Some simplest integral equalities equivalent to the Riemann hypothesis
UDC 511.3 We show that the following integral equalities are equivalent to the Riemann hypothesis for any real $a>0$ and any real $0<\epsilon<1,$ $\epsilon \neq 1$:\begin{gather*}\int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\rig...
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| author | Sekatskii, S. K. Beltraminelli, S. Sekatskii, S. K. Beltraminelli, S. |
| author_facet | Sekatskii, S. K. Beltraminelli, S. Sekatskii, S. K. Beltraminelli, S. |
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UDC 511.3
We show that the following integral equalities are equivalent to the Riemann hypothesis for any real $a>0$ and any real $0<\epsilon<1,$ $\epsilon \neq 1$:\begin{gather*}\int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{a+it}\,dt=-2\pi\ln\frac{a+\dfrac{1}{2}}{a}, \\ \int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{(a+it)^\epsilon}\,dt=-\frac{2\pi}{1-\epsilon}\left(\left(a+\frac{1}{2}\right)^{1-\epsilon}-a^{1-\epsilon}\right).\end{gather*} |
| doi_str_mv | 10.37863/umzh.v74i9.6222 |
| first_indexed | 2026-03-24T03:26:34Z |
| format | Article |
| fulltext |
DOI: 10.37863/umzh.v74i9.6222
UDC 511.3
S. K. Sekatskii1 (Laboratory of Biological Electron Microscopy, IPHYS, Ecole Polytechnique Fédérale de Lausanne,
Switzerland),
S. Beltraminelli (CERFIM, Research Center for Math. and Phys., Locarno, Switzerland)
SOME SIMPLEST INTEGRAL EQUALITIES EQUIVALENT
TO THE RIEMANN HYPOTHESIS
ДЕЯКI НАЙПРОСТIШI IНТЕГРАЛЬНI РIВНОСТI,
ЩО ЕКВIВАЛЕНТНI ГIПОТЕЗI РIМАНА
We show that the following integral equalities are equivalent to the Riemann hypothesis for any real a > 0 and any real
0 < \epsilon < 1, \epsilon \not = 1:
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a+ it
dt = - 2\pi \mathrm{l}\mathrm{n}
a+
1
2
a
,
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt = - 2\pi
1 - \epsilon
\Biggl( \biggl(
a+
1
2
\biggr) 1 - \epsilon
- a1 - \epsilon
\Biggr)
.
Показано, що iнтегральнi рiвностi
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a+ it
dt = - 2\pi \mathrm{l}\mathrm{n}
a+
1
2
a
,
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt = - 2\pi
1 - \epsilon
\Biggl( \biggl(
a+
1
2
\biggr) 1 - \epsilon
- a1 - \epsilon
\Biggr)
еквiвалентнi гiпотезi Рiмана для довiльного дiйсного a > 0 i довiльного дiйсного 0 < \epsilon < 1, \epsilon \not = 1.
1. Introduction. In recent papers [1, 2] (see also reprints [3]), we, together with Merlini, have
established and proved an infinite number of integral equalities equivalent to the Riemann hypothesis
(RH; see, e.g., [4] for the general discussion of the Riemann \zeta -function). In particular, we have
shown that Balazard – Saias – Yor equality [5] and Volchkov equality [6] are certain particular cases
of our general approach.
The aim of this short note is to establish and briefly discuss what we believe possibly are the
simplest integral equalities of this type.
