circle divisor problemsberndt/articles/bessel2... · 2014. 5. 23. · BRUCE C. BERNDT, SUN KIM, AND ALEXANDRU ZAHARESCU Abstract. On page 335 in his lost notebook, Ramanujan recorded

WEIGHTED DIVISOR SUMS AND BESSEL FUNCTION SERIES, III

BRUCE C. BERNDT, SUN KIM, AND ALEXANDRU ZAHARESCU

Abstract. On page 335 in his lost notebook, Ramanujan recorded without proofs two identities involving finitetrigonometric sums and doubly infinite series of Bessel functions. The two identities are intimately connected withthe classical circle and divisor problems, respectively. For each of Ramanujan’s identities, there are three possibleinterpretations for the double series. In two earlier papers, the authors proved the two identities under each of twopossible interpretations. Weighted (or twisted) divisor sums are central to the proofs. The ideas that the authors usedin the second paper are extended here to derive analogous Bessel series identities for finite sums of products of twotrigonometric (sine–sine; cosine–cosine; sine–cosine) functions.

1. INTRODUCTION

We begin by recalling the classical circle and divisor problems, which have remained unsolvedfor nearly two centuries.

Let r2(n) denote the number of representations of the positive integer n as a sum of two squares,where representations with different orders and different signs of the summands being squared areregarded as distinct. Write, for x > 0,∑

0≤n≤x

′r2(n) = πx+ P (x), (1.1)

where the prime ′ on the summation sign indicates that if x is an integer, then only 12r2(x) is

counted. The function P (x) is the “error term.” Gauss showed that P (x) = O(√x), as x → ∞;

finding the correct order of magnitude of the error term P (x) is the famous circle problem.Next, let d(n) denote the number of positive divisors of the positive integer n. Define the “error

term” ∆(x), for x > 0, by∑n≤x

′d(n) = x (log x+ (2γ − 1)) +

1

4+ ∆(x), (1.2)

where γ denotes Euler’s constant, and where the prime ′ on the summation sign on the left sideindicates that if x is an integer, then only 1

2d(x) is counted. Dirichlet proved that ∆(x) = O(

√x),

as x→∞. The famous Dirichlet divisor problem asks for the correct order of magnitude of ∆(x)

The first author’s research was partially supported by NSA grant MDA904-00-1-0015.The third author’s research was partially supported by NSF grant DMS-0901621.Key Words and Phrases: circle problem; divisor problem; Bessel function series; weighted divisor sums; trigono-

metric sums; character sums.2010 Mathematics Subject Classification: Primary, 11L03; Secondary, 11P21, 33C10.

1

2 BRUCE C. BERNDT, SUN KIM, AND ALEXANDRU ZAHARESCU

as x → ∞. As with the circle problem, the correct order of magnitude of the error term ∆(x) isunknown.

It is conjectured that P (x) = O(x1/4+ε) and that ∆(x) = O(x1/4+ε) as x → ∞, for eachε > 0. In 1915 and 1916, G.H. Hardy [7], [8], [9, pp. 243–263, 268–292], proved, respectively,that P (x) 6= O(x1/4) and ∆(x) 6= O(x1/4), as x tends to ∞. (Actually, Hardy proved slightlystronger results.)

Improvements on the elementary upper bounds established by Gauss and Dirichlet were notmade until early in the twentieth century. In 1904, G.F. Voronoı [15] established a representationfor ∆(x) in terms of Bessel functions with his now famous formula∑

n≤x

′d(n) = x (log x+ (2γ − 1)) +

1

4+∞∑n=1

d(n)(xn

)1/2I1(4π

√nx), (1.3)

where x > 0, the prime ′ on the summation sign has the same meaning as above, and I1(z) isdefined by

Iν(z) := −Yν(z)− 2

πKν(z). (1.4)

Here, Yν(z) is the Bessel function of the second kind of order ν [16, p. 64], and Kν(z) is themodified Bessel function of order ν [16, p. 78]. Voronoı employed (1.3) to prove that ∆(x) =O(x1/3+ε) as x → ∞, for each ε > 0, and since that time many improvements on the order ofmagnitude of ∆(x) use (1.3) as the starting point.

Two years later, W. Sierpinski [14] proved that P (x) = O(x1/3) as x → ∞. After the work ofSierpinski, improvements on the order of magnitude of P (x) have depended upon the identity∑

0≤n≤x

′r2(n) = πx+

∞∑n=1

r2(n)(xn

)1/2J1(2π

√nx), (1.5)

where the prime ′ on the summation sign on the left side has the same meaning as above, and Jν(x)is the ordinary Bessel function of order ν [16, Chap. II]. The identity (1.5) is generally called theHardy identity or the Hardy–Landau identity. To the best of our knowledge, (1.5) was first statedby Hardy [7], [9, p. 245] in 1915. However, in a footnote, he acknowledges that the identitywas suggested to him by Ramanujan. Also in 1915, E. Landau [10], [12, pp. 219–229] implicitlyderives (1.5), but he does not explicitly state or prove it. Landau himself [11, p. 189, Eq. (685)]refers to (1.5) as the “Hardyschen Identitat.” In conclusion, (1.5) might be more appropriatelynamed the “Ramanujan–Hardy identity.”

On page 335 in his lost notebook [13], Ramanujan offers two identities that are closely relatedto (1.5) and (1.3). For lengthy discussions on these relationships, please see the authors’ papers[4], [3], and [2]. To state Ramanujan’s claims, we need to first define

F (x) =

{[x], if x is not an integer,x− 1

2, if x is an integer,

(1.6)

where, as customary, [x] is the greatest integer less than or equal to x.

WEIGHTED DIVISOR SUMS AND BESSEL FUNCTION SERIES, III 3

Entry 1.1 (p. 335). If 0 < θ < 1 and x > 0, then∞∑n=1

F(xn

)sin(2πnθ) = πx

(1

2− θ)− 1

4cot(πθ)

+1

2

√x∞∑m=1

∞∑n=0

J1(

4π√m(n+ θ)x

)√m(n+ θ)

−J1

(4π√m(n+ 1− θ)x

)√m(n+ 1− θ)

. (1.7)

Entry 1.2. Let F (x) be defined by (1.6). Then, for x > 0 and 0 < θ < 1,∞∑n=1

F(xn

)cos(2πnθ) =

1

4− x log(2 sin(πθ))

+1

2

√x

∞∑m=1

∞∑n=0

I1(

4π√m(n+ θ)x

)√m(n+ θ)

+I1

(4π√m(n+ 1− θ)x

)√m(n+ 1− θ)

, (1.8)

where I1(z) is defined in (1.4).

Observe that the sums on the left-hand sides of (1.7) and (1.8) are finite. Entry 1.1 was provedin [4], but with the order of summation reversed. In [4], the authors employed Entry 1.1 to prove ageneral theorem about weighted divisor sums, and, as a corollary, formally established the follow-ing result.

Corollary 1.3. For any x > 0,

∑0≤n≤x

′r2(n) = πx+ 2

√x∞∑n=0

∞∑m=1

J1

(4π√m(n+ 1

4)x)

√m(n+ 1

4)

−J1

(4π√m(n+ 3

4)x)

√m(n+ 3

4)

. (1.9)

Note the similarity of the Bessel functions in formulas (1.5), (1.7), and (1.9).In [3], the present three authors reproved Entry 1.1 but now under the assumption that the double

sum (1.7) is not to be regarded as an iterated sum but as a double sum wherein the product mnof the indices tends to ∞. Thus, (1.7) has now been proved under two different interpretationsfor the sum on the right-hand side. Corollary 1.3 is a genuine corollary of Entry 1.1, under theinterpretation that the product of the indices m and n tends to infinity [3].

Entry 1.2 was established by us in [3] under two different interpretations, one with the orderof summation reversed, and the other with the product mn of the summation indices tending toinfinity. Thus, Entry 1.2 has been proved under two different interpretations, but with neitherinterpretation being that given by Ramanujan. The similar appearances of the Bessel functions in(1.3) and (1.8) are not accidental.

