(t) dt

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Analogy
L1(g) =
∞
−∞
ζ
1
2
+ it
2
g(t) dt
Integrating over all t ∈ R is analogous to summing over all
characters χ(mod pβ).

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Analogy
L1(g) =
∞
−∞
ζ
1
2
+ it
2
g(t) dt
Integrating over all t ∈ R is analogous to summing over all
characters χ(mod pβ).
By comparing ζ(u + it) with L(u, χ), we note that χ(n) plays
the role of n−it in Motohashi: ζ(u + it) = ∞
n=1
1
nunit and
L(u, χ) = ∞
n=1
χ(n)
nu .

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Results Part 1
P. X. Gallagher (1975): For q ≥ 2 and t ∈ R,
χ(mod q)
L
1
2
+ it, χ
2
= O ((q + |t|) log q (|t| + 2))

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Results Part 2
R. Balasubramanian (1980): For q ≥ 2 and t ≥ 3,
χ(mod q)
L
1
2
+ it, χ
2
=
ϕ2(q)
q
log qt + O q (log log q)2
+
O te10
√
log q
+ O q
1
2 t
2
3 e10
√
log q

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Results Part 3
D. R. Heath-Brown (1981):
χ(mod q)
L
1
2
, χ
2
=
ϕ(q)
q
k|q
µ
q
k
T(k),
where
T(k) = k log
k
8π
+ γ + 2ζ2 1
2
k
1
2 +
2N−1
n=0
cnk−n
2 + O k−N
for any integer N ≥ 1, numerical constants cn, and Euler’s
constant γ.

Outline
Introduction
Main Theorem
Abstract
Review
Motivation
Analogy
Results
Results Part 4
Zhang Wenpeng (1989 - 1991):
χ(mod q)
L
1
2
+ it, χ
2
=
ϕ2(q)
q



log
qt
2π
+ 2γ +
p|q
log p
p − 1



+
O qt−1/12
+ O t5/6
+ q1/2
t5/12
exp
2 log(qt)
log log(qt)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let p be a prime, β ∈ N, ν ∈ N such that 0 ≤ ν < β, and g be an
arbitrary real-valued weight function on (Z/pβZ)∗.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let F1(g) =
χ∈(Z/pβ Z)∗ L 1
2
, χ
2
g (χ) where g is a weight function.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let F1(g) =
2
, χ
2
Let C = 2γ +
Γ (1
2 )
Γ(1
2 )
, where Γ is the Gamma function and γ is the
Euler-Mascheroni constant.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let F1(g) =
2
, χ
2
Let C = 2γ +
Γ (1
2 )
Γ(1
2 )
Let τ be the Gauss sum. Let d be the divisor function and ∆ be the
Delta function.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let F1(g) =
2
, χ
2
Let C = 2γ +
Γ (1
2 )
Γ(1
2 )
Delta function.
Let f : C × R → C be deﬁned by f (u, x) = 2
∞
0
y−u
(y + 1)u−1
cos(2π xy).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1 of Statement
Let F1(g) =
2
, χ
2
Let C = 2γ +
Γ (1
2 )
Γ(1
2 )
Delta function.
Let f : C × R → C be deﬁned by f (u, x) = 2
∞
0
y−u
(y + 1)u−1
cos(2π xy).
Let Tg (z) = 1
ϕ(q) ψ∈(Z/pβ Z)∗
g(ψ)ψ(z) and ˜gβ−ν (y) = y(mod pβ−ν )
Tg (1 + pν
y)χ(y)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Off-Diagonal Sums
Part 2 of Statement
Let W1 be defined by
W1(b) = π−1/2
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s e−isπ/2
ds +
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s eisπ/2
ds
and W2 be defined by
W2(b) = π1/2
i
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s e−isπ/2
ds −
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s eisπ/2
ds .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 3 of Statement
Then, F1(g) = A1(g) + A2(g) + A3(g) + A4(g), where
A1(g) := pβ
1 −
1
p
2
χ∈(Z/pβZ)∗
g (χ)
log p
1 − p
+ log
pβ
2π
+ C+
4p−β
1 −
1
p
−1
=pβ,pβ−1
µ
pβ
f
1
2
, 1 − f
1
2
,
3
2
∆
3
2
+
(π )−1
∞
0
y3
(y + 1)
−1/2
sin (3π y)
1
2
− log
3
2
+ 2γ (y + 1)
−
∞
3/2
∆(t)
∂f 1
2, t
∂t
dt ,

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 4 of Statement
A2(g) :=
0≤ν<β
pν p − 1
p
Re(˜gβ−ν(χ0)) log p

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 5 of Statement
A3(g) := −
2
π
0≤ν<β
pν
∞
f =1
d(f )f −1/2
j=−1,−2
f j
Γ
1
2
− j
(−1)−j
Γ(1 − j)
p2νj
pβ
π
− 1
2 −j
∗
χ(mod pβ−ν )
Γ
1
2 −j+aχ
2
Γ
1
2 +j+aχ
2
Re i−aχ
τ(χ)˜gβ−ν(χ)χ(f ) +
β−ν
γ=1 χpγ ≡χ
π−1/2
Γ
aχpγ −j+ 1
2
2
Γ
aχpγ + 1
2 +j
2
Re i−aχpγ
τ(χpγ )˜gβ−ν(χpγ )χpγ (f ) ,
and