2. Integral equalities. Let us introduce the function g(z) =
i\biggl(
a+
\biggl(
z - 1
2
\biggr) \biggr) \epsilon , where a
is real positive, i =
\surd
- 1 and 1 \geq \epsilon > 0. We use our “standard” [1 – 3] contour C com-
1 Corresponding author, e-mail: serguei.sekatski@epfl.ch.
c\bigcirc S. K. SEKATSKII, S. BELTRAMINELLI, 2022
1256 ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
SOME SIMPLEST INTEGRAL EQUALITIES EQUIVALENT TO THE RIEMANN HYPOTHESIS 1257
Fig. 1. The contour C used for the calculation of the integrals. The cut made to exclude the simple pole of the
Riemann \zeta -function is shown, see [1 – 3] for details.
posed by the segments of straight lines
\biggl[
1
2
- iT,
1
2
+ iT
\biggr]
,
\biggl[
1
2
+ iT, T + iT
\biggr]
, [T + iT, T - iT ]
and
\biggl[
T - iT,
1
2
- iT
\biggr]
, where T is a real positive number larger than 1 (see Fig. 1), and consider a
contour integral
\int
C
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz. Consideration of similar integrals with \epsilon > 1 is given in [1].
Lemma. Assume RH. Then, for T \rightarrow +\infty ,
\int
C
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz =
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt = - 2\pi
1\int
1
2
dx\biggl(
a+
\biggl(
x - 1
2
\biggr) \biggr) \epsilon .
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
1258 S. K. SEKATSKII, S. BELTRAMINELLI
Proof. To calculate the contour integral at question we use the generalized Littlewood theorem
concerning contour integrals of the logarithm of an analytical function [1 – 3]. This theorem expresses
the contour integral value via the sum over residues given by the poles of the function g(z) lying
inside the contour, and the sum of the integrals
\int \sigma +it
1
2
+it
g(z)dz, where \sigma + it, \sigma >
1
2
, is a zero or the
pole (these contributions have different signs) of the Riemann \zeta -function; see [1 – 3] for details.
Inside the contour we have no poles of the function g(z) and, due to the fact that we assume RH,
also no zeros of the Riemann \zeta -function. Thus for any T we obtain
\int
C
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz = - 2\pi
1\int
1
2
dx\biggl(
a+
\biggl(
x - 1
2
\biggr) \biggr) \epsilon , (1)
where, of course, the integral in the right-hand side is the contribution of the simple pole of the
\zeta -function at z = 1.
From the other side, the contour integral at question is equal to
\int T
- T
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt plus
corresponding integrals taken over three remaining sides of the quadratic contour C. To finish the
proof of the lemma we need to show that the latter tend to zero. First we consider an integral\int T+iT
1
2
+iT
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz (consideration for
\int T - iT
1
2
- iT
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz is quite similar) and divide the
integration area into the segments
\biggl[
1
2
+ iT,N + iT
\biggr]
and [N + iT, T + iT ], where N \ll T is some
positive number depending on T (we can take it as large as we please; we will select its value later
on).
We use a “standard” definition of \mathrm{a}\mathrm{r}\mathrm{g}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
, that is, that where the argument starts
from zero value at z = b > 1, t = 0 (\mathrm{l}\mathrm{n}(\zeta (b)) > 0 here) and then is obtained by the continued
variation along the segments of straight lines [b, b+ it],
\biggl[
b+ it,
1
2
+ it
\biggr]
. With this definition, for the
segment
\biggl[
1
2
+ iT,N + iT
\biggr]
, we have | \mathrm{a}\mathrm{r}\mathrm{g}(\zeta (z))| = \scrO (\mathrm{l}\mathrm{n}T ) ([4], this is Backlund theorem [7]) and
| \mathrm{l}\mathrm{n}(\zeta (z))| = \scrO (\mathrm{l}\mathrm{n}T ) [4]. Hence\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
N+iT\int
1
2
+iT
g(z) \mathrm{l}\mathrm{n}(\zeta (z)) dz
\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| = \scrO
\bigl(
NT - \epsilon \mathrm{l}\mathrm{n}T
\bigr)
.
