Download - Hypergeometric Function
25 Hypergeometric function
IMRN International Mathematics Research NoticesVolume 2006, Article ID 41417, Pages 1–19
Integral Mean Values of Maass L-Functions
Qiao Zhang
1 Introduction
A standing topic in analytic number theory is to estimate integral mean values
∫T
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2m
dt, (1.1)
where f is an automorphic form over GL(n) and L(s, f) its associated L-function, nor-
malized so that the central point is at s = 1/2. Over the last ninety years, many au-
thors have worked in this field with fruitful results; see, for example, an excellent sur-
vey in [7]. In most cases, we have to face delicate discussions of the arithmetic nature
of Fourier coefficients, such as estimates of the “generalized additive divisor problems”∑n≤x λf(n)λf(n + r).
To avoid this difficulty, in this paper we consider instead the Dirichlet integral,
Zf(w) =
∫∞
1
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
t−wdt (�w� 1), (1.2)
and study its analytic properties, especially its meromorphic continuation beyond �w =
1 and its polar behavior. Our goal is to express Zf(w) as an inner product of fwith a cer-
tain kernel function, by directly exploring the symmetries satisfied by f itself, so that
Received 11 April 2005; Accepted 19 February 2006
Communicated by Dennis Hejhal
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
2 Qiao Zhang
it suffices to study the analytic properties of this kernel function alone. This realizes
and generalizes the ideas suggested by Good [3], and is applicable to cases where our
knowledge of the arithmetic nature of f is still limited. This is the main motivation of the
present work.
Along this line, in [9] we have considered modular forms with respect to Hecke
congruence subgroups, and studied the mean squares of their L-functions. The present
paper is devoted to the study of the corresponding results for Maass forms. In particular,
we have the following theorem.
Theorem 1.1. Let f(z) be an even Maass form for Γ0(N) with Laplacian eigenvalue 1/4+ν2
and nebentypus χ(modN). Then asymptotically,
∫T
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
dt ∼
8
Γ
(1
2+ iν
)
Γ
(1
2− iν
)‖f‖2
vol(Γ\H)· T log T
=8 coshπν
π
‖f‖2
vol(Γ\H)· T log T,
(1.3)
where the norm of f is given by
‖f‖2 =
∫∫
Γ0(N)\H
∣∣f(z)
∣∣2dxdy
y2. (1.4)
�
Remark 1.2. The mean squares of the L-functions associated to Maass forms were firstly
studied in 1981 by Kuznetsov [6], who used the method of approximate functional equa-
tions to obtain the asymptotic formula (1.3) for Maass forms with respect to SL(2,Z).
Along this line, in 1992 Muller considered nonholomorphic automorphic forms over
Fuchsian groups of the first kind with real weight. In 1997, Jutila [5] used the Laplace
transform method to study in a unified way the fourth moment of ζ(s) and the mean
squares of L-functions associated to both cusp forms and Maass forms with respect to
SL(2,Z). However, all these cases rely on delicate discussions of certain generalized ad-
ditive divisor problems.
Remark 1.3. As in [9], we can also obtain a much more accurate asymptotic formula for a
weighted mean value problem, but we will not give details here.
Remark 1.4. Using the renormalized integrals as introduced by Zagier [8], we may also
expect a corresponding result along this line for the fourth moment of the Riemann zeta
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 3
function or a Dirichlet L-function. However, due to the lack of space, we omit the discus-
sion here.
The arguments for Theorem 1.1 are essentially the same as those in [9], and the
main obstacle here is to establish an inner product representation for the Dirichlet in-
tegral (1.2), as now general hypergeometric functions, instead of the gamma functions,
come into the picture. Similar difficulties arise in the recent work of Beineke and Bump
[1] in which the authors take yet another approach to study the mean squares but get only
the upper bound
∫T
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
dt� T log T, (1.5)
where f is an even Maass form with respect to SL(2,Z).
To better illustrate the idea of our approach, now we present the proof for
Theorem 1.1 but leave the verification of some crucial estimates to the later sections.
Proof for Theorem 1.1. For simplicity, in this proof we write Γ = Γ0(N).