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 6 of Statement
A4(g) :=
1
πpβ
0≤ν<β
p2ν
∞
b=1
d(b) Im W1(b)
∗
χ(mod pβ−ν )
χ even
τ(χ)˜gβ−ν(χ)
χ(b) +
β−ν
γ=1 χpγ ≡χ
χpγ even
τ(χpγ )˜gβ−ν(χpγ )χpγ (b) + Re W2(b)
∗
χ(mod pβ−ν )
χ odd
τ(χ)˜gβ−ν(χ)χ(b) +
χpγ ≡χ
χpγ odd
τ(χpγ )˜gβ−ν(χpγ )χpγ (b) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Introducing F(u, v; g)
When dealing with F1(g) we can consider
F(u, v; g) =
χ∈(Z/pβZ)
∗
L(u, χ)L(v, χ)g (χ)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Expansion of F(u, v; g)
Since we are starting with Re(u), Re(v) > 1, we have that
F(u, v; g) =
χ∈(Z/pβZ)
∗
∞
m=1
χ(m)
mu
∞
n=1
χ(n)
nv
g (χ)
=
∞
m=1
∞
n=1
m−u
n−v
χ∈(Z/pβZ)
∗
χ
m
n
g (χ)
= ϕ(pβ
)
∞
m=1
∞
n=1
m−u
n−v
Tg
n
m
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Off-Diagonal Sums
Orthogonality, Diagonal, Near-diagonal, and Off-diagonal
Sums
By orthogonality and breaking up F(u, v; g) into diagonal, near-diagonal, and
off-diagonal sums, we get that F(u, v; g) is equal to
ϕ(pβ
)
m=n
(m,pβ
)=(n,pβ
)=1
+
m<n
m≡n(mod pβ
)
(m,pβ
)=(n,pβ
)=1
+
m>n
m≡n(mod pβ
)
(m,pβ
)=(n,pβ
)=1
+
m<n
m≡n(mod pβ
)
(m,pβ
)=(n,pβ
)=1
+
m>n
m≡n(mod pβ
)
(m,pβ
)=(n,pβ
)=1
m−u
n−v
Tg
n
m
. (Call each of these sums F1(u, v; g)(diagonal)
F2(u, v; g) (near-diagonal), F3(u, v; g)(near-diagonal), F4(u, v; g) (off-diagonal) and
F5(u, v; g)(off-diagonal), respectively.)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Big Picture
The ﬁrst three terms, diagonal and near-diagonal sums,
contribute to the main term, which is A1(g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Big Picture
Contour shifting and analytic continuation will be used.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Big Picture
The oﬀ-diagonal sums will contribute to the sum terms, A3(g)
and A4(g), and A2(g) as well.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Big Picture
The oﬀ-diagonal sums will contribute to the sum terms, A3(g)
and A4(g), and A2(g) as well.
Mellin transform and Mellin Inversion Formula are used for the
last two.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Evaluation of F1(u, v; g)
Notice that
F1(u, v; g) = L(u + v, χ0)
χ∈(Z/pβZ)
∗
g(χ)
by appealing directly from F(u, v; g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Expanding F2(u, v; g)
By appealing directly from F(u, v; g),
F2(u, v; g) =
χ∈(Z/pβZ)
∗
g(χ)
m
(m,pβ)=1
∞
j=1
m−u
(m + jpβ
)−v
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Relating F3(u, v; g) to F2(u, v; g)
Note that by utilizing the deﬁnitions of F2(u, v; g) and F3(u, v; g)
and complex conjugation we get that
F2(v, u; g) = F3(u, v; g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Introducing fpβ (u, v) and simplifying
F1(u, v; g) + F2(u, v; g) + F3(u, v; g)
Let
fpβ (u, v) =
m
(m,pβ)=1
∞
j=1
m−u
(m + jpβ
)−v
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Introducing fpβ (u, v) and simplifying
F1(u, v; g) + F2(u, v; g) + F3(u, v; g)
Let
fpβ (u, v) =
m
(m,pβ)=1
∞
j=1
m−u
(m + jpβ
)−v
.
Then, we have that
F1(u, v; g) + F2(u, v; g) + F3(u, v; g) = L(u + v, χ0)+
fpβ (u, v) + fpβ (v, u)
χ∈(Z/pβZ)
∗
g (χ) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Off-Diagonal Sums
Part 1: Taking fpβ (u, v) from R1 = Re(u) > 1, Re(v) > 1
to R2 = Re(u) < −1, Re(u + v) > 2
In the first step, we write fpβ (u, v) as double sum when
Re(u) > 1 and Re(v) > 1:
fpβ (u, v) =
|pβ
µ( )
∞
r=1
∞
j=1
( r)−u
( r + jpβ
)−v
,
where µ is the Möbius function.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Off-Diagonal Sums
Part 1: Taking fpβ (u, v) from R1 = Re(u) > 1, Re(v) > 1
to R2 = Re(u) < −1, Re(u + v) > 2
In the first step, we write fpβ (u, v) as double sum when
Re(u) > 1 and Re(v) > 1:
fpβ (u, v) =
|pβ
µ( )
∞
r=1
∞
j=1
( r)−u
( r + jpβ
)−v
,
where µ is the Möbius function.
In the second step, we rearrange the double sum and note
that we can do this when Re(v) > 1 and Re(u + v) > 2.
∞
r=1
∞
j=1
( r)−u
( r + jpβ
)−v
=
∞
j=1
∞
r=1
( r)−u
( r + jpβ
)−v
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 2: Taking fpβ (u, v) from R1 to R2
In the third step, we will use Poisson Summation to note that
when Re(u) < −1 and Re(u + v) > 2 that
∞
j=1
∞
r=1
( r)−u
( r + jpβ
)−v
=
∞
0
( x)−u
( x + jpβ
)−v
dx+
2
∞
r=1
∞
0
( x)−u
( x + jpβ
)−v
cos(2πrx) dx.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 3: Taking fpβ (u, v) from R1 to R2
In the fourth step, we will substitute z = x, simplify, and
then substitute y = z
jpβ to obtain when Re(u) < −1 and
Re(u + v) > 2 that
∞
0
( x)−u
( x + jpβ
)−v
dx + 2
∞
r=1
∞
0
( x)−u
( x + jpβ
)−v
cos(2πrx) dx = −1
(jpβ
)1−u−v
∞
0
y−u
(y + 1)−v
dy + 2
∞
r=1
∞
0
y−u
(y + 1)−v
cos 2πr
jpβy