For the point x+ iT with sufficiently large x belonging to the segment [N + iT, T + iT ], we have
| \mathrm{l}\mathrm{n}(\zeta (z))| = \scrO
\bigl(
2 - x
\bigr)
, which follows trivially from the definition of the Riemann \zeta -function, hence
T+iT\int
x+iT
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz = \scrO
\bigl(
2 - NT - \epsilon \mathrm{l}\mathrm{n}T
\bigr)
.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
SOME SIMPLEST INTEGRAL EQUALITIES EQUIVALENT TO THE RIEMANN HYPOTHESIS 1259
Thus it is enough to select, say, N = T
\epsilon
2 to see the disappearance of these two integrals in the limit
of large T. Due to the same | \mathrm{l}\mathrm{n}(\zeta (z))| = \scrO
\bigl(
2 - \Re (z)
\bigr)
, the disappearance of the integral taken along
the line [T + iT, T - iT ] in the limit of large T is evident, which finishes the proof of the lemma.
Thus on RH the integral
\int \infty
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt for 1 \geq \epsilon > 0 do exists. This should not
be surprising because “similar” integrals are known, such as, e.g., the following examples from
Gradshtein and Ryzhik book [8]. Example 4.421.1 gives
\int \infty
0
\mathrm{l}\mathrm{n}x \mathrm{s}\mathrm{i}\mathrm{n}(ax)
x
dx = - \pi
2
(\gamma + \mathrm{l}\mathrm{n} a);
more generally we have Example 4.422.1: for | \Re (\mu )| < 1, a > 0,
\infty \int
0
\mathrm{l}\mathrm{n}x \mathrm{s}\mathrm{i}\mathrm{n}(ax)
x1 - \mu
dx = - \Gamma (\mu )
a\mu
\mathrm{s}\mathrm{i}\mathrm{n}
\mu \pi
2
\Bigl(
\psi (\mu ) - \mathrm{l}\mathrm{n} a+
\pi
2
\mathrm{c}\mathrm{o}\mathrm{s}
\mu \pi
2
\Bigr)
,
and Example 4.422.2: for 0 < \Re (\mu ) < 1, a > 0,
\infty \int
0
\mathrm{l}\mathrm{n}x \mathrm{c}\mathrm{o}\mathrm{s}(ax)
x1 - \mu
dx = - \Gamma (\mu )
a\mu
\mathrm{c}\mathrm{o}\mathrm{s}
\mu \pi
2
\Bigl(
\psi (\mu ) - \mathrm{l}\mathrm{n} a - \pi
2
\mathrm{t}\mathrm{a}\mathrm{n}
\mu \pi
2
\Bigr)
.
Here \gamma is Euler – Mascheroni constant and \Gamma , \psi are gamma- and digamma-functions, respectively.
Our next aim is the proof of the following theorem.
Theorem 1. For any real a > 0 the integral equality
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a+ it
dt = - 2\pi \mathrm{l}\mathrm{n}
a+
1
2
a
(2)
is equivalent to the Riemann hypothesis.
Proof. Applying Eq. (1) for \epsilon = 1, we see that on RH Eq. (2) takes place. If RH fails and a
non-trivial Riemann zero \sigma k + itk with \sigma k >
1
2
exists, it contributes
- 2\pi i
\delta k\int
0
i
a+ x+ itk
dx = 2\pi \mathrm{l}\mathrm{n}
a+ \delta k + itk
a+ itk
,
where \delta k = \sigma k -
1
2
> 0, to the integral value. Pairing the contributions of zeros with \sigma k \pm itk, we
have the total contribution of 2\pi \mathrm{l}\mathrm{n}
(a+ \delta k)
2 + t2k
a2 + t2k
, always a purely real and strictly positive number.
Theorem 1 is proved.
For large tk we see that the contribution of non-trivial zeros \sigma k \pm itk is equal to
2\pi \mathrm{l}\mathrm{n}
(a+ \delta k)
2 + t2k
a2 + t2k
= \scrO
\biggl(
1
t2k
\biggr)
.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
1260 S. K. SEKATSKII, S. BELTRAMINELLI
The known counting function of the Riemann zeta-function zeros \scrO (T \mathrm{l}\mathrm{n}T ) [4] thus guaranties that
this sum converges, so that the integral
\int \infty
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a+ it
dt exists irrespective on the validity
of RH.
Similarly, we have the following theorem.