Write
f(z) =
∞∑n=−∞n�=0
a(n)√yKiν
(
2π|n|y)
e2πinx, (1.6)
then for �s > 1we have
∫∞
0
f(
reiθ)
rs−1/2dr
r=√
sin θ∞∑
n=−∞n�=0
an
∫∞
0
Kiν
(
2π|n|r sin θ)
e2πinr cos θrsdr
r
=2√
sin θ(2π sin θ)s
L(s, f)∫∞
0
Kiν(r) cos(r cot θ)rsdr
r.
(1.7)
By the integral formula (2.4), this gives that
∫∞
0
f(
reiθ)
rs−1/2dr
r=1
2(sin θ)1/2−sΛ(s, f) 2F1
(s + iν
2,s − iν
2;1
2; −(cot θ)2
)
, (1.8)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
4 Qiao Zhang
where
Λ(s, f) = π−sΓ
(s + iν
2
)
Γ
(s − iν
2
)
L(s, f) (1.9)
is the complete L-function for f, so by (2.2) we have the Mellin transform
∫∞
0
f(
reiθ)
rs−1/2dr
r=1
2(sin θ)1/2+iνΛ(s, f) 2F1
(s + iν
2,1 − s + iν
2;1
2; (cos θ)2
)
.
(1.10)
Hence the inverse Mellin transform gives
f(
reiθ)
=(sin θ)1/2+iν
4πi
∫(2)Λ(s, f) 2F1
(s + iν
2,1 − s + iν
2;1
2; (cos θ)2
)
r1/2−sds,
(1.11)
where we write∫
(c) to denote the integral∫c+i∞
c−i∞ .
Also, for �τ > 1 and �w ≥ 1 let
Pw,τ(z) =∑γ∈Γ
(�γz)τ
(�γz
|γz|
)w
(1.12)
be the nonholomorphic kernel function, then in [9] we have shown that Pw,τ(z) has a
meromorphic continuation up to �τ > −ε, �w > 1/2. In particular, Pw,0(z) is holomor-
phic up to �w = 1with a double pole atw = 1 and Laurent expansion
Pw,0(z) =4
vol(Γ\H)1
(w − 1)2+O
(1
|w − 1|
)
. (1.13)
Now assume that �τ > 1 and �w ≥ 1, and consider the inner product
Zf(w, τ) =⟨
Pw,τ, |f|2⟩
=
∫∫Γ\H
Pw,τ(z)∣∣f(z)
∣∣2dxdy
y2
=
∫∞
0
∫∞
−∞yτ
(y
|z|
)w∣∣f(z)
∣∣2dxdy
y2,
(1.14)
then the above analytic properties of Pw,τ(z) imply that Zf(w, τ) also has a meromorphic
continuation up to �τ > −ε, �w > 1/2, and that Zf(w, 0) is holomorphic up to �w = 1
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 5
with a double pole atw = 1 and Laurent expansion
Zf(w, 0) =4
vol(Γ\H)‖f‖2
(w − 1)2+O
(1
|w − 1|
)
. (1.15)
On the other hand, we may further assume that �τ > 2, then using the polar
coordinates in H and the inverse Mellin transform (1.11) gives
Zf(w, τ) =
∫∞
0
∫π
0
rτ(sin θ)w+τ−2f(
reiθ)
f(
reiθ)
dθdr
r
=1
4πi
∫(2)Λ(s, f)ds
∫π
0
2F1
(s + iν
2,1 − s + iν
2;1
2; (cos θ)2
)
(sin θ)3/2−w−τ−iνdθ
·∫∞
0
rτ+1/2−sf(
reiθ)dr
r,
(1.16)
and another application of the Mellin transform (1.10) now gives
Zf(w, τ) =1
4πi
∫(2)Λ(s, f)Λ(τ + 1 − s, f)I(s;w, τ)ds, (1.17)
where the function
I(s;w, τ)
=1
2
∫π
0
2F1
(s+iν
2,1−s+iν
2;1
2; (cos θ)2
)
2F1
(τ+1−s−iν
2,s−τ−iν
2;1
2; (cos θ)2
)
(sin θ)1−w−τdθ
=
∫π/2
0
2F1
(s+iν
2,1−s+iν
2;1
2; (cos θ)2
)
2F1
(τ+1−s−iν
2,s−τ−iν
2;1
2; (cos θ)2
)
(sin θ)1−w−τdθ
(1.18)
obviously has an analytic continuation up to �s ≥ 1/2, �(w + τ) ≥ 1. In particular, for
w > 1we have
Zf(w, 0) =1
4πi
∫(2)Λ(s, f)Λ(1 − s, f)I(s;w, 0)ds
=1
4πi
∫(1/2)
Λ(s, f)Λ(1 − s, f)I(s;w, 0)ds
=1
2π
∫∞
0
∣∣∣∣Λ
(1
2+ it, f
)∣∣∣∣
2
I
(1
2+ it;w, 0
)
dt.