dy .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Off-Diagonal Sums
Defining Gpβ (u, v)
Define
Gpβ (u, v) := 2(pβ
)1−u−v
|pβ
µ( ) −1
∞
j=1
j1−u−v
∞
r=1
∞
0
y−u
(1 + y)−v
cos 2πr
pβ
jy dy.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Determining fpβ (u, v) + fpβ (v, u)
Then, when Re(u) < −1 and Re(u + v) > 2, we have that
fpβ (u, v) + fpβ (v, u) = ϕ(pβ
)(pβ
)−u−v
ζ(u + v − 1)Γ(u + v − 1)
Γ(1 − u)
Γ(v)
+
Γ(1 − v)
Γ(u)
+ Gpβ (u, v) + Gpβ (v, u).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Determining fpβ (u, v) + fpβ (v, u)
Then, when Re(u) < −1 and Re(u + v) > 2, we have that
fpβ (u, v) + fpβ (v, u) = ϕ(pβ
)(pβ
)−u−v
ζ(u + v − 1)Γ(u + v − 1)
Γ(1 − u)
Γ(v)
+
Γ(1 − v)
Γ(u)
+ Gpβ (u, v) + Gpβ (v, u).
This is achieved by: substitution from previous evaluations for
fpβ (u, v), substituting jpβy for x, and utilizing that
|pβ µ( ) −1 = ϕ(pβ)
pβ , ∞
j=1 j−1−u−v = ζ(u + v − 1), and
∞
0 y−u(y + 1)−v dy = Γ(u + v − 1)Γ(1−u)
Γ(v) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Analytic Continuation of F1 + F2 + F3 from R1 to
R2
Notice that fpβ (u, v) − Gpβ (u, v) = ϕ(pβ) pβ −u−v
ζ(u + v − 1)Γ(u + v − 1)Γ(1−u)
Γ(v) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
R2
ζ(u + v − 1)Γ(u + v − 1)Γ(1−u)
Γ(v) .
Note that the right-hand side does in fact provide an analytic
continuation of fpβ (u, v) − Gpβ (u, v) to a meromorphic
function of u and v.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
R2
ζ(u + v − 1)Γ(u + v − 1)Γ(1−u)
Γ(v) .
Note that the right-hand side does in fact provide an analytic
continuation of fpβ (u, v) − Gpβ (u, v) to a meromorphic
function of u and v.
So, we have that fpβ (u, v) − Gpβ (u, v) + fpβ (v, u) − Gpβ (v, u)
= ϕ(pβ)(pβ)−u−v ζ(u + v − 1)Γ(u + v − 1) Γ(1−u)
Γ(v) + Γ(1−v)
Γ(u)
is meromorphic.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
R2
This provides an expression for fpβ (u, v) + fpβ (v, u) if this
sum can be continued analytically, but from F1(u, v; g)+
F2(u, v; g)+F3(u, v; g) = L(u+v, χ0)+fpβ (u, v)+fpβ (v, u)
χ∈(Z/pβZ)
∗ g (χ) it deﬁnitely can be.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Show that Gpβ (u, v) is analytic in a region
containing Re(u) = 1
2, Re(v) = 1
2
Recall that
Gpβ (u, v) = 2(pβ
)1−u−v
|pβ
µ( ) −1
∞
j=1
j1−u−v
∞
r=1
∞
0
y−u
(1 + y)−v
cos 2πr
pβ
jy dy.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Recall that
Gpβ (u, v) = 2(pβ
)1−u−v
|pβ
µ( ) −1
∞
j=1
j1−u−v
∞
r=1
∞
0
y−u
(1 + y)−v
cos 2πr
pβ
jy dy.
First, we examine the convergence of 2
∞
0
y−u
(y + 1)−v
cos 2πr pβ
jy dy.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Recall that
Gpβ (u, v) = 2(pβ
)1−u−v
|pβ
µ( ) −1
∞
j=1
j1−u−v
∞
r=1
∞
0
y−u
(1 + y)−v
cos 2πr
pβ
jy dy.
First, we examine the convergence of 2
∞
0
y−u
(y + 1)−v
cos 2πr pβ
jy dy.
This will become 2
∞
0
y−u
(y + 1)−v
cos (2πny) dy by letting n = rpβ
j
(pβ
/ is an integer due to summing over all that divide pβ
).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Contour integration and estimation can be used to show that
2
∞
0
y−u
(y + 1)−v
cos (2πny) dy = nu−1
i∞
0
y−u y
n
+ 1
−v
e2πiy
dy +
−i∞
0
y−u y
n
+ 1
−v
e−2πiy
dy
|n|Re(u)−1
|1 − Re(u)|
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We use the previous estimate to establish that when
Re(u) < 0, Re(v) > 1 and Re(u + v) > 0, Gpβ (u, v) converges
absolutely.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We use the previous estimate to establish that when
Re(u) < 0, Re(v) > 1 and Re(u + v) > 0, Gpβ (u, v) converges
absolutely.
This is achieved by noting that Gpβ (u, v)
2(pβ)1−u−v
|pβ µ( ) −1 ∞
j=1 j1−u−v ∞
n=1
|n|Re(u)−1
|1−Re(u)| .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Now, we will provide a continuation for Gpβ (u, v) from
Re(u) < 0, Re(v) > 1 and Re(u + v) > 0 to Re(u + v)
= 1 + δ, where |δ| < 1
2.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
= 1 + δ, where |δ| < 1
2.
We will soon see that the divisor function appears in analogy
with Motohashi.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
= 1 + δ, where |δ| < 1
2.
with Motohashi.
The divisor function will be multiplied by a smooth weight.
The smooth weight is dealt with by summation by parts.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
= 1 + δ, where |δ| < 1
2.
with Motohashi.
We use results on the divisor problem and advances in it are
why we ultimately end up succeeding.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
= 1 + δ, where |δ| < 1
2.
with Motohashi.
We use results on the divisor problem and advances in it are
why we ultimately end up succeeding.
We use the estimate: ∆(t) t1/3 log t.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We can write
Gpβ (u, 1 − u) = Gpβ,1(u) − Gpβ,2(u) + Gpβ,3(u) − Gpβ,4(u)
where

