to form $F_{x}'$ because it is then easy to ..... $T^{*}(s,z):= \sum_{\gamma}\infty=1\sum_{k=1}^{r}\frac{1}{r^{s}k^{z}(r+k)}=\frac{1}{\zeta(s+Z+1)}T(S ..... [12] Y. Matsuoka, On the values of a certain Dirichlet series at rational integers, Tokyo J. Math.
数理解析研究所講究録 958 巻 1996 年 14-22
14
Some Sums involving Farey fractions S. Kanemitsu (Univ. of Kinki, Kyushu) 金光滋
Introduction.
For any
fractions with denominators
$\leqq x$
,
$x\geqq 1$
$(vbc_{v})=1\}\gamma$ ’
$x$
$F_{x+1}’$
from
$F_{x}=F_{[x]}$
denote the sequence of all irreducible
$( \rho_{1}=\frac{1}{[x]},$
$\rho_{\Phi(X)-}1-\frac{1}{[x]}=1)$
$F_{x}’$
$p_{0}= \frac{0}{1}=\frac{b_{0}}{C_{0}}$
to
by just inserting all mediants
between them as long as
$c_{v}+c_{v+1}\leqq \mathrm{X}+1$
. E. . from
$F_{x}$
to form
$\frac{b_{v}+b_{v+1}}{C_{v}+C_{v+1}}$
$F_{x}’$
$x$
because it is then easy to
of successive terms
$F_{2}^{\mathrm{t}}= \{\frac{0}{1},\frac{1}{2},\frac{1}{1}\}$
$\mathrm{g}$
The number of terms in the Farey series of order
we form
stands for the Euler function
$\varphi(n)$
$\sum_{m\leq n}1$
$n$
By
,
, the
and is equal to the number of terms in $Q_{n}$
number of integers
$F_{x}\mathrm{w}\mathrm{h}\mathrm{o}\mathrm{S}\mathrm{e}$
$\mathrm{g}$
$,$
we denote the set of all pairs of consecutive terms in
.
$Q_{4}=\{(1,3),(3,2),(2,3),(3,1)\}$
$(\# Q_{n}=\# FX=\Phi(_{X}))$
$F_{x}’$
.
.
In [7] we considered the sums $(m\in \mathrm{N})$
$S_{m}(x):= \sum_{\chi}(cv’ C)v+1\in Q(Cvc_{v+}1)-m$
mainly in the special cases
$m=2,3$
.
,
in
$\frac{2}{3’}\frac{1}{1}\}$
.
$\leqq n$
denominator
$Q_{x}= \{(c_{v^{c_{v})|\}}},+1\frac{b_{v}}{C_{\mathcal{V}}}1$
,
Finally,
,
denotes the periodic Bemoulli polynomial of
).
Remark 1. Theorem 1 gives very precise description of asymptotic behavior of these sums. Theorem 1 refines fromer results of Mikolas, Hall, Lehner-Newman, Kanemitsu et al and provides
a direct relationship between the sums the summatory function of
$\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}^{\dagger}\mathrm{s}$
$S_{m}(n)$
involving Farey fractions and the error term $U(n)$ of
function.
Between Theorem 1 and the results of Maier there is a close connection as follows. We put $\delta_{v}=p_{\mathcal{V}}-\frac{v}{\Phi(n)}$
,
$v=0,\ldots,\Phi(n)$
16 Hence
$\delta_{\Phi(n)}=\delta_{0}=0,$
$\delta_{1}=\frac{1}{n}-\frac{v}{\Phi(n)}$
etc.
Also following Maier, we put $s_{\mathit{8}^{h}},= \Phi()-2\sum_{0v=}^{n}(\delta)vg(\delta)^{h}v+1$
Noting that
$\delta_{v+1^{-}}\delta_{v}=(c_{v}c_{v+1})-1-\Phi(n)^{1}-$
,
$v{\rm Re} u+1$
$1^{\mathrm{t}}$
.
For any
and any
$u\in \mathrm{C}$
$xarrow\infty$
,
let
$I_{u}(x):= \sum n^{u}n5x$
. Then for any
with
$\ell\in \mathrm{N}$
we have ’
$L_{u}(x)=\{$ $\frac{\mathrm{l}}{\log u+1x}x^{u+1}+\sigma(-u)+\mathit{7},$
In the special case where
$u=-u \neq-11+\sum_{=r1}^{\ell}\frac{(-1)^{r}}{r}B\Gamma(\chi)_{X}u+1-r+o(x^{{\rm Re} u^{-\ell}})$
$u\in \mathrm{N}\cup\{0\}$
,
we can take
$l=u+1$
.
without error $tem$, in which
case it is convenient to write the fomula in the $fom$ $I_{u}(X)= \frac{1}{u+1}\sum^{u+}\gamma=01(-1)^{r(\begin{array}{ll}u +\mathrm{l} r\end{array})u} \overline{B}\Gamma(X)X-r+\zeta+1(-u)$
For instance, for $k=0$ , Lemma 1’ just says
$b(x)= \chi-\overline{B}1(X)-\frac{1}{2}$
.
In the course of proof of Lemma 1, we prove the identity for any
(3)
.
$\ell\in \mathrm{N}\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}\ell>-{\rm Res}$
$\zeta(s)=A_{-}(Sl)=1+(-1)^{l+1(\begin{array}{l}-S\ell\end{array})-}\int_{1}\infty-S-\ell-\frac{1}{1-s}\overline{B_{\ell}}(t)tdt\sum_{\Gamma=0}\ell(1)\Gamma B_{r}$
$= \frac{1}{s-1}+\frac{1}{2}+\frac{1}{12}s+\cdots+(-1)\ell+1\int_{1}^{\infty}\overline{B}_{f}(t)t^{-s^{-}}\ell dt$
The Cauchy criterion shows that the the integral on the RHS represents an analytic function for ${\rm Res}>-l$
Since
,
so that for
$\ell\in \mathrm{N}$
${\rm Res}>-x,$
$\zeta(S)$
is analytic with the excception of a simple pole at
is arbitrary, this signifies that
the excception of a simple pole at
$s=1$
$\zeta(s)$
$s=1$
.
is extended analytically over the whole plane with
. We can, however, proceed still further to give a
conceptually the simplest proof of the functional equation of the Riemann zeta-fucntion. Actually,
although in a well-known proof one considers the case
$P=1$
of Formula (3) to use the Lebesgue
dominated convergence theorem on the grounds that the partial sums of the Fourier series of
$\overline{B_{1}}(t)$
are bounded, we can prove the following corollary by just employing absolutely convergent Fourier
19 series in the case
of Formula (3) without resorting such a rather high-brow theorem
$\ell=2$
Corollary.
The Riemann zeta-fi4nction satisfies thefunctional equation $\zeta(s)=2(2\pi)s-1\mathrm{r}(1-S)\sin\frac{\pi}{2}S\sigma(1-s)$
The defining Dirichlet series
gives an analytic continuation of
Proof.
In the case
$l=2$
$\zeta(s)$
being absolutely convergent for
$\mathrm{f}\mathrm{o}\mathrm{r}\zeta(1-s)$
into
.
${\rm Res}