数理解析研究所講究録
1091 巻 1999 年 189-198

189

RAPIDLY CONVERGENT SERIES REPRESENTATIONS

FOR $\zeta(2n+1)$ AND THEIR$\chi$-ANALOGUE

MASANORI KATSURADA
Department of Mathematics and Computer Science, Kagoshima University

1.

$\mathrm{I}\mathrm{N}\mathrm{T}\dot{\mathrm{R}}$ODUCTION

Let $s=\sigma+it$ be a complex variable. The Riemann zeta-function$\zeta(s)=\sum_{m=1}m^{-S}\infty$$({\rm Re} s=\sigma>1)$$\zeta(s)$is defined by,and its meromorphic continuation over the whole -plane, whose only singularity is a simple
pole with the residue one.
$s$

For the specific values of(1.1)$\zeta(s)$at positive even integers, the formula$\zeta(2n)=(-1)^{n-}1\frac{(2\pi)^{2n}}{2(2n)!}B_{2}n$$(n=1,2,3, \ldots)$ ,due to Euler, is classically known. Here $B_{n}(n\geq 0)$ is the Bernoulli number defined by
the Taylor series expansion$\frac{z}{e^{z}-1}=\sum_{n=0}^{\infty}\frac{B_{n}}{n!}z^{n}$$(|z|<2\pi)$Closed form evaluations like (1.1), however, for the values ofhave been unknown up to the present time.

.
$\zeta(s)$at positive odd integers

It is the purpose of this paper to study rapidly convergent series representations for thevalues ofat positive odd integers. We shall prove certain transformation formulae for$\zeta(s)$1991 Mathematics Subject Classification. Primary llM06; Secondary llM35.
Key words and phrases. Riemann zeta-function, Dirichlet -function, Mellin-Barnes integral, series$L$

representation.

The author was supported in part by Grant-in-Aid for Scientific Research (No. 90224485), the Ministryof Education, Science, Sports and Culture in Japan.
Typeset by$A_{\Lambda 1^{S_{-}\mathrm{q}}\mathrm{r}}\mathrm{x}$190

MASANORI KATSURADA

at positive even integers in their coefficientsthe power series including the values ofcase of each of these formulae implies
particularA
below).
given
2
1
and
(see Theorems
atthe previously known rapidly convergent series representations for the values of-analogueofA
recentlyfound.)
otherispositive odd integers. (One is classic, and theour transformation formulae will also be given in Theorem 3.
$\zeta(s)$$\zeta(s)$$\chi$

It was found by Euler in 1772 (see Ayoub
infinite series representation(1.2)$\lceil \mathrm{A}\mathrm{y}$, p.1080, Section 7]) that$\zeta(3)=\frac{1}{7}\pi\{21-4\sum_{k=1}\frac{\zeta(\mathit{2}k)}{(\mathit{2}k+1)(\mathit{2}k+2)2^{2k}}\infty\}$$\zeta(3)$has an.

This formula was rediscovered by Ramaswami [Ra] and (more recently) by Ewell [Ewl]. In
fact, Euler’s formula (1.2) was reproduced by Srivastava [Srl, p.7, Equation (2.23)] fromthe work of Ramaswami [Ra]. Inspired by Ewell’s rediscovery of (1.2), and his subsequent
result [Ew2], Yue and Williams [YW] established a generalization of (1.2), which gives,though complicated, an exact series representation for $\zeta(2n+1)$ with any nonnegative
integer . The formula of Yue and Williams was considerably simplified by Cvijovi\v{c} andKlinowski [CK, Theorem ], who proved that$n$$\mathrm{A}$(1.3)$\zeta(\mathit{2}n+1)=(-1)^{n}\frac{(\mathit{2}\pi)^{2n}}{n(2^{2n+1}-1)}\{^{n}\sum_{k=1}^{-1}(-1)^{k-1_{\frac{k\zeta(2k+1)}{(\mathit{2}n-\mathit{2}k)!\pi^{2k}}}}$$+ \sum_{k=0}^{\infty}\frac{(\mathit{2}k)!\zeta(\mathit{2}k)}{(2n+2k)!2^{2k}}\}$