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We can write
where
Gpβ,1(u) = 2p−β
|pβ µ pβ
m≤M d(m)f (u, m),

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We can write
where
Gpβ,1(u) = 2p−β
|pβ µ pβ
m≤M d(m)f (u, m),
Gpβ,2(u) = 2p−β
|pβ µ pβ
f u, M + 1
2 ∆ M + 1
2 ,

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We can write
where
Gpβ,1(u) = 2p−β
|pβ µ pβ
m≤M d(m)f (u, m),
Gpβ,2(u) = 2p−β
|pβ µ pβ
f u, M + 1
2 ∆ M + 1
2 ,
Gpβ,3(u) = 2p−β
|pβ µ pβ ∞
M+1
2
f (u, t)(log t + 2γ) dt,

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
We can write
where
Gpβ,1(u) = 2p−β
|pβ µ pβ
m≤M d(m)f (u, m),
Gpβ,2(u) = 2p−β
|pβ µ pβ
f u, M + 1
2 ∆ M + 1
2 ,
Gpβ,3(u) = 2p−β
|pβ µ pβ ∞
M+1
2
f (u, t)(log t + 2γ) dt,
Gpβ,4(u) = 2p−β
|pβ µ pβ ∞
M+1
2
∆(t)∂f (u,t)
∂t dt, M ∈ N,
and f (u, x) = 2
∞
0 yu(y + 1)u−1 cos(2π xy) dy.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Each of the four previous summands analytically continue Gpβ (u, 1 − u) from
Re(u) > 0 to Re(u) < 2
3
and Re(v) = Re(1 − u) > 1
3
which contains Re(u) = 1
2
,
Re(v) = 1
2
(due to the 1
3
’s coming from the bounds from the divisor problem)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
2, Re(v) = 1
2
Each of the four previous summands analytically continue Gpβ (u, 1 − u) from
Re(u) > 0 to Re(u) < 2
3
and Re(v) = Re(1 − u) > 1
3
which contains Re(u) = 1
2
,
Re(v) = 1
2
(due to the 1
3
’s coming from the bounds from the divisor problem)
Thus, by substitution, we have that
Gpβ (u, 1 − u) = 2p−β
|pβ
µ
pβ
f (u, 1) − f u,
3
2
∆
3
2
− log
3
2
+
2γ (π )−1
∞
0
y−u−1
(y + 1)u−1
sin (3π y) dy + (π u)−1
∞
0
y−u−1
(y + 1)u
sin (3π y) dy −
∞
3
2
∆(t)
∂f (u, t)
∂t
dt .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Taking F1 + F2 + F3 from R2 = Re(u) < −1, Re(u + v) > 2
to R3 = Re(u + v) = 1 + δ where |δ| < 1
2
Write 1 + δ in place of u + v