Readers will naturally ask how Ramanujan might have discovered Entries 1.1 and 1.2. We can-not definitively answer such a question, but the motivations of the circle and divisor problems andthe success of the methods employed by the authors in this paper and [3] suggest that Ramanujan’sarguments might have been developed under a similar umbrella.


There exist broad classes of arithmetical functions generated by Dirichlet series satisfying func-tional equations involving the gamma function. For example, see [5]. The arithmetic functionr2(n) is generated by a Dirichlet series satisfying a functional equation with a simple gamma fac-tor Γ(s), while d(n) is generated by the square of the Riemann zeta function ζ(s), which satisfiesa functional equation involving Γ2(1

2s). It is therefore natural to ask if Entries 1.1 and 1.2 are

isolated examples or if they point to a path of further identities of this sort. It is the purpose of thispaper to provide three additional theorems in which double series of Bessel functions appear. Onthe “left sides” are sums of products of trigonometric functions. We are unaware of any theoremsin the literature of this sort. However, we can derive them only under the assumption that theproduct mn of the summation indices m and n tends to infinity. In the sequel, and, in particular,in the statements of all of our theorems below, we always make the assumption that, in thedouble series of Bessel functions, the product mn of the series indices tends to infinity. Itseems very difficult to prove corresponding theorems wherein the double series are iterated. It ishoped that these three theorems (Theorems 2.1–2.3), which we offer in the next section, will havebroader connections, just as Ramanujan’s two theorems have intimate relationships with the circleand divisor problems.

2. NOTATION AND STATEMENTS OF THEOREMS

For arithmetic functions f and g, we define∑n≤x

′f(n) =

{∑n≤x f(n), if x is not an integer,∑n≤x f(n)− 1

2f(x), if x is an integer,

and ∑nm≤x

′f(n)g(m) =

{∑nm≤x f(n)g(m), if x is not an integer,∑nm≤x f(n)g(m)− 1

2

∑nm=x f(n)g(m), if x is an integer.

Theorem 2.1. Let I1(x) be defined by (1.4). If 0 < θ, σ < 1 and x > 0, then∑nm≤x

′cos(2πnθ) cos(2πmσ) (2.1)

=1

4+

√x

4

∑n,m≥0

{I1(4π

√(n+ θ)(m+ σ)x)√

(n+ θ)(m+ σ)+I1(4π

√(n+ 1− θ)(m+ σ)x)√

(n+ 1− θ)(m+ σ)

+I1(4π

√(n+ θ)(m+ 1− σ)x)√

(n+ θ)(m+ 1− σ)+I1(4π

√(n+ 1− θ)(m+ 1− σ)x)√

(n+ 1− θ)(m+ 1− σ)

}.

Theorem 2.2. Let Jν(x) denote the ordinary Bessel function of order ν. If 0 < θ, σ < 1 andx > 0, then∑

nm≤x

′cos(2πnθ) sin(2πmσ) (2.2)


= −cot(πσ)

4+

√x

4

∑n,m≥0

{J1(4π

√(n+ θ)(m+ σ)x)√

(n+ θ)(m+ σ)+J1(4π

√(n+ 1− θ)(m+ σ)x)√

(n+ 1− θ)(m+ σ)

−J1(4π

√(n+ θ)(m+ 1− σ)x)√

(n+ θ)(m+ 1− σ)−J1(4π

√(n+ 1− θ)(m+ 1− σ)x)√

(n+ 1− θ)(m+ 1− σ)

}.

Theorem 2.3. If 0 < θ, σ < 1 and x > 0, then∑nm≤x

′nm sin(2πnθ) sin(2πmσ) (2.3)

=x√x

4

∑n,m≥0

{T 3

2

(4π2(n+ θ)(m+ σ)x

)√(n+ θ)(m+ σ)

−T 3

2

(4π2(n+ 1− θ)(m+ σ)x

)√(n+ 1− θ)(m+ σ)

−T 3

2

(4π2(n+ θ)(m+ 1− σ)x

)√(n+ θ)(m+ 1− σ)

+T 3

2

(4π2(n+ 1− θ)(m+ 1− σ)x

)√(n+ 1− θ)(m+ 1− σ)

}.

Here

T32(x) =

∫ ∞0

J 12(u)J 3

2(x)du,

which can be evaluated in terms of Bessel functions by (5.12), (5.10), and (1.4).

We now give a definition and elementary lemmas. Define, for Dirichlet characters χ1 modulo pand χ2 modulo q,

dχ1,χ2(n) =∑d|n

χ1(d)χ2(n/d). (2.4)

Lemma 2.4. If χ is a non-principal even primitive character modulo q, then

χ(n) =1

τ(χ)

q−1∑h=1

χ(h) cos(2πnh/q),

where τ(χ) denotes the Gauss sum

τ(χ) :=

q−1∑h=1

χ(h)e2πih/q.

If χ is an odd primitive character modulo q, then

χ(n) =i

τ(χ)

q−1∑h=1

χ(h) sin(2πnh/q).

Proof. For a non-principal even primitive character χ modulo q,

1

τ(χ)

q−1∑h=1

χ(h) cos(2πnh/q) =1

2τ(χ)

q−1∑h=1

χ(h)(e2πinh/q + e−2πinh/q

)


=1

2

(χ(n) + χ(−n)

)= χ(n),

where we used the formula [6, p. 65]

χ(n)τ(χ) =

q∑h=1

χ(h)e2πinh/q, (2.5)

for any primitive character χ modulo q.Similarly, for an odd primitive character χ modulo q, we have

i

τ(χ)

q−1∑h=1

χ(h) sin(2πnh/q) =1

2τ(χ)

q−1∑h=1

χ(h)(e2πinh/q − e−2πinh/q

)=

1

2

(χ(n)− χ(−n)

)= χ(n).

�

Lemma 2.5. If (n, q) = (a, q) = 1, then

cos(2πna/q

)=

1

φ(q)

∑χ mod qχ even

χ(a)τ(χ)χ(n),

and

sin(2πna/q

)=

1

iφ(q)

∑χ mod qχ odd

χ(a)τ(χ)χ(n).

Proof. Noting that (2.5) holds for any character if (n, q) = 1, and using∑χ mod q

χ(a1)χ(a2) =

{φ(q), if a1 ≡ a2 (mod q) and (a1, q) = 1,

0, otherwise,(2.6)

we find that

e2πina/q =1

φ(q)

q∑h=1

e2πinh/q∑

χ mod q

χ(a)χ(h)

=1

φ(q)

∑χ mod q

χ(a)

q∑h=1

χ(h)e2πinh/q

=1

φ(q)

∑χ mod q

χ(a)τ(χ)χ(n). (2.7)

So, we have

cos(2πna/q

)=

1

2

(e2πina/q + e−2πina/q

)


=1

2φ(q)

∑χ mod q

χ(a)τ(χ)(χ(n) + χ(−n)

)=

1

φ(q)

∑χ mod qχ even

χ(a)τ(χ)χ(n).

Similarly, we obtain

sin(2πna/q

)=

1

2i

(e2πina/q − e−2πina/q

)=

1

2iφ(q)

∑χ mod q

χ(a)τ(χ)(χ(n)− χ(−n)

)=

1

iφ(q)

∑χ mod qχodd

χ(a)τ(χ)χ(n).

�

In Sections 3–5, we give proofs of Theorems 2.1–2.3 by proving three equivalent formulationsinvolving (2.4). In Section 6, we note some curious differential equations satisfied by either theleft- or right-hand sides of the identities of our three theorems.

3. PROOF OF THEOREM 2.1

We begin with an outline of our argument. First, an identity for sums of (2.4) is established inTheorem 3.1 below. Second, we indicate that it suffices to prove Theorem 2.1 for rational valuesθ = a/p and σ = b/q, where p and q are primes and 0 < a < p and 0 < b < q.We then reformulateTheorem 2.1 for these rational numbers in Theorem 3.2 below. Lastly, we employ Theorem 3.1 toprove Theorem 3.2. At the end of Section 3, we establish that Theorems 2.1 and 3.1 are equivalentby showing that the former theorem implies the latter theorem.