Theorem 2. For any real a > 0 and any 0 < \epsilon < 1, the integral equality
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt = - 2\pi
1 - \epsilon
\Biggl( \biggl(
a+
1
2
\biggr) 1 - \epsilon
- a1 - \epsilon
\Biggr)
(3)
is equivalent to the Riemann hypothesis.
Proof. On RH, Eq. (3) is an immediate consequence of Eq. (1). If there is a non-trivial Riemann
zero
1
2
+ \delta k + itk, we take tk > 0 here, it contributes - 2\pi i
\int \delta k
0
i
(a+ x+ itk)\epsilon
dx to the integral
value. Two complex conjugate zeros contribute
2\pi
\delta k\int
0
(a+ x+ itk)
\epsilon + (a+ x - itk)
\epsilon \bigl(
(a+ x)2 + t2k
\bigr) \epsilon dx = 2\pi
\delta k\int
0
2 \mathrm{c}\mathrm{o}\mathrm{s}(\epsilon \phi k)\bigl(
(a+ x)2 + t2k
\bigr) \epsilon
2
dx, (4)
where \phi k = \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
tk
a+ x
. The expression under the integral sign in the right-hand side here is
certainly positive if | \epsilon \phi k| <
\pi
2
so that if
\bigm| \bigm| \bigm| \bigm| \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n} tk
a+ x
\bigm| \bigm| \bigm| \bigm| < \pi
2\epsilon
. This is always true because 0 <
< \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
tk
a+ x
<
\pi
2
and 0 < \epsilon < 1.
Using \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
tk
a+ x
=
\pi
2
- \mathrm{a}\mathrm{r}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}
a+ x
tk
we see that for large tk asymptotically \mathrm{c}\mathrm{o}\mathrm{s}(\epsilon \phi k) \sim =
\sim = \mathrm{c}\mathrm{o}\mathrm{s}
\epsilon \pi
2
= \scrO (1). Hence, as it follows from Eq. (4), asymptotically each pair of complex conjugate
non-trivial Riemann zeros contributes \scrO
\bigl(
\delta kt
- \epsilon
k
\bigr)
. Thus the existence of the integral
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt
depends on the unknown density of the non-trivial Riemann zeros. If, for example, the density of
non-trivial Riemann zeros with \delta \geq \delta 0 >
1
2
in a limit of large T is constant, the sum over such
zeros diverges and the integral does not exist. Of course, this makes the (not easy due to the slow
convergence) numerical study of the integral
\int N
- N
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt with 0 < \epsilon < 1 for large N
especially interesting from the viewpoint of the RH testing.
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
SOME SIMPLEST INTEGRAL EQUALITIES EQUIVALENT TO THE RIEMANN HYPOTHESIS 1261
3. Discussion and numerical testing. For the rest of this section, we assume RH. Then, for
a =
1
2
, we have
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
1
2
+ it
dt = - 2\pi \mathrm{l}\mathrm{n} 2 = - 4.35517 . . . . (5)
This equation has been tested numerically with Mathematica. We used the built-in function NInte-
grate. After some testing we found out that a good trade-off between precision and computational
time was to choose GlobalAdaptive as the integration strategy for NIntegrate (in general it has better
performance than local adaptive strategy) and to set the value of the parameter MaxRecursion to 100
and that of MaxErrorIncrease to 10\prime 000. In addition, we explicitly specified the singularities (i.e.,
where \zeta (s) = 0) in the range of integration. The convergence of an integral is quite slow (as it
should be following the proof of Lemma 1), it takes about 4 and a half hours to calculate the integral
on a standard desktop PC but for
\int N
- N
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
1
2
+ it
dt with N = 9878 (i.e., integrating over the
region including the first 10\prime 000 non-trivial Riemann zeros), we succeed to obtain the “reasonable”
value - 4.35518 . . . .
The value of the integral (3) for \epsilon =
1
2
, a =
1
2
, that is,
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
\sqrt{}
1
2
+ it
dt = - 4\pi
\Biggl(
1 -
\surd
2
2
\Biggr)
= - 3.6806 . . . (6)
also has been tested numerically with the same limits obtaining - 3.683 . . . .