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
to R3 = Re(u + v) = 1 + δ where |δ| < 1
2
Use the functional equation for the Riemann zeta function
and Euler’s reﬂection formula

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
to R3 = Re(u + v) = 1 + δ where |δ| < 1
2
Writing the Riemann zeta function with Big O notation:
ζ(1 + δ) = 1
δ + γ + O(|δ|) (permissible due to ζ(1 + δ) being
near 1 + δ = 1)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
to R3 = Re(u + v) = 1 + δ where |δ| < 1
2
Writing the Riemann zeta function with Big O notation:
ζ(1 + δ) = 1
δ + γ + O(|δ|) (permissible due to ζ(1 + δ) being
near 1 + δ = 1)
Taylor expansions of each of the terms and expanding the
products

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
When δ → 0...
Note that as δ → 0, F1(u, v; g) + F2(u, v; g) + F3(u, v; g) in
Re(u + v) = 1 + δ, where |δ| < 1
2, approaches
χ∈(Z/pβZ)∗
g (χ)
ϕ(pβ)
pβ
p|pβ
log p
1 − p
+ 2γ + log
pβ
2π
+
1
2
Γ (1 − u)
Γ(1 − u)
+
1
2
Γ (u)
Γ(u)
+ Gpβ (u, 1 − u) + Gpβ (1 − u, u) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Evaluating F1
1
2, 1
2; g + F2
1
2, 1
2; g + F3
1
2, 1
2; g
Thus, we have that
(F1 + F2 + F3)
1
2
,
1
2
; g = 1 −
1
p
χ∈(Z/pβ Z)∗
g (χ)
log p
1 − p
+ log
pβ
2π
+ C + 4p−β
1 −
1
p
−1
=pβ ,pβ−1
µ
pβ
f
1
2
, 1 − f
1
2
,
3
2
∆
3
2
+ (π )−1
∞
0
y3
(y + 1)
−1/2
sin(3π y)
1
2
− log
3
2
+ 2γ (y + 1) dy −
∞
3/2
∆(t)
∂f 1
2
, t
∂t
dt ,
which is precisely A1(g) (depends only on
χ∈(Z/pβ Z)∗
g (χ), not on individual values
of g(χ)).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Reviewing F4 and F5 and Relating Them to Each Other
When Re(u) > 1 and Re(v) > 1, recall that
F4(u, v; g) := ϕ(pβ
)
m<n
m≡n(mod pβ)
(m,pβ)=(n,pβ)=1
m−u
n−v
Tg
n
m
and
F5(u, v; g) := ϕ(pβ
)
m>n
m≡n(mod pβ)
(m,pβ)=(n,pβ)=1
m−u
n−v
Tg
n
m
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Reviewing F4 and F5 and Relating Them to Each Other
When Re(u) > 1 and Re(v) > 1, recall that
F4(u, v; g) := ϕ(pβ
)
m<n
m≡n(mod pβ)
(m,pβ)=(n,pβ)=1
m−u
n−v
Tg
n
m
and
F5(u, v; g) := ϕ(pβ
)
m>n
m≡n(mod pβ)
(m,pβ)=(n,pβ)=1
m−u
n−v
Tg
n
m
.
As from before, we have that F4(v, u; g) = F5(u, v; g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Taking F4 from R1 to R4 = Re(u + v) > 3
First, we will further simplify F4 to obtain
ϕ(pβ
)
m,n>0
m≡n+m(mod pβ)
(m,pβ)=(n+m,pβ)=1
m−u−v n
m
+ 1
−v
Tg
n
m
+ 1 .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Taking F4 from R1 to R4 = Re(u + v) > 3
First, we will further simplify F4 to obtain
ϕ(pβ
)
m,n>0
m≡n+m(mod pβ)
(m,pβ)=(n+m,pβ)=1
m−u−v n
m
+ 1
−v
Tg
n
m
+ 1 .
Let n = pνn1 where n1 ∈ N, (n1, p) = 1, and 0 ≤ ν < β.
Note that by the Inverse Mellin Transform,
n
m + 1
−v
= 1
2πi (σ) F(s, v) pν n1
m
−s
ds, where
F(s, v) = R+ (x + 1)−v xs−1 dx and Re(σ) > 0.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 2: Taking F4 from R1 to R4
Next, write Tg
n
m + 1 in terms of the Mellin Transform for
Finite Abelian Groups and the Mellin Inversion Formula for
Finite Abelian Group to get Tg
n
m + 1 = Tg 1 + pν n1
m =
1
ϕ(pβ−ν
) χ(mod pβ−ν ) ˜gβ−ν(χ) χ n1
m .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Next, write Tg
n
m + 1 in terms of the Mellin Transform for
Finite Abelian Groups and the Mellin Inversion Formula for
Finite Abelian Group to get Tg
n
m + 1 = Tg 1 + pν n1
m =
1
ϕ(pβ−ν
) χ(mod pβ−ν ) ˜gβ−ν(χ) χ n1
m .
Substitution and simpliﬁcation give
0≤ν<β
pν
2πi (σ)
F(s, v)p−νs
χ(mod pβ−ν )
˜gβ−ν(χ)
n1
(n1,pβ)=1
χ(n1)
ns
1
(m,pβ)=1
χ(m)
mu+v−s
ds.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Applying the deﬁnition of the beta function and L-functions,
and taking σ = 2 will yield
0≤ν<β
pν
2πi
1
Γ(v) (2)
Γ(s)Γ(v − s)p−νs
χ(mod pβ−ν )
˜gβ−ν(χ)L(s, χ)
L(u + v − s, χ) ds.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Moving the Integral and Providing a Continuation
from R4 to D = {u , v ∈ C | |u |, |v | < B}
Next, we will move the integral from Re(s) = 2 to
Re(s) = 3B where B is a suﬃciently large real number. This
will give a meromorphic continuation of the previous equation
for F4 to D.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
from R4 to D
By using the residue theorem (F4,1 comes from the residue at s = u + v − 1,
F4,2 comes from the residues when there are nonpositive arguments for the
Gamma function, and F4,3 comes from the vertical strip Re(s) = 3B) we get
that F4(u, v; g) = − 0≤ν<β
pν
Γ(v)
F4,1(u, v; g) + F4,2(u, v; g) − F4,3(u, v; g)
where