for any positive integer , where the finite sum on the right-hand side is to be regarded as.
null if $n=1$ . Since $\zeta(0)=-1/2$ , we see that (1.3) reduces to (1.2)
Srivastava [Sr2] recently found the existence of? families of rapidly convergent
and Klinowski’s formula (1.3) belongs to one.
series representations forof these families, while another family includes classical Wilton’s [Wi] formula$n$$\mathrm{W}\}_{1\mathrm{e}\mathrm{n}n}=1$$\mathrm{c}\mathrm{e}\mathrm{r}\mathrm{t}$$\zeta(\mathit{2}n+1)$(1.4)$\mathrm{C}\mathrm{v}\mathrm{i}\mathrm{j}_{0}\mathrm{v}\mathrm{i}_{\acute{\mathrm{C}}}$$\zeta(2n+1)=(-1)^{n}-1\pi^{2n\{}\frac{1}{(\mathit{2}n+1)!}(\sum_{m=1}^{+1}\frac{1}{m}-\log\pi 2n)$$+ \sum_{k=1}^{1}(-1)^{k}\frac{\zeta(2k+1)}{(2n-\mathit{2}k+1)!\pi^{2k}}n-+\mathit{2}\sum_{k=1}^{\infty}\frac{(2k-1)!\zeta(\mathit{2}k)}{(2n+2k+1)!\mathit{2}^{2k}}\}$.

appearing in [Sr2],
From the observation of various series representations forthat haveformulae
of
the
(1.3)
is
one
we may say that Cvijovi\v{c} and Klinowski’s formula
the simplest figure among these families. It is in fact possible to show that (1.3) is aparticular case of a general transformation formula, that is$\zeta(\mathit{2}n+1)$

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RAPIDLY CONVERGENT SERIES REPRESENTATIONS FOR $\zeta(2n+1)$

Theorem 1. Let
(1.5)

$n$

be a positive integer, and$x$a real$var\dot{i}able$with $|x|\leq 1$ . Then we have$n \zeta(2n+1)-n\sum_{1\iota=}^{\infty}\frac{\cos(\mathit{2}\pi l_{X})}{l^{2n+1}}-\pi X\sum_{=\mathrm{t}1}^{\infty}\frac{\sin(2\pi l_{X})}{t^{2n}}$$=(-1)^{n}(2 \pi x)^{2}n\{_{k=}^{n-}\sum^{1}(-1)^{k-1_{\frac{k\zeta(\mathit{2}k+1)}{(\mathit{2}n-2k)!(\mathit{2}\pi X)2k}}}1$$+ \sum_{k=0}^{\infty}\frac{(\mathit{2}k)!\zeta(2k)}{(\mathit{2}n+2k)!}X2k\}$.

Remark. Since$\sum_{l=1}^{\infty}\frac{(-1)^{l}}{l^{2n+1}}=(2^{-}2n-1)\zeta(2n+1)$,

we see that the case $x=1/2$ of Theorem 1 implies (1.3).
For the proof of Theorem 1 we treat the infinite sum on the right-hand side of (1.5),
based on Mellin transform technique (see (2.1) and (2.2) below). This technique has theadvantage of heuristic treatments, particularly for the infinite sums of the type mentioned
above. Studies on certain power series and asymptotic series associated with the Riemannzeta and allied zeta-functions, based on this technique, were recently made by the author(see [Kal] [Ka2] [Ka3]). The same technique also yields here another transformation
formula, which includes Wilton’s formula (1.4) as a particular case.
Theorem 2. Let
(1.6)

$n$

be a positive integer, and$x$a real variable with $|x|\leq 1$ . Then we have$\zeta(2n+1)+\frac{1}{\mathit{2}\pi x}\sum_{=l1}^{\infty}\frac{\sin(\mathit{2}\pi lx)}{l^{2n+2}}$$=(-1)^{n-1}(2 \pi X)^{2}n\{\frac{1}{(2n+1)!}(^{21}m\sum\frac{1}{m}-\mathrm{l}\mathrm{o}n+=12\mathrm{g}\pi X)$$+ \sum_{k=1}^{n-}(-11)^{k_{\frac{\zeta(2k+1)}{(\mathit{2}n-2k+1)!(2\pi X)2k}+\sum\frac{(2k-1)!\zeta(2k)}{(\mathit{2}n+2k+1)!}x^{2}}}2k=\infty 1k\}$.