Theorem 3.1. If χ1 and χ2 are non-principal even primitive characters modulo p and q, respec-tively, then ∑

n≤x

′dχ1,χ2(n) =

τ(χ1)τ(χ2)√pq

∞∑n=1

dχ1,χ2(n)(xn

) 12I1

(4π

√nx

pq

).

Proof. We show that Theorem 3.1 is a special case of [1, p. 351, Theorem 2; p. 356, Theorem 4].We first recall that if χ is a non-principal even primitive character of modulus q, then the DirichletL-function L(x, χ) satisfies the functional equation [6, p. 69]

(π/q)−sΓ(s)L(2s, χ) =τ(χ)√q

(π/q)−(12−s)Γ(1

2− s)L(1− 2s, χ). (3.1)

So, we have

(π2/(pq))−sΓ2(s)L(2s, χ1)L(2s, χ2)


=τ(χ1)τ(χ2)√

pq(π2/(pq))−(

12−s)Γ2(1

2− s)L(1− 2s, χ1)L(1− 2s, χ2).

Note that

L(2s, χ1)L(2s, χ2) =∞∑n=1

χ1(n)

n2s

∞∑m=1

χ2(m)

m2s=∞∑n=1

dχ1,χ2(n)

n2s.

In the notation of the aforementioned theorems from [1], let q = 0, r = 12, m = 2, λn = µn =

π2n2/(pq), a(n) = dχ1,χ2(n), and b(n) = τ(χ1)τ(χ2)dχ1,χ1(n)/√pq. Also, Jα(x) denotes the

ordinary Bessel function of order α. Then, we have∑λn≤x

′dχ1,χ2(n) =

τ(χ1)τ(χ2)√pq

∞∑n=1

dχ1,χ2(n)( xµn

) 14K1/2(4

√µnx;−1

2; 2) +Q0(x), (3.2)

where

Kν(x;µ; 2) =

∫ ∞0

uν−µ−1Jµ(u)Jν(x/u)du (3.3)

and

Q0(x) =1

2πi

∫C

(π2/pq)−sL(2s, χ1)L(2s, χ2)xs

sds, (3.4)

where C is a positively oriented, closed curve encircling the poles of the integrand.Next, recall that [16, p. 54]

J−1/2(z) =

√2

πzcos z and J1/2(z) =

√2

πzsin z.

Using a formula obtained from [16, p. 184, formula (3)] by differentiating with respect to x andmaking a change of variable, we can easily derive that∫ ∞

0

cosu sin(y2u

)du = −y

(π2Y1(2y) +K1(2y)

). (3.5)

We now replace x by π2x2/(pq) in (3.2). Thus, from (3.3), (3.5), and (1.4), we find that

K1/2(4π2nx/pq;−1

2; 2) =

1

π2

√pq

nx

∫ ∞0

cosu sin

(4π2nx

pqu

)du

= − 1

π2

√pq

nx2π

√nx

pq

(π2Y1(4π

√nx/(pq)) +K1(4π

√nx/(pq))

)= I1(4π

√nx/(pq)).

Also, from (3.4),

Q0

(π2x2

pq

)=

1

2πi

∫C

L(2s, χ1)L(2s, χ2)x2s

sds = 0,

since L(s, χ1) and L(s, χ2) are entire functions and L(0, χ1) = L(0, χ2) = 0.


Thus, from (3.2), we deduce that∑n≤x

′dχ1,χ2(n) =

τ(χ1)τ(χ2)√pq

∞∑n=1

dχ1,χ2(n)(xn

) 12I1

(4π

√nx

pq

),

which completes the proof. �

A slight modification of the analysis from [1, 354–356], in particular, Lemma 14, shows thatthe series on the right-hand side of (2.1) converges uniformly with respect to θ on any interval 0 <θ1 ≤ θ ≤ θ2 < 1 and uniformly with respect to σ, for any interval 0 < σ1 ≤ σ ≤ σ2 < 1. (Thereis a misprint in (3.5) of Theorem 4 in [1]; read b(n)/µ

σ−1/2mn for b(n)µ

σ−1/2mn .) The sums are

therefore continuous functions of θ and σ on their respective intervals given above. By continuity,it therefore suffices to prove Theorem 2.1 for rational numbers θ = a/p and σ = b/q, where p andq are primes, and 0 < a < p and 0 < b < q. Thus, Theorem 2.1 is equivalent to the followingresult.

Theorem 3.2. If p and q are primes, and 0 < a < p and 0 < b < q, then∑nm≤x

′cos(2πna/p) cos(2πmb/q)

=1

4+

√x

4

∑n,m≥0

{I1(4π

√(n+ a/p)(m+ b/q)x)√

(n+ a/p)(m+ b/q)+I1(4π

√(n+ 1− a/p)(m+ b/q)x)√

(n+ 1− a/p)(m+ b/q)

+I1(4π

√(n+ a/p)(m+ 1− b/q)x)√

(n+ a/p)(m+ 1− b/q)+I1(4π

√(n+ 1− a/p)(m+ 1− b/q)x)√

(n+ 1− a/p)(m+ 1− b/q)

}

=1

4+

√pqx

4

∞∑n,m=0

n≡±a mod pm≡±b mod q

I1(4π√nmx/pq)√nm

. (3.6)

Our next goal is to show that Theorem 3.1 implies Theorem 3.2. For any Dirichlet character χ,set

dχ(n) =∑d|n

χ(d).

We need a special case of Lemma 6 from [4].

Lemma 3.3. If q is prime and 0 < a < q, then∞∑n=1

F(xn

)cos(2πna

q

)=

∑1≤n≤x/q

′d(n) +

1

φ(q)

∑χ mod qχ even

χ(a)τ(χ)∑

1≤n≤x

′dχ(n),

where, as before, d(n) is the divisor function.

We also need the following lemma.


Lemma 3.4. [3, Theorem 4] If q is prime and 0 < a < q, then

∞∑n=1

F(xn

)cos(2πna

q

)− 1

4+ x log(2 sin (πa/q)) =

√qx

2

∞∑m=1

∞∑r=0

r≡±a mod q

I1(4π√mrx/q

)√mr

.

Proof : Theorem 3.1⇒ Theorem 3.2. Let p and q be primes, and 0 < a < p and 0 < b < q. Then∑nm≤x

′cos(2πna

p

)cos(2πmb

q

)=∑nm≤xp-n,q-m

′cos(2πna

p

)cos(2πmb

q

)+

∑nm≤x/p

′cos(2πmb

q

)+

∑nm≤x/q

′cos(2πna

p

)−

∑nm≤x/pq

′1

=∑nm≤xp-n,q-m

′cos(2πna

p

)cos(2πmb

q

)+∑m≤x/p

F( x

pm

)cos(2πmb

q

)

+∑n≤x/q

F( xqn

)cos(2πna

p

)−∑

n≤x/pq

′d(n)

=∑nm≤xp-n,q-m

′cos(2πna

p

)cos(2πmb

q

)+∞∑m=1

F( x

pm

)cos(2πmb

q

)

+∞∑n=1

F( xqn

)cos(2πna

p

)−∑

n≤x/pq

′d(n). (3.7)

Using Lemma 2.5 and the fact that τ(χ0) = −1, where χ0 is a principal character modulo anyprime, we have∑

nm≤xp-n,q-m

′cos(2πna

p

)cos(2πmb

q

)

=1

φ(p)φ(q)

∑nm≤x

′ ∑χ1 mod pχ1 even

χ1(a)τ(χ1)χ1(n)∑

χ2 mod qχ2 even

χ2(b)τ(χ2)χ2(m)

=1

φ(p)φ(q)

∑χ1 mod pχ1 even

∑χ2 mod qχ2 even

χ1(a)χ2(b)τ(χ1)τ(χ2)∑n≤x

′dχ1,χ2(n)

=1

φ(p)φ(q)

{ ∑χ1 mod pχ1 6=χ0, even

∑χ2 mod qχ2 6=χ0, even


′dχ1,χ2(n)