Now we want to make the following remarks. First, nothing prevents us to take a = 0 in Eq. (3)
obtaining
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(it)\epsilon
dt = - 2\epsilon \pi
1 - \epsilon
. (7)
In particular,
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
\surd
it
dt = - 2
\surd
2\pi . (8)
Of course, integrals (7), (8) are to be understood as Cauchy principal values.
Next, if in the same contour integral
\int
C
g(z) \mathrm{l}\mathrm{n}(\zeta (z))dz as above we use g(z) =
i
- a+
\biggl(
z - 1
2
\biggr) ,
we get a simple pole of the function g(z) with the residue i at z = a +
1
2
inside the contour. This
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
1262 S. K. SEKATSKII, S. BELTRAMINELLI
gives the contribution - 2\pi \mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
a+
1
2
\biggr) \biggr)
which should be summed up with the contribution of
the simple pole of \zeta (z), - 2\pi \mathrm{l}\mathrm{n}
a - 1
2
a
, hence
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
- a+ it
dt = - 2\pi \mathrm{l}\mathrm{n}
\left( \zeta \biggl( 1
2
+ a
\biggr) a - 1
2
a
\right) .
Taking a =
1
2
+ \delta , \delta \rightarrow 0, we have - 2\pi \mathrm{l}\mathrm{n} \delta - 2\pi \mathrm{l}\mathrm{n} 2 for the contribution of a simple pole of g(z)
and 2\pi \mathrm{l}\mathrm{n} \delta +\scrO (\delta ) for the contribution of a simple pole of \zeta (z), that is, a total of - 2\pi \mathrm{l}\mathrm{n} 2 +\scrO (\delta ).
Thus we get
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
- 1
2
+ it
dt = - 2\pi \mathrm{l}\mathrm{n} 2. (9)
Certainly,
1
1
4
+ t2
=
1
1
2
+ it
- 1
- 1
2
+ it
, thus combining (5) and (9) we obtain again the
Balazard – Saias – Yor theorem stating that RH is equivalent to the integral equality [5]
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
1
4
+ t2
dt = 0.
Similarly, combining the values of the integrals
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a+ it
dt and
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
- a+ it
dt
given above for a \not = 1
2
, we restore the Theorem 3 from [1]: the equality
a
\pi
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
a2 + t2
dt = \mathrm{l}\mathrm{n}
\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
\zeta
\biggl(
a+
1
2
\biggr) \biggl(
a - 1
2
\biggr)
a+
1
2
\bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm| \bigm|
is equivalent to the Riemann hypothesis. Numerous examples of this equality were tested numerically.
This is worthwhile to note that if \epsilon \rightarrow 0, expression (3) has a finite limit (of course) independent
on a, which we formulate as the following proposition.
Proposition. Assume RH. Then for any real positive a
\mathrm{l}\mathrm{i}\mathrm{m}
\epsilon \rightarrow 0
\infty \int
- \infty
\mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
(a+ it)\epsilon
dt = - \pi .
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
SOME SIMPLEST INTEGRAL EQUALITIES EQUIVALENT TO THE RIEMANN HYPOTHESIS 1263
Thus, in a certain sense, on RH the function \mathrm{l}\mathrm{n}
\biggl(
\zeta
\biggl(
1
2
+ it
\biggr) \biggr)
is indeed “almost periodic”.
References
1. S. K. Sekatskii, S. Beltraminelli, D. Merlini, On equalities involving integrals of the logarithm of the Riemann
\zeta -function and equivalent to the Riemann hypothesis, Ukr. Math. J., 64, № 2, 218 – 228 (2012).
2. S. K. Sekatskii, S. Beltraminelli, D. Merlini, On equalities involving integrals of the logarithm of the Riemann \zeta -
function with exponential weight which are equivalent to the Riemann hypothesis, Int. J. Anal., 2015, Article 980728
(2015).