(1.19)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
6 Qiao Zhang
Now we quote the estimates (3.5) and (3.28), to be proved later in Propositions 3.2 and
3.3 in Section 3, that for t → ∞ we have
I
(1
2+ it;w, 0
)
=
2w−1πΓ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)∣∣∣∣Γ
(1
4+it + iν
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4+it − iν
2
)∣∣∣∣
2t−w +O
(
eπtt1/2−w)
,
(1.20)
where theO-constant is uniform forw ∈ [1, 3/2]. Plugging this into (1.19) gives
Zf(w, 0) =
Γ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)2w−2
π
∫∞
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
t−wdt +G(w)
(1.21)
for some function G(w) analytic for �w > 1/2.
Finally, we combine this formula with (1.15), then asw → 1we have
∫∞
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
t−wdt ∼
24−wπΓ(w)
Γ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
‖f‖2
vol(Γ\H)1
(w − 1)2
∼
8π
Γ
(1
2
)2
Γ
(1
2+ iν
)
Γ
(1
2− iν
)
‖f‖2
vol(Γ\H)1
(w − 1)2
=8
Γ
(1
2+ iν
)
Γ
(1
2− iν
)‖f‖2
vol(Γ\H)1
(w − 1)2.
(1.22)
Hence by classical complex Tauberian arguments we have
∫T
0
∣∣∣∣L
(1
2+ it, f
)∣∣∣∣
2
dt ∼
8
Γ
(1
2+ iν
)
Γ
(1
2− iν
)‖f‖2
vol(Γ\H)· T log T
=8 coshπν
π
‖f‖2
vol(Γ\H)· T log T.
(1.23)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 7
Note that we have applied the well-known formula
Γ
(1
2+ z
)
Γ
(1
2− z
)
= π sec(πz). (1.24)
This completes the proof of Theorem 1.1. �
2 Hypergeometric functions
For reference, in this section we summarize some results on hypergeometric functions
that we will use in sequel.
Let a, b, c ∈ C, then the hypergeometric function 2F1(a, b; c; z) is defined by the
infinite series,
2F1(a, b; c; z) =
∞∑n=0
Γ(a + n)Γ(a)
Γ(b + n)Γ(b)
Γ(c)Γ(c + n)
zn
n!
(
|z| < 1)
, (2.1)
with meromorphic continuation in z to the whole complex plane.
Now we quote some well-known basic formulas for hypergeometric functions.
Lemma 2.1. There exists
2F1(a, b; c; z) = (1 − z)−a2F1
(
a, c − b; c;z
z − 1
)
.
(2.2)
�
Lemma 2.2. Assume that 1 − c, b − a, and c − a − b are not integers. Then
2F1(a, b; c; z) =Γ(c)Γ(c − a − b)Γ(c − a)Γ(c − b) 2F1(a, b;a + b − c + 1; 1 − z)
+Γ(c)Γ(a + b − c)
Γ(a)Γ(b)2F1(c − a, c − b; c − a − b + 1; 1 − z)
(1 − z)a+b−c.
(2.3)
�
Lemma 2.3. Assume that �a = 0 and that �s > |�ν|. Then
∫∞
0
Kν(x) cos(ax)xsdx
x= 2s−2Γ
(s + ν
2
)
Γ
(s − ν
2
)
2F1
(s + ν
2,s − ν
2;1
2; −a2
)
.