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
from R4 to D
that F4(u, v; g) = − 0≤ν<β
pν
Γ(v)
F4,1(u, v; g) + F4,2(u, v; g) − F4,3(u, v; g)
where
F4,1(u, v; g) :=
ϕ(pβ−ν
)
pβ−ν p−ν(u+v−1)Γ(u+v −1)Γ(1−u)L(u+v −1, χ0)˜gβ−ν (χ0),

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
from R4 to D
that F4(u, v; g) = − 0≤ν<β
pν
Γ(v)
F4,1(u, v; g) + F4,2(u, v; g) − F4,3(u, v; g)
where
F4,1(u, v; g) :=
ϕ(pβ−ν
)
F4,2(u, v; g) := j∈Z
Re(v−3B)≤j≤0
Γ(v − j)
(−1)−j
Γ(1−j)
p−ν(v−j)
χ(mod pβ−ν ) L(v −
j, χ)L(u + j, χ)˜gβ−ν (χ), and

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
from R4 to D
that F4(u, v; g) = − 0≤ν<β
pν
Γ(v)
F4,1(u, v; g) + F4,2(u, v; g) − F4,3(u, v; g)
where
F4,1(u, v; g) :=
ϕ(pβ−ν
)
F4,2(u, v; g) := j∈Z
Re(v−3B)≤j≤0
Γ(v − j)
(−1)−j
Γ(1−j)
p−ν(v−j)
χ(mod pβ−ν ) L(v −
j, χ)L(u + j, χ)˜gβ−ν (χ), and
F4,3(u, v; g) := 1
2πi (3B) Γ(s)Γ(v − s)p−νs
χ(mod pβ−ν ) L(s, χ)L(u + v − s, χ)
˜gβ−ν (χ) ds.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Break up F4,3 into Two Summands
Let u, v ∈ D. Then, F4(u, v; g) =
− 0≤ν<β
pν
Γ(v)
(F4,1(u, v; g)+F4,2(u, v; g)−F4,3,1(u, v; g)−F4,3,2(u, v; g)),
where

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
− 0≤ν<β
pν
Γ(v)
where
F4,3,1(u, v; g) := 1
2πi (3B)
Γ(s)Γ(v − s)p−νs ∗
χ(mod pβ−ν )
τ(χ)
iaχ π1/2
pβ−ν
π
s−u−v
˜gβ−ν (χ)Γ
s−u−v+aχ+1
2
Γ
u+v−s+aχ
2
−1
∞
b=1
χ(b)
bs σu+v−1(b) ds
and

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
− 0≤ν<β
pν
Γ(v)
where
F4,3,1(u, v; g) := 1
2πi (3B)
χ(mod pβ−ν )
τ(χ)
iaχ π1/2
pβ−ν
π
s−u−v
˜gβ−ν (χ)Γ
s−u−v+aχ+1
2
Γ
u+v−s+aχ
2
−1
∞
b=1
χ(b)
bs σu+v−1(b) ds
and
F4,3,2(u, v; g) := 1
2πi (3B)
Γ(s)Γ(v − s)p−νs β−ν
γ=1 χpγ ≡χ
τ(χpγ )
i
aχpγ
π1/2
pβ−ν
π
s−u−v
˜gβ−ν (χpγ )Γ
s−u−v+aχpγ +1
2
Γ
u+v−s+aχpγ
2
−1
∞
c=1
χpγ (c)
cs
σu+v−1(c) ds.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
− 0≤ν<β
pν
Γ(v)
where
F4,3,1(u, v; g) := 1
2πi (3B)
χ(mod pβ−ν )
τ(χ)
iaχ π1/2
pβ−ν
π
s−u−v
˜gβ−ν (χ)Γ
s−u−v+aχ+1
2
Γ
u+v−s+aχ
2
−1
∞
b=1
χ(b)
bs σu+v−1(b) ds
and
F4,3,2(u, v; g) := 1
2πi (3B)
Γ(s)Γ(v − s)p−νs β−ν
γ=1 χpγ ≡χ
τ(χpγ )
i
aχpγ
π1/2
pβ−ν
π
s−u−v
˜gβ−ν (χpγ )Γ
s−u−v+aχpγ +1
2
Γ
u+v−s+aχpγ
2
−1
∞
c=1
χpγ (c)
cs
σu+v−1(c) ds.
Note that F4,3,1 uses the case that χ is primitive and F4,3,2 uses the case
that χ is not primitive. Then, use the functional equation to get the
result.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Symmetry of D
Since D = {u , v ∈ C | |u |, |v | < B} is symmetric we can take u = v = α for
0 < α < 1 to obtain
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
F4,3,1(α, α; g) − F4,3,2(α, α; g)).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Take The Integrals from Re(s) = 3B to Re(s) = 2
Let 0 < α < 1. Then,
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
F4,3,1,1(α, α; g) − F4,3,1,2(α, α; g) − F4,3,2,1(α, α; g) − F4,3,2,2(α, α; g)),
where