3. S. K. Sekatskii, S. Beltraminelli, D. Merlini, A few equalities involving integrals of the logarithm of the Riemann
\zeta -function and equivalent to the Riemann hypothesis I, II, III; arXiv: 0806.1596v1, 0904.1277v1, 1006.0323v1.
4. E. C. Titchmarsh, E. R. Heath-Brown, The theory of the Riemann Zeta-function, Clarendon Press, Oxford (1988).
5. M. Balazard, E. Saias, M. Yor, Notes sur la fonction de Riemann, Adv. Math., 143, 284 – 287 (1999).
6. V. V. Volchkov, On an equality equivalent to the Riemann hypothesis, Ukr. Math. J., 47, № 4, 491 – 493 (1995).
7. R. J. Backlund, Über die Nullstellen der Riemannschen Zetafunktion, Acta Math., 41, 345 – 375 (1918).
8. I. S. Gradshtein, I. M. Ryzhik, Tables of integrals, series and products, Acad. Press, New York (1990).
Received 10.07.20
ISSN 1027-3190. Укр. мат. журн., 2022, т. 74, № 9
|
| id | umjimathkievua-article-6222 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T03:26:34Z |
| publishDate | 2022 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/5b/219affd240096245387b7bc71d59855b.pdf |
| spelling | umjimathkievua-article-62222023-01-07T13:45:33Z Some simplest integral equalities equivalent to the Riemann hypothesis Some simplest integral equalities equivalent to the Riemann hypothesis Sekatskii, S. K. Beltraminelli, S. Sekatskii, S. K. Beltraminelli, S. Riemann hypothesis Integral equality analytical number theory UDC 511.3 We show that the following integral equalities are equivalent to the Riemann hypothesis for any real $a&gt;0$ and any real $0&lt;\epsilon&lt;1,$ $\epsilon \neq 1$:\begin{gather*}\int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{a+it}\,dt=-2\pi\ln\frac{a+\dfrac{1}{2}}{a}, \\ \int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{(a+it)^\epsilon}\,dt=-\frac{2\pi}{1-\epsilon}\left(\left(a+\frac{1}{2}\right)^{1-\epsilon}-a^{1-\epsilon}\right).\end{gather*} УДК 511.3 Деякі найпростіші інтегральні рівності, що еквівалентні гіпотезі Рімана Показано, що інтегральні рівності \begin{gather*}\int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{a+it}\,dt=-2\pi\ln\frac{a+\dfrac{1}{2}}{a}, \\\int\limits_{-\infty}^{\infty}\frac{\ln\left(\zeta\left(\dfrac{1}{2}+it\right)\right)}{(a+it)^\epsilon}\,dt=-\frac{2\pi}{1-\epsilon}\left(\left(a+\frac{1}{2}\right)^{1-\epsilon}-a^{1-\epsilon}\right).\end{gather*} еквівалентні гіпотезі Рімана для довільного дійсного $a&gt;0$ і довільного дійсного $0&lt;\epsilon&lt;1,$ $\epsilon \neq 1$. Institute of Mathematics, NAS of Ukraine 2022-11-08 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/6222 10.37863/umzh.v74i9.6222 Ukrains’kyi Matematychnyi Zhurnal; Vol. 74 No. 9 (2022); 1256 - 1263 Український математичний журнал; Том 74 № 9 (2022); 1256 - 1263 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/6222/9300 Copyright (c) 2022 Sergey Sekatskii, Stefano Beltraminelli |
| spellingShingle | Sekatskii, S. K. Beltraminelli, S. Sekatskii, S. K. Beltraminelli, S. Some simplest integral equalities equivalent to the Riemann hypothesis |
| title | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_alt | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_full | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_fullStr | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_full_unstemmed | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_short | Some simplest integral equalities equivalent to the Riemann hypothesis |
| title_sort | some simplest integral equalities equivalent to the riemann hypothesis |
| topic_facet | Riemann hypothesis Integral equality analytical number theory |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/6222 |
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