(2.4)�
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
8 Qiao Zhang
The main ingredient of this paper is the following asymptotic estimate for hyper-
geometric functions.
Lemma 2.4 (see [4]). Assume that |z| ≤ 1 but z = −1, then as |λ| → ∞
2F1
(
a + λ, b − λ; c;1 − z
2
)
= Γ(c)2(a+b−1)/2 (z + 1)(c−a−b−1)/2
(z − 1)c/2
(sinh ζζ
)1/2
α1−c(
ζIc−1(αζ) + ε(ζ))
,
(2.5)
where we have put z = cosh ζwith �ζ ≥ 0 and |�ζ| ≤ π, and
α =a − b
2+ λ. (2.6)
Furthermore, the error term ε(ζ) is bounded by
ε(ζ) � |ζ|
|α|2∣∣Kc(αζ)
∣∣, (2.7)
where the �-constant depends on a, b, c only. �
Lemma 2.5 (see [4]). In the notation of the above lemma, the error term ε(ζ) can be ob-
tained as follows. Consider the sequence of functions {εn(ζ)} with ε0(ζ) = 0,
ε1(ζ) = 2ζ
∫ζ
0
tIc−1(αt)(
Ic−1(αζ)Kc−1(αt) − Ic−1(αt)Kc−1(αζ)){tB0(t)
} ′dt,
εn+1(ζ)−εn(ζ)=ζ
∫ζ
0
(
Ic−1(αζ)Kc−1(αt)−Ic−1(αt)Kc−1(αζ))(
εn(t)−εn−1(t))
ψ(t)dt,
(2.8)
where
B0(ζ) =1
2ζ
((
(c − 1)2 −1
4
)(1
ζ− coth ζ
)
+(c − 1)2 − (a + b − c)2
2tanh
ζ
2
)
,
ψ(ζ) =
(
(c − 1)2 −1
4
)(1
sinh2ζ
−1
ζ2
)
+(c − 1)2 − (a + b − c)2
4 cosh2 ζ
2
.
(2.9)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 9
Then
ε(ζ) = limn→∞ εn(ζ) =
∞∑n=0
(
εn+1(ζ) − εn(ζ))
. (2.10)�
3 Propositions and their proofs
Now we are ready to prove the estimate (1.20) of I(1/2 + it;w, 0) that we quoted in the
proof of Theorem 1.1.
Proposition 3.1. Letw ∈ [1, 3/2] and write s = 1/2 + it, then as t → ∞,
∫π/2
log t/2t
2F1
(s + iν
2,1 − s + iν
2;1
2; (cos θ)2
)
2F1
(s − iν
2,1 − s − iν
2;1
2; (cos θ)2
)
(sin θ)1−wdθ
� eπtt1/2−w,
(3.1)
where the �-constant is uniform inw. �
Proof. Applying the estimate (2.5) with z = − cos 2θ and recalling the well-known for-
mula
I−1/2(x) =
√
2
πxcosh x, (3.2)
we have, as t → ∞, that
2F1
(1
4+iν
2+it
2,1
4+iν
2−it
2;1
2; (cos θ)2
)
∼
et(π/2−θ) + e−t(π/2−θ)
2(sin θ)1/2+iν, (3.3)
so forw ≥ 1 the integral in (3.1) is asymptotically equal to
∫π/2
log t/2t
(sin θ)w−2
(et(π/2−θ) + e−t(π/2−θ)
2
)2
dθ
� eπt
∫π/2
log t/2t
θw−2e−2θtdθ� eπtt1−w
∫∞
(1/2) log t
θw−2e−2θdθ
� eπtt1−w
∫∞
(1/2) log t
e−θdθ� eπtt1/2−w.
(3.4)
This completes the proof. �
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
10 Qiao Zhang
To prove the estimate (1.20), we consider the cases ν = 0 and ν = 0 separately, as
the latter requires some additional treatment.
Proposition 3.2. Letw ∈ [1, 3/2]. Assume that ν = 0, then
I
(1
2+ it;w, 0
)
=
2w−1πΓ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)∣∣∣∣Γ
(1
4+it + iν
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4+it − iν
2
)∣∣∣∣
2t−w +O
(
eπtt1/2−w)
,
(3.5)�
where the �-constant is uniform inw.
