|
§4
in
Linnik
[13]
,
we
have
that
S
(
α
) =
Γ(1
/
)
z
1
/
−
1
ρ
z
−
ρ/
Γ
ρ
−
ζ
ζ
(0)
−
1
2
πi
(
−
√
3
/
2)
ζ
ζ
(
w
)Γ(
w
)
z
−
w
d
w.
(5)
Now
we
estimate
the
integral
in
(5)
.
Writing
w
=
−
√
3
/
2
+
it
,
we
have
|
(
ζ
/ζ
)(
w
)
|
log(
|
t
|
+ 2),
z
−
w
=
|
z
|
√
3
/
2
exp(
t
arg(
z
)),
where
|
arg(
z
)
|
≤
π/
2.
Furthermore
the
Stirling
formula
implies
that
Γ(
w
)
|
t
|
−
(
√
3+1)
/
2
exp(
−
π
|
t
|
/
2).
Hence
(
−
√
3
/
2)
ζ
ζ
(
w
)Γ(
w
)
z
−
w
d
w
|
z
|
√
3
/
2
1
0
log(
t
+ 2) d
t
+
|
z
|
√
3
/
2
∞
1
log(
t
+ 2)
t
−
(
√
3+1)
/
2
exp
(arg(
z
)
−
π
2
)
t
d
t
|
z
|
√
3
/
2
+
|
z
|
√
3
/
2
∞
1
log(
t
+ 2)
t
−
(
√
3+1)
/
2
d
t
|
z
|
√
3
/
2
.
This
is
1 as
stated
since
z
1 by
(2)
.
Hence
the
lemma
is
proved.
2
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
49
We
explicitly
remark
that
Lemma 2
is
stronger
than
the
corresponding
Lemma
1
of
[9]
(or
Lemma
1
of
[7]
)
because
in
this
case
α
is
bounded.
The
second
lemma
is
an
L
2
-estimate
of
the
remainder
term
in
(4)
which
generalizes
a
result
of
Languasco
and
Perelli
[5]
;
we
will
follow
their
proof
inserting
many
details
since
the
presence
of
changes
the
shape
of
the
involved
estimates
at
several
places.
In
fact
we
will
use
Lemma 3
just
for
= 1
,
2 but
we
take
this
occasion
to
describe
the
more
general
case
since
it
may
be
useful
for
future
works.
Lemma
3.
Assume
RH.
Let
≥
1
be
an
integer
and
N
be
a
sufficiently
large
integer.
For
0
≤
ξ
≤
1
/
2
,
we
have
ξ
−
ξ
S
(
α
)
−
Γ(1
/
)
z
1
/
2
d
α
N
1
/
ξL
2
.
Proof.
Since
z
−
ρ/
=
|
z
|
−
ρ/
exp
−
i
(
ρ/
)
arctan 2
πN α
,
by
RH
and
Stirling’s
formula
we
have
that
1
ρ
z
−
ρ/
Γ
ρ
ρ
|
z
|
−
1
/
(2
)
|
γ
|
(1
−
)
/
(2
)
exp
γ
arctan 2
πN α
−
π
2
|
γ
|
.
If
γα
≤
0 or
|
α
|
≤
1
/N
we
get
ρ
z
−
ρ/
Γ(
ρ/
)
N
1
/
(2
)
,
where,
in
the
first
case,
ρ
runs
over
the
zeros
with
γα
≤
0.
Hence
I
(
N, ξ,
) :=
ξ
−
ξ
S
(
α
)
−
Γ(1
/
)
z
1
/
2
d
α
N
1
/
ξ
(6)
if
0
≤
ξ
≤
1
/N
,
and
I
(
N, ξ,
)
ξ
1
/N
γ>
0
z
−
ρ/
Γ
ρ
2
d
α
+
−
1
/N
−
ξ
γ<
0
z
−
ρ/
Γ
ρ
2
d
α
+
N
1
/
ξ
(7)
if
ξ >
1
/N
.