−∑

χ2 mod qχ2 even

χ2(b)τ(χ2)∑nm≤xp-n

′χ2(m)−


χ1(a)τ(χ1)∑nm≤xq-m

′χ1(n)−

∑nm≤xp-n,q-m

′1

}

=1

φ(p)φ(q)




′dχ1,χ2(n)

−∑

χ2 mod qχ2 even

χ2(b)τ(χ2)(∑nm≤x

′χ2(m)−

∑nm≤x/p

′χ2(m)

)−


χ1(a)τ(χ1)(∑nm≤x

′χ1(n)−

∑nm≤x/q

′χ1(n)

)

−∑nm≤x

′1 +

∑nm≤x/p

′1 +

∑nm≤x/q

′1−

∑nm≤x/pq

′1

}

=1

φ(p)φ(q)




′dχ1,χ2(n)

−∑

χ2 mod qχ2 even

χ2(b)τ(χ2)(∑n≤x

′dχ2(n)−

∑n≤x/p

′dχ2(n)

)−


χ1(a)τ(χ1)(∑n≤x

′dχ1(n)−

∑n≤x/q

′dχ1(n)

)

−∑n≤x

′d(n) +

∑n≤x/p

′d(n) +

∑n≤x/q

′d(n)−

∑n≤x/pq

′d(n)

}. (3.8)

Using Lemma 3.3, we see that∑χ2 mod qχ2 even

χ2(b)τ(χ2)(∑n≤x

′dχ2(n)−

∑n≤x/p

′dχ2(n)

)(3.9)

= φ(q)( ∞∑n=1

F(xn

)cos(2πnb

q

)−∞∑n=1

F( xpn

)cos(2πnb

q

)−∑n≤x/q

′d(n) +

∑n≤x/pq

′d(n)

),

and ∑χ1 mod pχ1 even

χ1(a)τ(χ1)(∑n≤x

′dχ1(n)−

∑n≤x/q

′dχ1(n)

)(3.10)


= φ(p)( ∞∑n=1

F(xn

)cos(2πna

p

)−∞∑n=1

F( xqn

)cos(2πna

p

)−∑n≤x/p

′d(n) +

∑n≤x/pq

′d(n)

).

Thus, putting (3.9) and (3.10) in (3.8), and then putting (3.8) in (3.7), after simplification, wefind that∑

nm≤x

′cos(2πna

p

)cos(2πmb

q

)(3.11)

=1

φ(p)φ(q)

( ∑χ1 mod pχ1 6=χ0, even



′dχ1,χ2(n)

)

− 1

φ(p)

∞∑n=1

F(xn

)cos(2πnb

q

)− 1

φ(q)

∞∑n=1

F(xn

)cos(2πna

p

)+

p

φ(p)

∞∑n=1

F( xpn

)cos(2πnb

q

)+

q

φ(q)F( xqn

)cos(2πna

p

)− 1

φ(p)φ(q)

( ∑1≤n≤x

′d(n)− q

∑1≤n≤x/q

′d(n)− p

∑1≤n≤x/p

′d(n) + pq

∑1≤n≤x/pq

′d(n)

).

Next, we examine the right-hand side of the equation in Theorem 3.2. Using (2.6), we have√pqx

4

∞∑n,m=0


{I1(4π

√nmx/pq)√nm

}

=

√pq

φ(p)φ(q)

∞∑n=1

∞∑m=1

√x

nmI1(4π√nmx/(pq)

) ∑χ1 mod pχ1 even

χ1(a)χ1(n)∑

χ2 mod qχ2 even

χ2(b)χ2(m)

=

√pq

φ(p)φ(q)



χ1(a)χ2(b)∞∑n=1

dχ1,χ2(n)

√x

nI1(4π√nx/(pq)

)

=

√pq

φ(p)φ(q)



χ1(a)χ2(b)∞∑n=1

dχ1,χ2(n)

√x

nI1(4π√nx/(pq)

)

+∑

χ2 mod qχ2 even

χ2(b)∞∑n=1p-n

∞∑m=1

χ2(m)

√x

nmI1(4π√nmx/(pq)

)

+∑

χ1 mod pχ1 even

χ1(a)∞∑n=1

∞∑m=1q-m

χ1(n)

√x

nmI1(4π√nmx/(pq)

)


−∞∑n=1p-n

∞∑m=1q-m

√x

nmI1(4π√nmx/(pq)

)}.

Using Lemma 3.4, we find that√pq

φ(p)φ(q)


χ2(b)∞∑n=1p-n

∞∑m=1

χ2(m)

√x

nmI1(4π√nmx/(pq)

)

=

√pq

φ(p)φ(q)


χ2(b)( ∞∑n=1

∞∑m=1

χ2(m)

√x

nmI1(4π√nmx/(pq)

)

−∞∑n=1

∞∑m=1

χ2(m)

√x

pnmI1(4π√nmx/q

))=

√pq

2φ(p)

∞∑n=1

∞∑m=1

m≡±b mod q

(√ x

nmI1(4π√nmx/pq

)−√

x

pnmI1(4π√nmx/q

))

=p

φ(p)

( ∞∑n=1

F( xpn

)cos(2πnb

q

)− 1

4− x

plog(2 sin(πb/q))

)− 1

φ(p)

( ∞∑n=1

F(xn

)cos(2πnb

q

)− 1

4− x log(2 sin(πb/q))

)=

p

φ(p)

∞∑n=1

F( xpn

)cos(2πnb

q

)− 1

φ(p)

∞∑n=1

F(xn

)cos(2πnb

q

)− 1

4.

Similarly, we can see that√pq

φ(p)φ(q)


χ1(a)∞∑n=1

∞∑m=1q-m

χ1(n)

√x

nmI1(4π√nmx/(pq)

)

=q

φ(q)

∞∑n=1

F( xqn

)cos(2πna

p

)− 1

φ(q)

∞∑n=1

F(xn

)cos(2πna

p

)− 1

4.

Also, using (1.3), we find that√pq

φ(p)φ(q)

∞∑n=1p-n

∞∑m=1q-m

√x

nmI1(4π√nmx/(pq)

)

=

√pq

φ(p)φ(q)

{∞∑n=1

d(n)

√x

nI1(4π√nx/(pq)

)−∞∑n=1

d(n)

√x

pnI1(4π√nx/q

)


−∞∑n=1

d(n)

√x

qnI1(4π√nx/p

)+∞∑n=1

d(n)

√x

pqnI1(4π√nx)}

=1

φ(p)φ(q)

{pq

∑1≤n≤x/pq

′d(n)− q

∑1≤n≤x/q

′d(n)− p

∑1≤n≤x/p

′d(n) +

∑1≤n≤x

′d(n)

}− 1

4.

Thus, we obtain

1

4+

√pqx

4

∞∑n,m=0


{I1(4π

√nmx/pq)√nm

}(3.12)

=

√pq

φ(p)φ(q)

∑χ1 mod pχ1 6=χ0, even


χ1(a)χ2(b)∞∑n=1

dχ1,χ2(n)

√x

nI1(4π√nx/(pq)

)

+p

φ(p)

∞∑n=1

F( xpn

)cos(2πnb

q

)− 1

φ(p)

∞∑n=1

F(xn

)cos(2πnb

q

)+

q

φ(q)

∞∑n=1

F( xqn

)cos(2πna

p

)− 1

φ(q)

∞∑n=1

F(xn

)cos(2πna

p

)− 1

φ(p)φ(q)

{pq

∑1≤n≤x/pq

′d(n)− q

∑1≤n≤x/q

′d(n)− p

∑1≤n≤x/p

′d(n) +

∑1≤n≤x

′d(n)

}.

Hence, by Theorem 3.1, (3.11) and (3.12), we complete the proof of Theorem 3.2. �

As promised at the beginning of this section, we now prove a converse theorem.

Proof : Theorem 2.1⇒ Theorem 3.1. Let θ = h/p and σ = k/q in (2.1), and let χ1 and χ2 benon-principal even primitive characters modulo p and q, respectively.