Proof. The integral over [log t/2t, π/2] has been estimated in Proposition 3.1. For 0 ≤ θ ≤log t/2t, since ν = 0, by (2.3) we have
2F1
(1
4+iν
2+it
2,1
4+iν
2−it
2;1
2; (cos θ)2
)
=
Γ
(1
2
)
Γ(−iν) 2F1
(1
4+iν
2+it
2,1
4+iν
2−it
2; 1 + iν; (sin θ)2
)
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
+
Γ
(1
2
)
Γ(iν) 2F1
(1
4−iν
2+it
2,1
4−iν
2−it
2; 1 − iν; (sin θ)2
)
(sin θ)2iνΓ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
)
=
√π
(sin θ)iν
∑ε=±1
Γ(−iεν) 2F1
(1
4+iεν
2+it
2,1
4+iεν
2−it
2; 1 + iεν; (sin θ)2
)
(sin θ)−iενΓ
(1
4−iεν
2+it
2
)
Γ
(1
4−iεν
2−it
2
)
=
√π
(sin θ)iνF(t; θ),
(3.6)
say, then by symmetry we also have
2F1
(1
4−iν
2+it
2,1
4−iν
2−it
2;1
2; (cos θ)2
)
=√π(sin θ)iνF(t; θ), (3.7)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 11
so
I
(1
2+ it;w, 0
)
= π
∫ log t/2t
0
(sin θ)w−1F(t; θ)2dθ +O(
eπtt1/2−w)
. (3.8)
Applying (2.5) with z = cos 2θ, we have
(sin θ)iν
Γ(−iν) 2F1
(1
4+iν
2+it
2,1
4+iν
2−it
2; 1 + iν; (sin θ)2
)
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
=νΓ(iν)Γ(−iν)
2Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
(2
t
)iν2iθIiν(tθ) + ε(2iθ)
(θ sin θ)1/2.
(3.9)
Now (2.7), together with the well-known asymptotic formula
Ks(x) ∼
√π
2xe−x (x −→ ∞), (3.10)
shows that the error term ε(2iθ) is bounded by
ε(2iθ) � θ
t21
∣∣K1+iν(tθ)
∣∣� θ
t2
(
1 +√tθetθ
) � θtε−3/2, (3.11)
so Stirling’s formula gives
(sin θ)iν
Γ(−iν) 2F1
(1
4+iν
2+it
2,1
4+iν
2−it
2; 1 + iν; (sin θ)2
)
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
=
iνΓ(iν)Γ(−iν)(2
t
)iν
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
(θ
sin θ
)1/2
Iiν(tθ) +O(
e(π/2)ttε−1)
.
(3.12)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
12 Qiao Zhang
By symmetry, we also have
(sin θ)−iν
Γ(iν) 2F1
(1
4−iν
2+it
2,1
4−iν
2−it
2; 1 − iν; (sin θ)2
)
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
)
= −
iνΓ(iν)Γ(−iν)(2
t
)−iν
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
)
(θ
sin θ
)1/2
I−iν(tθ) +O(
e(π/2)ttε−1)
,
(3.13)
so
F(t; θ) =
(θ
sin θ
)1/2 ∑ε=±1
iενΓ(iν)Γ(−iν)(2
t
)iεν
Iiεν(tθ)
Γ
(1
4−iεν
2+it
2
)
Γ
(1
4−iεν
2−it
2
) +O(
e(π/2)ttε−1)
.