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
where
F4,3,1,1(α, α; g) consists of an integral and a sum over primitive characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
where
F4,3,1,2(α, α; g) consists of the residues and a sum over primitive
characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
where
characters
F4,3,2,1(α, α; g) consists of an integral and a sum over non-primitive
characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2(α, α; g)−
where
characters
F4,3,2,1(α, α; g) consists of an integral and a sum over non-primitive
characters
F4,3,2,2(α, α; g) consists of the residues and a sum over non-primitive
characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
The Functional Equation and F2 Plus Simpliﬁcation to F4
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2,1,1(α, α; g)+
F4,2,2,1(α, α; g) − F4,3,1,1(α, α; g) − F4,3,2,1(α, α; g)),

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2,1,1(α, α; g)+
F4,2,2,1(α, α; g) − F4,3,1,1(α, α; g) − F4,3,2,1(α, α; g)),
where F4,2,1,1 = F4,2,1 − F4,3,1,2 and F4,2,1 contains 2 more residues than
F4,3,1,2 and a sum over primitive characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
F4(α, α; g) = −
0≤ν<β
pν
Γ(α)
(F4,1(α, α; g) + F4,2,1,1(α, α; g)+
F4,2,2,1(α, α; g) − F4,3,1,1(α, α; g) − F4,3,2,1(α, α; g)),
F4,3,1,2 and a sum over primitive characters
F4,3,2,2 and a sum over non-primitive characters

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Determining F4 + F5 as α → 1
2
Since F5(α, α; g) = F4(α, α; g), α ∈ R, and g is real-valued, it follows that
F4(α, α; g) + F5(α, α; g) = 2 Re F4(α, α; g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,1
Note that
lim
α→ 1
2
2 Re F4,1(α, α; g) = π1/2 1 − p
p
Re (˜gβ−ν (χ0)) log p.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,1
Note that
lim
α→ 1
2
2 Re F4,1(α, α; g) = π1/2 1 − p
p
Re (˜gβ−ν (χ0)) log p.
We used Γ(2α − 1) = 1
2α−1
+ γ + O(|2α − 1|), l’Hˆopital’s rule, and
L(2α − 1, χ0) = ζ(2α − 1)
p|pβ−ν
p prime
1 − p1−2α
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,2,1,1(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,2,1,1(α, α; g) = 2
∞
d0=1
d(d0)d
−1/2
0
j=−1,−2
dj
0Γ
1
2
− j
(−1)−j
Γ(1 − j)
p−ν(1
2
−j) ∗
χ(mod pβ−ν )
π−1/2
Re i−aχ
τ(χ)˜gβ−ν (χ)χ(d0)
Γ
1
2
− j + aχ
2
Γ
1
2
+ j + aχ
2
−1
.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,2,1,1(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,2,1,1(α, α; g) = 2
∞
d0=1
d(d0)d
−1/2
0
j=−1,−2
dj
0Γ
1
2
− j
(−1)−j
Γ(1 − j)
p−ν(1
2
−j) ∗
χ(mod pβ−ν )
π−1/2
Re i−aχ
τ(χ)˜gβ−ν (χ)χ(d0)
Γ
1
2
− j + aχ
2
Γ
1
2
+ j + aχ
2
−1
.
We checked for convergence and then we were able to take the limit
inside of the sum.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,2,2,1(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,2,2,1(α, α; g) =
∞
f =1
2d(f )f −1/2
j=−2,−1
f j
Γ
1
2
− j
(−1)−j
Γ(1 − j)
p−ν(1
2
−j) pβ−ν
π
− 1
2
−j β−ν
γ=1 χpγ ≡χ
π−1/2
Γ
aχpγ − j + 1
2
2
Γ
aχpγ + 1
2
+ j
2
−1
Re i
−aχpγ
τ(χpγ )˜gβ−ν (χpγ )χpγ (f ) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,2,2,1(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,2,2,1(α, α; g) =
∞
f =1
2d(f )f −1/2
j=−2,−1
f j
Γ
1
2
− j
(−1)−j
Γ(1 − j)
p−ν(1
2
−j) pβ−ν
π
− 1
2
−j β−ν
γ=1 χpγ ≡χ
π−1/2
Γ
aχpγ − j + 1
2
2
Γ
aχpγ + 1
2
+ j
2
−1
Re i
−aχpγ
τ(χpγ )˜gβ−ν (χpγ )χpγ (f ) .
We checked for convergence and then we were able to take the limit
inside of the sum.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Introduce F4,3,1,1,even and F4,3,1,1,odd
Let
F4,3,1,1,even(α, α; g) =
1
2πi
∞
b=1
σ2α−1(b)
(2)
b−s pβ−ν
π
s−2α
Γ(s)
Γ(α − s)p−νs
Γ
s − 2α + 1
2
Γ
2α − s
2
−1
ds
∗
χ(mod pβ−ν
)
χ even
τ(χ)
π1/2
˜gβ−ν (χ)χ(b)
and