We multiply both sides of (2.1) by χ1(h)χ2(k)/τ(χ1)τ(χ2), and sum on h and k with 1 ≤ h < pand 1 ≤ k < q.

First, by Lemma 2.4, we observe that

1

τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k)∑nm≤x

′cos(2πnh/p) cos(2πmk/q)

=∑nm≤x

′χ1(n)χ2(m) =

∑n≤x

′∑d|n

χ1(d)χ2(n/d) =∑n≤x

′dχ1,χ2(n), (3.13)


Now, we examine the right-hand side of (2.1). Since χ1 and χ2 are non-principal, we can seethat the contribution of 1/4 is

1

4τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k) =1

4τ(χ1)τ(χ2)

p−1∑h=1

χ1(h)

q−1∑k=1

χ2(k) = 0. (3.14)

Also, using (3.6) with a replaced by h and b replaced by k, we find that the contribution of thesecond expression on the right-hand side of (2.1) is

√xpq

4τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k)∞∑

n,m=0n≡±h mod pm≡±k mod q

{I1(4π

√nmx/(pq))√nm

}

=

√xpq

4τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

{I1(4π

√nmx/(pq))√nm

}p−1∑h=1

h≡±n mod q

χ1(h)

q−1∑k=1

k≡±m mod q

χ2(k)

=

√xpq

τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

χ1(n)χ2(m)

{I1(4π

√nmx/(pq))√nm

}

=τ(χ1)τ(χ2)√

pq

∞∑n=1

dχ1,χ2(n)(xn

) 12I1

(4π√nx/(pq)

), (3.15)

where we used τ(χ1)τ(χ1) = p and τ(χ2)τ(χ2) = q. Putting together (3.13), (3.14), and (3.15),we complete the proof. �


We first establish the following result.

Theorem 4.1. If χ1 is a non-principal even primitive character modulo p and χ2 is an odd primitivecharacter modulo q, then∑

n≤x

′dχ1,χ2(n) = −iτ(χ1)τ(χ2)√

pq

∞∑n=1

dχ1,χ2(n)(xn

) 12J1

(4π

√nx

pq

).

Proof. We show that Theorem 4.1 is a special case of [1, p. 351, Theorem 2; p. 356, Theorem 4].Recall that if χ is an odd primitive character of modulus q, then the Dirichlet L-function L(x, χ)satisfies the functional equation [6, p. 71](π

q

)−(2s+1)/2

Γ(s+

1

2

)L(2s, χ) = −iτ(χ)

√q

(πq

)−(1−s)Γ(1− s)L(1− 2s, χ). (4.1)

Using (3.1) and (4.1), we deduce that

π−2s−1/2

p−sq−s−1/2Γ(s)Γ

(s+

1

2

)L(2s, χ1)L(2s, χ2)


= −iτ(χ1)τ(χ2)√pq

π2s−3/2

ps−1/2qs−1Γ(1

2− s)

Γ(1− s)L(1− 2s, χ1)L(1− 2s, χ2). (4.2)

If we apply the duplication formula for the gamma function,

Γ(2s)√π = 22s−1Γ(s)Γ

(s+

1

2

),

then we can rewrite (4.2) asπ−2s

p−sq−sΓ(2s)

22s−1 L(2s, χ1)L(2s, χ2)

= −iτ(χ1)τ(χ2)π2s−1

psqsΓ(1− 2s)

2−2sL(1− 2s, χ1)L(1− 2s, χ2).

Replacing s by s/2, we have( 2π√pq

)−sΓ(s)L(s, χ1)L(s, χ2) = −iτ(χ1)τ(χ2)√

pq

( 2π√pq

)s−1Γ(1− s)L(1− s, χ1)L(1− s, χ2).

In the notation of Theorem 2 of [1], let q = 0, r = m = 1, λn = µn = 2πn/√pq, a(n) = dχ1,χ2(n),

b(n) = −iτ(χ1)τ(χ2)dχ1,χ2(n)/√pq, and K1(2

√µnx; 0; 1) = J1(2

√µnx). Then,∑

λn≤x

′dχ1,χ2(n) = −iτ(χ1)τ(χ2)√

pq

∞∑n=1

dχ1,χ2(n)

(x

µn

)1/2

J1(2√µnx) +Q0(x),

where

Q0(x) =1

2πi

∫C

(2π/√pq)−sL(s, χ1)L(s, χ2)x

s

sds,

whereC is a positively oriented, closed curve encircling the poles of the integrand. We now replacex by 2πx/

√pq to find that∑

n≤x

′dχ1,χ2(n) = −iτ(χ1)τ(χ2)√

pq

∞∑n=1

dχ1,χ2(n)(xn

) 12J1

(4π

√nx

pq

)+Q0

( 2πx√pq

).

Since L(0, χ1) = 0, and L(s, χ1) and L(s, χ2) are entire functions,

Q0

( 2πx√pq

)= 0.

Using this in the line above, we complete the proof of Theorem 4.1. �

As in the proof of Theorem 2.1, it suffices to prove Theorem 2.2 for rational numbers θ = a/pand σ = b/q, where p and q are primes, and 0 < a < p and 0 < b < q. Thus, Theorem 2.2 isequivalent to the following theorem.

Theorem 4.2. If p, q are primes, and 0 < a < p and 0 < b < q, then∑nm≤x

′cos(2πna/p) sin(2πmb/q) = −1

4cot

(bπ

q

)


+

√x

4

∑n,m≥0

{J1(4π

√(n+ a/p)(m+ b/q)x)√

(n+ a/p)(m+ b/q)+J1(4π

√(n+ 1− a/p)(m+ b/q)x)√

(n+ 1− a/p)(m+ b/q)

−J1(4π

√(n+ a/p)(m+ 1− b/q)x)√

(n+ a/p)(m+ 1− b/q)−J1(4π

√(n+ 1− a/p)(m+ 1− b/q)x)√

(n+ 1− a/p)(m+ 1− b/q)

}

= −1

4cot

(bπ

q

)+

√pqx

4

{∞∑

n,m=0n≡±a mod pm≡b mod q

J1(4π√nmx/pq)√nm

−∞∑

n,m=0n≡±a mod pm≡−b mod q


}.

Next, we show that Theorem 4.2 follows from Theorem 4.1. We first state two lemmas.

Lemma 4.3. [3, Lemma 11] If q is prime and 0 < a < q, then

∞∑n=1

F(xn

)sin(2πna

q

)=−iφ(q)

∑χ mod qχ odd

χ(a)τ(χ)∑

1≤n≤x

′dχ(n).

Lemma 4.4. [3, Theorem 8] If q is prime and 0 < a < q, then

∞∑n=1

F(xn

)sin(2πna

q

)− πx

(1

2− a

q

)+

1

4cot(aπq

)

=

√qx

2

∞∑m=1

∞∑r=1

r≡a mod q

J1(4π√mrx/q

)√mr

−∞∑m=1

∞∑r=1

r≡−a mod q

J1(4π√mrx/q

)√mr

.

Proof : Theorem 4.1⇒ Theorem 4.2. Let p and q be primes, and 0 < a < p and 0 < b < q. First,observe that ∑

nm≤x

′cos(2πna

p

)sin(2πmb

q

)(4.3)

=∑nm≤xp-n

′cos(2πna

p

)sin(2πmb

q

)+

∑nm≤x/p

′sin(2πmb

q

)

=∑nm≤xp-n

′cos(2πna

p

)sin(2πmb

q

)+∑m≤x/p

F( x

pm

)sin(2πmb

q

)

=∑nm≤xp-n

′cos(2πna

p

)sin(2πmb

q

)+∞∑m=1

F( x

pm

)sin(2πmb

q

).