(3.14)
As we recall that
I−s(z) = Is(z) +2 sin(sπ)
πKs(z), (3.15)
the above implies that
F(t; θ) = iνΓ(iν)Γ(−iν)(
θ
sin θ
)1/2
Iiν(tθ)∑
ε=±1
ε
(2
t
)iεν
Γ
(1
4−iεν
2+it
2
)
Γ
(1
4−iεν
2−it
2
)
+
(θ
sin θ
)1/2 2ν sinh(πν)Γ(iν)Γ(−iν)(2
t
)−iν
Kiν(tθ)
πΓ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
) +O(
e(π/2)ttε−1)
= iνΓ(iν)Γ(−iν)(
θ
sin θ
)1/2
Iiν(tθ)∑
ε=±1
ε
(2
t
)iεν
Γ
(1
4−iεν
2+it
2
)
Γ
(1
4−iεν
2−it
2
)
+
(θ
sin θ
)1/2 2
(2
t
)−iν
Kiν(tθ)
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
) +O(
e(π/2)ttε−1)
,
(3.16)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 13
where we have applied the well-known formula
Γ(iν)Γ(−iν) =π
ν sinh(πν). (3.17)
By Stirling’s formula, we have (see, e.g., [2, formula 1.18.4])
Γ(z + α)Γ(z + β)
= zα−β
(
1 +(α − β)(α + β − 1)
2z+O
(
z−2))
(−π < arg z < π), (3.18)
so
(2
t
)iν
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
) −
(2
t
)−iν
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
)
=
(2
t
)iν
⎛
⎜⎜⎝1 −
(2
t
)−2iν Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
)
⎞
⎟⎟⎠
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
� t−2−�(iν)∣∣∣∣Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)∣∣∣∣
� e(π/2)tt−3/2.
(3.19)
It is also well known that, for θ ∈ [0, log t/2t],
Iiν(tθ) � min
{1,etθ
√tθ
}� t1/2
√
log t� t1/2, (3.20)
so we have
F(t; θ) =
(θ
sin θ
)1/2 2
(2
t
)−iν
Kiν(tθ)
Γ
(1
4+iν
2+it
2
)
Γ
(1
4+iν
2−it
2
) +O(
eπ/2tε−1) � e(π/2)tt1/2.
(3.21)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
14 Qiao Zhang
By symmetry, we also have
F(t; θ) =
(θ
sin θ
)1/2 2
(2
t
)iν
Kiν(tθ)
Γ
(1
4−iν
2+it
2
)
Γ
(1
4−iν
2−it
2
)
+O(
eπ/2tε−1) � e(π/2)tt1/2.
(3.22)
Therefore
∫ log t/2t
0
(sin θ)w−1F(t; θ)2dθ
=4
∣∣∣∣Γ
(1
4+iν
2+it
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4−iν
2+it
2
)∣∣∣∣
2
∫ log t/2t
0
(sin θ)w−2θKiν(tθ)2dθ
+O
(
eπttε−1/2
∫ log t/2t
0
(sin θ)w−1dθ
)
=4
∣∣∣∣Γ
(1
4+iν
2+it
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4−iν
2+it
2
)∣∣∣∣
2
∫ log t/2t
0
(sin θ)w−2θKiν(tθ)2dθ
+O(
eπtt−w)
.
(3.23)
Furthermore, we have
∫ log t/2t
0
(sin θ)w−2θKiν(tθ)2dθ
=
(
1 +O
((log t)2
t2
)) ∫ log t/2t
0
θw−1Kiν(tθ)2dθ
= t−w
(
1 +O
((log t)2
t2
)) ∫∞
0
θw−1Kiν(θ)2dθ +O(1)
=
2w−3Γ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)t−w +O(1),
(3.24)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 15
where we have applied the well-known formula
∫∞
0
Ka(y)Kb(y)ysd×y =
Γ
(s + a + b
2
)
Γ
(s + a − b
2
)
Γ
(s − a + b
2
)
Γ
(s − a − b
2
)
23−sΓ(s),
(3.25)
so
∫ log t/2t
0
(sin θ)w−1F(t; θ)2dθ
=
2w−1Γ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)∣∣∣∣Γ
(1
4+iν
2+it
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4−iν
2+it
2
)∣∣∣∣
2t−w +O
(
eπtt−w)
.
(3.26)
Finally, we combine the above estimate with Proposition 3.1, and this gives
I
(1
2+ it;w, 0
)
= π
∫ log t/2t
0
(sin θ)w−1F(t; θ)2dθ +O(
eπtt1/2−w)
=
2w−1πΓ
(w
2
)2
Γ
(w
2+ iν
)
Γ
(w
2− iν
)
Γ(w)∣∣∣∣Γ
(1
4+iν
2+it
2
)∣∣∣∣
2∣∣∣∣Γ
(1
4−iν
2+it
2
)∣∣∣∣
2t−w +O
(
eπtt1/2−w)
.