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 2: Introduce F4,3,1,1,even(α, α; g) and
F4,3,1,1,odd(α, α; g)
F4,3,1,1,odd(α, α; g) =
1
2πi
∞
b=1
σ2α−1(b)
(2)
b−s pβ−ν
π
s−2α
Γ(s)
Γ(α − s)p−νs
Γ
s − 2α + 2
2
Γ
2α − s + 1
2
−1
ds
∗
χ(mod pβ−ν
)
χ odd
τ(χ)
iπ1/2
˜gβ−ν (χ)χ(b).
Then, we have that F4,3,1,1(α, α; g) = F4,3,1,1,even(α, α; g) + F4,3,1,1,odd(α, α; g).

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,3,1,1,even(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,3,1,1,even(α, α; g) =
1
π1/2pβ−ν
∞
b=1
d(b) Im W1(b)
∗
χ(mod pβ
)
χ even
τ(χ)˜gβ−ν (χ)χ(b)

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,3,1,1,even(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,3,1,1,even(α, α; g) =
1
π1/2pβ−ν
∞
b=1
d(b) Im W1(b)
∗
χ(mod pβ
)
χ even
τ(χ)˜gβ−ν (χ)χ(b)
where
W1(b) = π−1/2
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s e−isπ/2
ds +
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s eisπ/2
ds .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,3,1,1,odd(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,3,1,1,odd(α, α; g) =
1
π1/2pβ−ν
∞
b=1
d(b) Re W2(b)
∗
χ(mod pβ
)
χ odd
τ(χ)˜g(χ)χ(b) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,3,1,1,odd(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,3,1,1,odd(α, α; g) =
1
π1/2pβ−ν
∞
b=1
d(b) Re W2(b)
∗
χ(mod pβ
)
χ odd
τ(χ)˜g(χ)χ(b) .
where
W2(b) = π1/2
i
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s e−isπ/2
ds −
(2)
2bπ
pβ−ν
−s
Γ2
(s)Γ
1
2
− s eisπ/2
ds .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
limα→1
2
2 Re F4,3,2,1(α, α; g)
Note that
lim
α→ 1
2
2 Re F4,3,2,1(α, α; g) =
1
π1/2pβ−ν
∞
c=1
d(c) Im W1(c)
β−ν
γ=1
χpγ ≡χ
χpγ even
τ(χpγ )˜gβ−ν (χpγ )χpγ (c) + Re W2(c)
β−ν
γ=1 χpγ ≡χ
χpγ odd
τ(χpγ )
˜gβ−ν (χpγ )χpγ (c) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Part 1: Bring The Last Three Summands of the Main
Theorem Together
Note that
A2(g) = −
0≤ν<β
pν
π1/2
lim
α→ 1
2
2 Re F4,1(α, α; g),

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Theorem Together
Note that
A2(g) = −
0≤ν<β
pν
π1/2
lim
α→ 1
2
2 Re F4,1(α, α; g),
A3(g) = −
0≤ν<β
pν
π1/2
lim
α→ 1
2
2 Re F4,2,1,1(α, α; g)+ lim
α→ 1
2
2 Re F4,2,2,1(α, α; g) ,
and

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Theorem Together
Note that
A2(g) = −
0≤ν<β
pν
π1/2
lim
α→ 1
2
2 Re F4,1(α, α; g),
A3(g) = −
0≤ν<β
pν
π1/2
lim
α→ 1
2
2 Re F4,2,1,1(α, α; g)+ lim
α→ 1
2
2 Re F4,2,2,1(α, α; g) ,
and
A4(g) = −
0≤ν<β
pν
π1/2
− lim
α→ 1
2
2 Re F4,3,1,1(α, α; g)− lim
α→ 1
2
2 Re F4,3,2,1(α, α; g) .

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Closing Thoughts
We have a formula which consists of a main term from the diagonal and
near-diagonal terms as well as three lesser terms from the oﬀ-diagonal
terms which can be expressed in terms of residues and integrals.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Closing Thoughts
There is a strong correspondence with Motohashi’s formula in which the
divisor function and transforms appear in similar locations.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Closing Thoughts
Now, I am currently investigating the asymptotics for particular test
functions and seeing how these correspond with the asymptotics in
Motohashi’s formula.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Closing Thoughts
Now, I am currently investigating the asymptotics for particular test
functions and seeing how these correspond with the asymptotics in
Motohashi’s formula.
In the future, it might be of interest to determine what would happen for
other power moments.

Outline
Introduction
Main Theorem
Five Summands
Diagonal Sum
Near-Diagonal Sums
Oﬀ-Diagonal Sums
Thank you
Thank you very much for listening!

PresentationCopyPhiladelphiaAreaNumberTheorySeminarTalk12716

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