Using Lemma 2.5, we have

∑nm≤xp-n

′cos(2πna

p

)sin(2πmb

q

)

=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

−∑

χ2 mod qχ2 odd

χ2(b)τ(χ2)∑nm≤xp-n

′χ2(m)

}

=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

−∑

χ2 mod qχ2 odd

χ2(b)τ(χ2)∑nm≤x

′χ2(m) +

∑χ2 mod qχ2 odd

χ2(b)τ(χ2)∑

nm≤x/p

′χ2(m)

}

=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

−∑

χ2 mod qχ2 odd

χ2(b)τ(χ2)∑n≤x

′dχ2(n) +

∑χ2 mod qχ2 odd

χ2(b)τ(χ2)∑n≤x/p

′dχ2(n)

}

=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

− 1

φ(p)

∞∑m=1

F( xm

)sin(2πmb

q

)+

1

φ(p)

∞∑m=1

F( x

pm

)sin(2πmb

q

), (4.4)

where we used Lemma 4.3. Thus, using (4.4) in (4.3), we see that

∑nm≤x

′cos(2πna

p

)sin(2πmb

q

)(4.5)


=1

iφ(p)φ(q)


∑χ2 mod qχ2 odd


′dχ1,χ2(n)

− 1

φ(p)

∞∑m=1

F( xm

)sin(2πmb

q

)+

p

φ(p)

∞∑m=1

F( x

pm

)sin(2πmb

q

).

Next, we examine the right-hand side of the equation in Theorem 4.2. By (2.6), we obtain√pqx

4

{∞∑



−∞∑



}

=

√pqx

2φ(p)φ(q)

{∞∑

n,m=0



χ1(a)χ1(n)∑

χ2 mod q

χ2(b)χ2(m)

−∞∑

n,m=0



χ1(a)χ1(n)∑

χ2 mod q

χ2(−b)χ2(m)

}

=

√pqx

φ(p)φ(q)

∞∑n,m=0



∑χ2 mod qχ2 odd

χ1(a)χ2(b)χ1(n)χ2(m)

=

√pqx

φ(p)φ(q)


∑χ2 mod qχ2 odd

χ1(a)χ2(b)∞∑n=0

dχ1,χ2(n)J1(4π

√nx/pq)√n

=

√pqx

φ(p)φ(q)


∑χ2 mod qχ2 odd

χ1(a)χ2(b)∞∑n=0

dχ1,χ2(n)J1(4π

√nx/pq)√n

+

√pqx

φ(p)φ(q)

∑χ2 mod qχ2 odd

χ2(b)∞∑

n,m=0p-n

χ2(m)J1(4π

√nmx/pq)√nm

. (4.6)

Using Lemma 4.4, we find that√pqx

φ(p)φ(q)

∑χ2 mod qχ2 odd

χ2(b)∞∑

n,m=0p-n

χ2(m)J1(4π

√nmx/pq)√nm

=

√pqx

φ(p)φ(q)

∑χ2 mod qχ2 odd

χ2(b)

{∞∑

n,m=0

χ2(m)J1(4π

√nmx/pq)√nm

−∞∑

n,m=0

χ2(m)J1(4π

√nmx/q)

√pnm

}


=

√pqx

2φ(p)

{∞∑

n,m=0m≡b mod q


−∞∑

n,m=0m≡−b mod q


−∞∑

n,m=0m≡b mod q

J1(4π√nmx/q)

√pnm

+∞∑

n,m=0m≡−b mod q

J1(4π√nmx/q)

√pnm

}

=p

φ(p)

∞∑n=1

F( xpn

)sin(2πnb

q

)− 1

φ(p)

∞∑m=1

F( xm

)sin(2πmb

q

)+

1

4cot(bπq

). (4.7)

Therefore, using (4.7) in (4.6), we obtain

− 1

4cot(bπq

)+

√pqx

4

{∞∑



−∞∑



}

=

√pqx

φ(p)φ(q)


∑χ2 mod pχ2 odd

χ1(a)χ2(b)∞∑n=0

dχ1,χ2(n)J1(4π

√nx/pq)√n

+p

φ(p)

∞∑n=1

F( xpn

)sin(2πnb

q

)− 1

φ(p)

∞∑m=1

F( xm

)sin(2πmb

q

), (4.8)

which completes the proof by Theorem 4.1 and (4.5). �

We now establish a converse theorem.

Proof : Theorem 2.2⇒ Theorem 4.1. Let θ = h/p and σ = k/q in (2.2). Let χ1 be a non-principaleven primitive character modulo p and χ2 an odd primitive character modulo q. We multiply bothsides of (2.2) by χ1(h)χ2(k)/τ(χ1)τ(χ2), and sum on h and k with 1 ≤ h < p and 1 ≤ k < q.Then, using Lemma 2.4, we have

1

τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1


′cos(2πnh/p) sin(2πmk/q)

= −i∑nm≤x

′χ1(n)χ2(m) = −i

∑n≤x

′∑d|n

χ1(d)χ2(n/d) = −i∑n≤x

′dχ1,χ2(n). (4.9)

On the other hand,

1

4τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k) cot(kπ/q)

=1

4τ(χ1)τ(χ2)

p−1∑h=1

χ1(h)

q−1∑k=1

χ2(k) cot(kπ/q) = 0, (4.10)


since χ1 is non-principal. Also, the contribution of the second expression on the right-hand side of(2.2) is√xpq

4τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k)

{∞∑

n,m=0n≡±h mod pm≡k mod q


−∞∑

n,m=0n≡±h mod pm≡−k mod q


}

=

√xpq

2τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

χ1(n)( q−1∑

k=1k≡m mod q

χ2(k) +

q−1∑k=1

k≡−m mod q

χ2(−k))J1(4π√nmx/(pq))√

nm

=

√xpq

τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

χ1(n)χ2(m)

{J1(4π

√nmx/(pq))√nm

}

= −τ(χ1)τ(χ2)√pq

∞∑n=1

dχ1,χ2(n)(xn

) 12J1

(4π√nx/(pq)

), (4.11)

since τ(χ1)τ(χ1) = p and τ(χ2)τ(χ2) = −q. Hence, (4.9)–(4.11) imply Theorem 4.1, and wecomplete the proof. �


We begin by proving the following theorem.

Theorem 5.1. If χ1 and χ2 are odd primitive characters modulo p and q, respectively, then∑n≤x

′ndχ1,χ2(n) = −τ(χ1)τ(χ2)√

pq

∞∑n=1

ndχ1,χ2(n)(xn

) 32K 3

2

(4π2nx

pq;1

2; 2), (5.1)

where Kν(x;µ; 2) is defined by (3.3).

Proof. Using (4.1), we have

π−(2s+1)

(pq)−(s+1/2)Γ2(s+

1

2

)L(2s, χ1)L(2s, χ2) (5.2)

= −τ(χ1)τ(χ2)√pq

π−(2−2s)

(pq)−(1−s)Γ2(1− s)L(1− 2s, χ1)L(1− 2s, χ2).

Replacing s+ 1/2 by s, we can rewrite (5.2) as

π−2s

(pq)−sΓ2(s)L(2s− 1, χ1)L(2s− 1, χ2)

= −τ(χ1)τ(χ2)√pq

π−(3−2s)

(pq)−(3/2−s)Γ2(3

2− s)L(2− 2s, χ1)L(2− 2s, χ2).


We note that

L(2s− 1, χ1)L(2s− 1, χ2) =∞∑n=1

χ1(n)

n2s−1

∞∑m=1

χ2(m)

m2s−1 =∞∑n=1

ndχ1,χ2(n)

n2s.

In the notation of Theorem 2 of [1], let q = 0, r = 3/2, m = 2, λn = µn = π2n2/(pq),a(n)dχ1,χ2(n), and b(n) = −τ(χ1)τ(χ2)ndχ1,χ2(n)/

√pq. Also, replacing x by π2x2/(pq), we

obtain∑n≤x

′ndχ1,χ2(n) = −τ(χ1)τ(χ2)√

pq

∞∑n=1

ndχ1,χ2(n)(xn

) 32K 3

2

(4π2nx

pq;1

2; 2)

+Q0

(π2x2

pq

).