(3.27)
This completes the proof. �
Proposition 3.3. Letw ∈ [1, 3/2] and write s = 1/2 + it. Assume that ν = 0, then
I
(1
2+ it;w, 0
)
=
2w−1πΓ
(w
2
)4
Γ(w)∣∣∣∣Γ
(1
4+it
2
)∣∣∣∣
4t−w +O
(
eπtt1/2−w)
, (3.28)
where the �-constant is uniform inw. �
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
16 Qiao Zhang
Proof. The integral over [log t/2t, π/2] has been estimated in Proposition 3.1. For 0 ≤ θ ≤log t/2t, we may apply (2.3) by taking the limit as ν → 0, and this gives
2F1
(1
4+it
2,1
4−it
2;1
2; (cos θ)2
)
= −
Γ
(1
2
)(
2γ +Γ ′
Γ
(1
4+it
2
)
+Γ ′
Γ
(1
4−it
2
))
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
Γ
(1
4+it
2
)
Γ
(1
4−it
2
)
−
Γ
(1
2
)(
2 log sin θ +∂
∂a+∂
∂b+ 2
∂
∂c
)
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
Γ
(1
4+it
2
)
Γ
(1
4−it
2
) ,
(3.29)
where ∂/∂a, ∂/∂b, and ∂/∂c denote the partial derivatives with respect to the first, the
second, and the third parameters of the hypergeometric function, respectively.
As before, the asymptotic formula (2.5) gives
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)(θ
sin θ
)1/2
I0(tθ) +O(
t−3/2) � t1/2. (3.30)
Also, Stirling’s formula implies that
Γ ′
Γ(s) = log s +O
(
|s|−1)
, (3.31)
so the above formula can be simplified as
2F1
(1
4+it
2,1
4−it
2;1
2; (cos θ)2
)
= −
Γ
(1
2
)(
2γ + 2 logt
2+ 2 log sin θ
)
Γ
(1
4+it
2
)
Γ
(1
4−it
2
)
(θ
sin θ
)1/2
I0(tθ)
−
Γ
(1
2
)(∂
∂a+∂
∂b+ 2
∂
∂c
)
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
Γ
(1
4+it
2
)
Γ
(1
4−it
2
) +O(
e(π/2)t).
(3.32)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 17
As for the partial derivatives, by (2.5) we have
∂
∂a2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
=(θ sin θ)−1/2
2i
(
− log cos θ(
2iθI0(tθ) + ε(2iθ))
− 2θ2I ′0(tθ) +∂
∂aε(2iθ)
)
,
∂
∂b2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
=(θ sin θ)−1/2
2i
(
− log cos θ(
2iθI0(tθ) + ε(2iθ))
+ 2θ2I ′0(tθ) +∂
∂bε(2iθ)
)
,
∂
∂c2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
=
(Γ ′
Γ(1) + log cos θ − log sin θ − log
t
2
)2iθI0(tθ) + ε(2iθ)2i(θ sin θ)1/2
+1
2i(θ sin θ)1/2
(
2iθ∂I0(tθ)∂c
+∂
∂cε(2iθ)
)
=
(
− γ + log cos θ − log sin θ − logt
2
)2iθI0(tθ) + ε(2iθ)2i(θ sin θ)1/2
−
(θ
sin θ
)1/2
K0(tθ) +1
2i(θ sin θ)1/2
∂
∂cε(2iθ).
(3.33)
Note that here we have applied the formula that
∂
∂cIc(x)
∣∣c=0
= −K0(x). (3.34)
Hence
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
=2√π
Γ
(1
4+it
2
)
Γ
(1
4−it
2
)
(θ
sin θ
)1/2
K0(tθ) +O(
e(π/2)ttε−1)
+O
(
e(π/2)tt1/2θ−1
(∣∣∣∣
∂
∂aε(2iθ)
∣∣∣∣+
∣∣∣∣
∂
∂bε(2iθ)
∣∣∣∣+
∣∣∣∣
∂
∂cε(2iθ)
∣∣∣∣
))
.