We
will
treat
only
the
first
integral
on
the
right
hand
side
of
(7)
,
the
second
being
completely
similar.
Clearly
ξ
1
/N
γ>
0
z
−
ρ/
Γ
ρ
2
d
α
=
K
k
=1
2
η
η
γ>
0
z
−
ρ/
Γ
ρ
2
d
α
+
O
(1)
(8)
where
η
=
η
k
=
ξ/
2
k
,
1
/N
≤
η
≤
ξ/
2 and
K
is
a
suitable
integer
satisfying
K
=
O
(
L
).
Writing
arctan 2
πN α
=
π/
2
−
arctan(1
/
2
πN α
) and
using
the
Saffari–Vaughan
technique
we
have
50
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
2
η
η
γ>
0
z
−
ρ/
Γ
ρ
2
d
α
≤
2
1
2
δη
δη/
2
γ>
0
z
−
ρ/
Γ
ρ
2
d
α
d
δ
=
γ
1
>
0
γ
2
>
0
Γ
ρ
1
Γ
ρ
2
e
π
2
(
γ
1
+
γ
2
)
·
J,
(9)
say,
where
J
=
J
(
N, η, γ
1
, γ
2
) =
2
1
2
δη
δη/
2
f
1
(
α
)
f
2
(
α
) d
α
d
δ,
w
=
1
+
i
(
γ
1
−
γ
2
)
,
f
1
(
α
) =
|
z
|
−
w
and
f
2
(
α
) = exp
−
γ
1
+
γ
2
arctan
1
2
πN α
.
Now
we
proceed
to
the
estimation
of
J
.
Integrating
twice
by
parts
and
denoting
by
F
1
a
primitive
of
f
1
and
by
G
1
a
primitive
of
F
1
,
we
get
J
=
1
2
η
G
1
(4
η
)
f
2
(4
η
)
−
G
1
(2
η
)
f
2
(2
η
)
−
2
η
G
1
(
η
)
f
2
(
η
)
−
G
1
η
2
f
2
η
2
−
2
2
1
G
1
(2
δη
)
f
2
(2
δη
)d
δ
+ 2
2
1
G
1
δη
2
f
2
δη
2
d
δ
+
2
1
2
δη
δη/
2
G
1
(
α
)
f
2
(
α
) d
α
d
δ.
(10)
If
α >
1
/N
we
have
f
2
(
α
)
1
α
γ
1
+
γ
2
N α
f
2
(
α
)
f
2
(
α
)
1
α
2
γ
1
+
γ
2
N α
+
γ
1
+
γ
2
N α
2
f
2
(
α
)
,
hence
from
we
get
J
1
η
max
α
∈
[
η/
2
,
4
η
]
|
G
1
(
α
)
|
1 +
γ
1
+
γ
2
N η
2
exp
−
c
γ
1
+
γ
2
N η
,
(11)
where
c
=
c
(
)
>
0 is
a
suitable
constant.
In
order
to
estimate
G
1
(
α
) we
use
the
substitution
u
=
u
(
α
) =
1
N
2
+ 4
π
2
α
2
1
/
2
,
(12)
thus
getting
F
1
(
α
) =
1
2
π
u
1
−
w
(
u
2
−
N
−
2
)
1
/
2
d
u.
A. Languasco, A. Zaccagnini / Journal of Number Theory 159 (2016) 45–58
51
By
partial
integration
we
have
F
1
(
α
) =
1
2
π
(2
−
w
)
u
2
−
w
(
u
2
−
N
−
2
)
1
/
2
+
u
3
−
w
(
u
2
−
N
−
2
)
3
/
2
d
u
.
(13)
From
(12)
and
(13)
we
get
G
1
(
α
) =
1
2
π
(2
−
w
)
A
(
α
) +
B
(
α
) d
α
,
(14)
where
A
(
α
) =
1
2
π
u
3
−
w
u
2
−
N
−
2
d
u
and
B
(
α
) =
u
3
−
w
(
u
2
−
N
−
2
)
3
/
2
d
u.
Again
by
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