Since L(2s− 1, χ1) and L(2s− 1, χ2) are entire functions and they are zero at s = 0, we have

Q0

(π2x2

pq

)=

1

2πi

∫C

L(2s− 1, χ1)L(2s− 1, χ2)x2s

sds = 0,


Next, we examine

K 32

(4π2nx

pq;1

2; 2)

=

∫ ∞0

J 12(u)J 3

2

(4π2nx

pqu

)du.

Recall that [16, p. 54, formula (3)]

J1/2(z) =

√2

πzsin z and J3/2(z) =

√2

πz

(sin z

z− cos z

).

So, with y = 2π√nx/(pq), we have

K 32

(4π2nx

pq;1

2; 2)

=

∫ ∞0

J 12(u)J 3

2

(y2u

)du

=2

πy

∫ ∞0

( uy2

sinu sin(y2u

)− sinu cos

(y2u

))du. (5.3)

By differentiating a formula from [16, p. 184, formula (3)] with respect to y, or by making a changeof variable in (3.5), we can derive that∫ ∞

0

1

u2sinu cos

y2

udu =

π

2yI1(2y). (5.4)

We also use the formula [16, p. 184, formula (4)]∫ ∞0

1

ucosu cos

y2

udu = −π

2Y0(2y) +K0(2y). (5.5)

Subtracting (5.5) from (5.4), we find that∫ ∞0

( 1

u2sinu cos

y2

u− 1

ucosu cos

y2

u

)du =

π

2

(1

yI1(2y) + Y0(2y)− 2

πK0(2y)

). (5.6)


Next, we differentiate both sides of (5.6) with respect to y. Then, the left-hand side of (5.6) yields

2y

∫ ∞0

(− 1

u3sinu sin

y2

u+

1

u2cosu sin

y2

u

)du

=2

y

∫ ∞0

(− t

y2sin

y2

tsin t+ cos

y2

tsin t

)dt, (5.7)

where we made a change of variable y2/u = t. On the other hand, differentiation of the right-handside of (5.6) gives

π

2

(− I1(2y)

y2+

2I ′1(2y)

y+ 2Y ′0(2y)− 4

πK ′0(2y)

)= π

(− I1(2y)

2y2+I ′1(2y)

y− Y1(2y) +

2

πK1(2y)

). (5.8)

Equating (5.7) with (5.8) and using the result in (5.3), we deduce that

K 32

(4π2nx

pq;1

2; 2)

=I1(2y)

2y2− I ′1(2y)

y+ Y1(2y)− 2

πK1(2y). (5.9)

Using formulas from [16, p. 66, formula (4), p. 79, formula (4)], we find that

I ′1(2y) = −I1(2y)

2y+R0(2y),

where

Rν(x) := −Yν(x) +2

πKν(x). (5.10)

Using the last two equalities in (5.9), we arrive at

K 32

(4π2nx

pq;1

2; 2)

=I1(2y)

y2− R0(2y)

y−R1(2y). (5.11)

As before, it suffices to prove Theorem 2.3 for rational arguments of θ and σ. For brevity, set

K 32

(x;

1

2; 2)

:= T 32(x). (5.12)

Theorem 5.2. If p and q are primes, and 0 < a < p and 0 < b < q, then∑nm≤x

′nm sin(2πna/p) sin(2πmb/q)

=x√x

4

∑n,m≥0

{T 3

2

(4π2(n+ a/p)(m+ b/q)x

)√(n+ a/p)(m+ b/q)

−T 3

2

(4π2(n+ 1− a/p)(m+ b/q)x

)√(n+ 1− a/p)(m+ b/q)

−T 3

2

(4π2(n+ a/p)(m+ 1− b/q)x

)√(n+ a/p)(m+ 1− b/q)

+T 3

2

(4π2(n+ 1− a/p)(m+ 1− b/q)x

)√(n+ 1− a/p)(m+ 1− b/q)

}.

Next, we show that Theorem 5.1 implies Theorem 5.2.


Proof : Theorem 5.1⇒ Theorem 5.2. Let p and q be primes, and 0 < a < p and 0 < b < q. UsingLemma 2.5, we see that∑

nm≤x

′nm sin(2πna/p) sin(2πmb/q)

=−1

φ(p)φ(q)

∑χ1 mod pχ1 odd

∑χ2 mod qχ2 odd


′ndχ1,χ2(n). (5.13)

On the other hand, by (2.6), we obtain

x√x

4

∑n,m≥0

{T 3

2

(4π2(n+ a/p)(m+ b/q)x

)√(n+ a/p)(m+ b/q)

−T 3

2

(4π2(n+ 1− a/p)(m+ b/q)x

)√(n+ 1− a/p)(m+ b/q)

−T 3

2

(4π2(n+ a/p)(m+ 1− b/q)x

)√(n+ a/p)(m+ 1− b/q)

+T 3

2

(4π2(n+ 1− a/p)(m+ 1− b/q)x

)√(n+ 1− a/p)(m+ 1− b/q)

}

=x√pqx

4

{∞∑

n,m=0n≡a mod pm≡b mod q

T 32

(4π2nmx/pq

)√nm

−∞∑

n,m=0n≡−a mod pm≡b mod q

T 32(4π2nmx/pq)√nm

−∞∑

n,m=0n≡a mod pm≡−b mod q


+∞∑

n,m=0n≡−a mod pm≡−b mod q


}

=x√pqx

φ(p)φ(q)

∞∑n,m=0


∑χ1 mod pχ1 odd

∑χ2 mod qχ2 odd

χ1(a)χ2(b)χ1(n)χ2(m)

=x√pqx

φ(p)φ(q)

∑χ1 mod pχ1 odd

∑χ2 mod qχ2 odd

χ1(a)χ2(b)∞∑n=0

dχ1,χ2(n)T 3

2(4π2nx/pq)√n

=

√pq

φ(p)φ(q)

∑χ1 mod pχ1 odd

∑χ2 mod qχ2 odd

χ1(a)χ2(b)∞∑n=0

ndχ1,χ2(n)(xn

) 32T 3

2(4π2nx/pq). (5.14)

Hence, by (5.13), (5.14), and Theorem 5.1 we complete the proof. �

As in the previous sections, we can prove a converse theorem.

Proof : Theorem 2.3⇒ Theorem 5.1. Let θ = h/p and σ = k/q in Theorem 2.3. Let χ1 and χ2 beodd primitive characters modulo p and q, respectively. Then, we multiply by χ1(h)χ2(k)/τ(χ1)τ(χ2),


and sum on h and k with 1 ≤ h < p and 1 ≤ k < q. By Lemma 2.4, we deduce that

1

τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1


′nm sin(2πnh/p) sin(2πmk/q)

= −∑nm≤x

′nmχ1(n)χ2(m) = −

∑n≤x

′ndχ1,χ2(n). (5.15)

On the other hand, we observe that

x√pqx

4τ(χ1)τ(χ2)

p−1∑h=1

q−1∑k=1

χ1(h)χ2(k)

{∞∑

n,m=0n≡h mod pm≡k mod q

T 32

(4π2nmx/pq

)√nm

−∞∑

n,m=0n≡−h mod pm≡k mod q


−∞∑

n,m=0n≡h mod pm≡−k mod q


+∞∑

n,m=0n≡−h mod pm≡−k mod q


}

=x√pqx

4τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

(χ1(n)− χ1(−n))(χ2(m)− χ2(−m))T 3

2(4π2nmx/pq)√nm

=x√pqx

τ(χ1)τ(χ2)

∞∑n=1

∞∑m=1

χ1(n)χ2(m)T 3

2(4π2nmx/pq)√nm

(5.16)

=x√xτ(χ1)τ(χ2)√

pq

∞∑n=1

dχ1,χ2(n)T 3

2(4π2nx/pq)√n

=τ(χ1)τ(χ2)√

pq

∞∑n=1

ndχ1,χ2(n)(xn

) 32T 3

2(4π2nx/pq). (5.17)

Taking (5.15) and (5.17) together in Theorem 2.3, we complete the proof. �

We briefly indicate some special cases of Theorem 5.1. Let χ be an odd primitive character modq, and set χ1 = χ2 = χ. Then,

dχ1,χ2(n) = dχ,χ(n) =∑d|n

χ(d)χ(n/d) =∑d|n

χ(n) = χ(n)d(n).