(3.35)
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
18 Qiao Zhang
To estimate the above partial derivatives for ε(2iθ), we can in fact follow the ar-
guments in [4] for Lemma 2.5 line by line with only minor modifications. The discussion
is lengthy but routinic, so we omit the details here but only state the estimates
∂
∂aε(2iθ) � θ
t2K1(tθ)� θtε−3/2,
∂
∂bε(2iθ) � θ
t2K1(tθ)� θtε−3/2,
∂
∂cε(2iθ) � θ
(
1 +∣∣ log(tθ)
∣∣)
t2K1(tθ)� θtε−3/2| log θ|.
(3.36)
Hence
2F1
(1
4+it
2,1
4−it
2; 1; (sin θ)2
)
=2√π
∣∣∣∣Γ
(1
4+it
2
)∣∣∣∣
2
(θ
sin θ
)1/2
K0(tθ) +O(
e(π/2)ttε−1| log θ|) � e(π/2)tt1/2.
(3.37)
Combining the above estimate with Proposition 3.1, as in the proof of Proposition 3.2 we
have
I
(1
2+ it,w; 0
)
=
∫ log t/2t
0
(sin θ)w−12F1
(1
4+it
2;1
2; (cos θ)2
)2
dθ +O(
eπtt1/2−w)
=4π
∣∣∣∣Γ
(1
4+it
2
)∣∣∣∣
4
∫ log t/2t
0
(sin θ)w−2θK0(tθ)2dθ +O(
eπtt1/2−w)
=4π
∣∣∣∣Γ
(1
4+it
2
)∣∣∣∣
4t−w
∫∞
0
θw−1K0(θ)2dθ +O(
eπtt1/2−w)
=
2w−1πΓ
(w
2
)4
Γ(w)∣∣∣∣Γ
(1
4+it
2
)∣∣∣∣
4t−w +O
(
eπtt1/2−w)
.
(3.38)
This completes the proof. �
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from
Integral Mean Values of Maass L-Functions 19
References
[1] J. Beineke and D. Bump, Moments of the Riemann zeta function and Eisenstein series. II, Jour-
nal of Number Theory 105 (2004), no. 1, 175–191.
[2] A. Erdelyi, W. Magnus, F. Oberhettinger, and F. G. Tricomi, Higher Transcendental Functions.
Vols. I, II, McGraw-Hill, New York, 1953, based in part, on notes left by Harry Bateman.
[3] A. Good, The convolution method for Dirichlet series, The Selberg Trace Formula and Related
Topics (Brunswick, Maine, 1984), Contemporary Mathematics, vol. 53, American Mathematical
Society, Rhode Island, 1986, pp. 207–214.
[4] D. S. Jones, Asymptotics of the hypergeometric function, Mathematical Methods in the Applied
Sciences 24 (2001), no. 6, 369–389.
[5] M. Jutila, Mean values of Dirichlet series via Laplace transforms, Analytic Number Theory,
London Mathematical Society, Lecture Note Series, vol. 247, Cambridge University Press, Cam-
bridge, 1997, pp. 169–207.
[6] N. V. Kuznetsov, Mean value of the Hecke series of a cusp form of weight zero, Differential ge-
ometry, Lie groups and mechanics, IV, Zapiski Nauchnykh Seminarov Leningradskogo Otde-
leniya Matematicheskogo Instituta 109 (1981), 93–130, 181, 183.
[7] K. Matsumoto, Recent developments in the mean square theory of the Riemann zeta and other
zeta-functions, Number Theory, Trends Math., Birkhauser, Basel, 2000, pp. 241–286.
[8] D. Zagier,The Rankin-Selberg method for automorphic functions which are not of rapid decay,
Journal of the Faculty of Science. University of Tokyo. Section IA. Mathematics 28 (1981), no. 3,
415–437 (1982).
[9] Q. Zhang, Integral mean values of modular L-functions, Journal of Number Theory 115 (2005),
no. 1, 100–122.
Qiao Zhang: Department of Mathematics, Johns Hopkins University, Baltimore, MD 21218, USA
E-mail address: [email protected]
at Shandong University on M
arch 29, 2014http://im
rn.oxfordjournals.org/D
ownloaded from