Then if χ denotes, respectively, the Legendre symbol modulo 3 and the simple primitive charactermodulo 4, we obtain identities for∑

n≤xn≡1 (mod 3)

′nd(n)−

∑n≤x

n≡2 (mod 3)

′nd(n)


and ∑n≤x

n≡1 (mod 4)

′nd(n)−

∑n≤x

n≡3 (mod 4)

′nd(n).

Since the identities are readily obtained from Theorem 5.1, we forego stating them.

6. SOME DIFFERENTIAL EQUATIONS

In this section we show that certain double integrals on the right sides of either (2.1), (2.2), or(2.3) satisfy a partial differential equation. To that end, define, for x > 0,

U(θ, σ, x) (6.1)

=x2

8+

1

4

∫ x

0

∫ v

0

√u∑n,m≥0

{I1(4π

√(n+ θ)(m+ σ)u)√

(n+ θ)(m+ σ)+I1(4π

√(n+ 1− θ)(m+ σ)u)√

(n+ 1− θ)(m+ σ)

+I1(4π

√(n+ θ)(m+ 1− σ)u)√

(n+ θ)(m+ 1− σ)+I1(4π

√(n+ 1− θ)(m+ 1− σ)u)√

(n+ 1− θ)(m+ 1− σ)

}dudv.

By Theorem 2.1, U(θ, σ, x) is well defined. Using Theorems 2.2 and 2.3, we can similarly defineV (θ, σ, x) and W (θ, σ, x). For brevity, we do not write out these definitions.

Theorem 6.1. For 0 < θ < 1, 0 < σ < 1, and x > 0, let the double integrals of the right-handsides of (2.1), (2.2), and (2.3) be denoted by U(θ, σ, x), V (θ, σ, x), and W (θ, σ, x), respectively,as illustrated above in (6.1). Then, if G(θ, σ, x) is any one of these three functions and x is not aninteger,

G− x∂G∂x

+1

2x2∂2G

∂x2=

1

25π4

∂6G

∂x2∂θ2∂σ2. (6.2)

Proof. We prove Theorem 6.1 for the function G(θ, σ, x) = U(θ, σ, x); the proofs in the remainingtwo cases are almost identical.

By Theorem 2.1, we can rewrite U(θ, σ, x) in the form

U(θ, σ, x) =

∫ x

0

∫ v

0

∑nm≤u

′cos(2πnθ) cos(2πmσ)dudv

=

∫ x

0

∑nm≤v

′∫ v

nm

cos(2πnθ) cos(2πmσ)dudv

=

∫ x

0

∑nm≤v

(v − nm) cos(2πnθ) cos(2πmσ)dv

=∑nm≤x

cos(2πnθ) cos(2πmσ)

∫ x

nm

(v − nm)dv

=1

2

∑nm≤x

(x− nm)2 cos(2πnθ) cos(2πmσ). (6.3)


The function U(θ, σ, x) is infinitely many times differentiable with respect to θ and σ, while withrespect to x, U(θ, σ, x) is only once differentiable at points (θ, σ, x), where x is an integer, andinfinitely many times differentiable at points (θ, σ, x), where x is not an integer. Thus, for allx > 0,

∂U

∂x=∑nm≤x

(x− nm) cos(2πnθ) cos(2πmσ) (6.4)

and, for all non-integral x,

∂2U

∂x2=∑nm≤x

cos(2πnθ) cos(2πmσ). (6.5)

Furthermore, differentiating (6.5) twice with respect to both θ and σ, we easily find that

∂6U

∂x2∂θ2∂σ2= (2π)4

∑nm≤x

n2m2 cos(2πnθ) cos(2πmσ). (6.6)

Hence, from (6.3)–(6.6),

U − x∂U∂x

+1

2x2∂2U

∂x2=∑nm≤x

((x− nm)2

2− x(x− nm) +

x2

2

)cos(2πnθ) cos(2πmσ)

=1

2

∑nm≤x

n2m2 cos(2πnθ) cos(2πmσ)

=1

25π4

∂6U

∂x2∂θ2∂σ2, (6.7)


We remark that if 0 < θ < 1, 0 < σ < 1, and x is any positive integer k, then (6.2) remains valid,provided that the higher order derivatives in (6.2) are appropriately replaced. We illustrate withU(θ, σ, x); the necessary replacements for the remaining two functions, V (θ, σ, x) and W (θ, σ, x)are analogous. Thus, at k, ∂2U/∂x2 needs to be replaced by

limε→0

1

2

(∂2U

∂x2(θ, σ, k − ε) +

∂2U

∂x2(θ, σ, k + ε)

)=∑mn≤k

′cos(2πnθ) cos(2πmσ),

and similarly, at x = k, ∂6G/∂x2∂θ2∂σ2 should be replaced by

limε→0

1

2

(∂6U

∂x2∂θ2∂σ2(θ, σ, k − ε) +

∂6U

∂x2∂θ2∂σ2(θ, σ, k + ε)

)= (2π)4

∑nm≤k

′n2m2 cos(2πnθ) cos(2πmσ).

Acknowledgements. The authors are grateful to the referee for helpful suggestions and correc-tions, and to Andrzej Schinzel for information on the work of Sierpinski.


REFERENCES

[1] B.C. Berndt, Identities involving the coefficients of a class of Dirichlet series. I, Trans. Amer. Math. Soc. 137(1969), 345–359.

[2] B.C. Berndt, O.–Y. Chan, S.–G. Lim, and A. Zaharescu, Questionable claims found in Ramanujan’s lost note-book, in Tapas in Experimental Mathematics, T. Amdeberhan and V.H. Moll, eds., Contemp. Math., vol. 457,American Mathematical Society, Providence, RI, 2008, pp. 69–98.

[3] B.C. Berndt, S. Kim, and A. Zaharescu, Weighted divisor sums and Bessel function series, II, Adv. Math., toappear.

[4] B.C. Berndt and A. Zaharescu, Weighted divisor sums and Bessel function series, Math. Ann. 335 (2006), 249–283.

[5] K. Chandrasekharan and R. Narasimhan, Hecke’s functional equation and arithmetical identities, Ann. Math. (2)74 (1961), 1–23.

[6] H. Davenport, Multiplicative Number Theory, 3rd ed., Springer-Verlag, New York, 2000.[7] G.H. Hardy, On the expression of a number as the sum of two squares, Quart. J. Math. (Oxford) 46 (1915),

263–283.[8] G.H. Hardy, On Dirichlet’s divisor problem, Proc. London Math. Soc. (2) 15 (1916), 1–25.[9] G.H. Hardy, Collected Papers, Vol. II, Oxford University Press, Oxford, 1967.

[10] E. Landau, Uber die Gitterpunkte in einem Kreise (Zweite Mitteilung), Gottingen Nachr. (1915), 161–171.[11] E. Landau, Vorlesungen uber Zahlentheorie, Erster Band, Zweiter Teil, Chelsea, New York, 1947.[12] E. Landau, Collected Works, Vol. 6, Thales Verlag, Essen, 1985.[13] S. Ramanujan, The Lost Notebook and Other Unpublished Papers, Narosa, New Delhi, 1988.[14] W. Sierpinski, O pewnem zagadnieniu z rachunku funkeyj asymptotycznych, Prace Mat. Fiz. 17 (1906), 77–118.[15] G.F. Voronoı, Sur une fonction transcendante et ses applications a la sommation de quelques series, Ann. Ecole

Norm. Sup. (3) 21 (1904), 207–267, 459–533.[16] G.N. Watson, Theory of Bessel Functions, 2nd ed., University Press, Cambridge, 1966.

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF ILLINOIS, 1409 WEST GREEN STREET, URBANA, IL61801, USA

E-mail address: [email protected]

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circle divisor problemsberndt/articles/bessel2... · 2014. 5. 23. · BRUCE C. BERNDT, SUN KIM, AND ALEXANDRU ZAHARESCU Abstract. On page 335 in his lost notebook, Ramanujan recorded