Remark. The formula which has a similar nature to (1.6) was proved in a quite different
way by Ewell [Ew3, Throrem 1]. His formula yields a determinantial expression of $\zeta(2n+1)$ ,from which he derived exact infinite series representaions forwith $n=1,2$ and3.
$\zeta(\mathit{2}n+1)$

Furthermore, the proof of Theorem 1 suggests that a -analogue of (1.5) exists. Let
be a positive integer, a Dirichlet character of modulus . We denote by $L(s, x)$ theDirichet -function attached to , andGauss’ sum defined by$\chi$$q$$\chi$$L$$q$$\chi$$\tau(\chi)$$\tau(\chi)=\sum\chi(a=1qa)e2\pi ia/q$

.

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MASANORI KATSURADA

Theorem 3. Let be a positive integer, and a real variable with $|x|\leq 1$ . For any
primitive character of modulus , we have the following formulae.
(i) If is an even character $(\dot{i}.e., \chi(-1)=1)$ ,
$x$

$n$

$q$

$\chi$$\chi$(1.7)$nL(2n+1, x)-n \sum\frac{\chi(l)\cos(\mathit{2}\pi lX/q)}{l^{2n+1}}-\iota=\infty 1\pi x\sum^{\infty}\frac{\chi(l)\sin(2\pi lx/q)}{l^{2n}}l=1$$=(-1)^{n}( \frac{2\pi x}{q})^{2n}\{_{k}^{n-}\sum^{1}(-1)^{k-1_{\frac{kL(2k+1,\chi)}{(2n-2k)!(2\pi x/q)2k}}}=1$$+ \frac{\tau(\chi)}{q}\sum_{k=0}^{\infty}\frac{(2k)!L(2k,\overline{\chi})}{(\mathit{2}n+\mathit{2}k)!}x^{2}k\}$(ii) If(1.8)$\chi$;

is an odd character $(\dot{i}.e., \chi(-1)=-1)$ ,$L(2n, \chi)-\sum\frac{\chi(l)\cos(\mathit{2}\pi lX/q)}{l^{2n}}l=\infty 1$$=(-1)^{n}( \frac{2\pi x}{q})2n-1\{^{n}\sum_{k=1}^{-}(-11)^{k_{\frac{L(\mathit{2}k,x)}{(2n-2k)!(2\pi X/q)2k-1}}}$$+ \mathit{2}i\frac{\tau(\chi)}{q}\sum_{=k0}^{\infty}\frac{(2k)!L(2k+1,\overline{\chi})}{(2n+2k)!}X^{2}k+1\}$.

Remark. The shape of the left-hand side of (1.8) shows that this formula is rather aanalogue of (1.6).
The author would like to express his sincere gratitude to Professor H. M. Srivastava forkindly sending the paper [Sr2].
We shall prove Theorem 1 in the next section. Theorem 2 will be shown in Section 3.
The last section will be devoted to the proof of Theorem 3.
$\chi-$

2. PROOF OF THEOREM 1
Let

$n$

(2.1)

be a fixed positive integer,$x$a real variable, and set$I(x)= \frac{1}{4_{\dot{i}}}\int_{(\sigma_{0})}\cot(\frac{1}{2}\pi s)\zeta(s)\frac{x^{s}}{(_{S}+1)(_{S}+\mathit{2})\cdots(_{S+}\mathit{2}n)}ds$$(|x|\leq 1)$,

denotes the vertical straightis a constant fixed with $-1/2<\sigma_{0}<0$ , and
where. The integral in (2.1) converges absolutely, because the ordertoline from, when $tarrow\pm\infty$ , with an arbitrary smallof the integrand is bouded as
for $\sigma<0$ (cf. Titchmarsh [Ti, p.95,, by the vertical estimate
Chapter V]).
$(\sigma_{0})$$\sigma_{0}$$\sigma_{0}-i\infty$$\sigma_{0}+\dot{i}\infty$$O(|t| \frac{1}{2}-\sigma 0-2n+\epsilon)$$\epsilon>0$$\zeta(s)=O(|t|\frac{1}{2}-\sigma+\epsilon)$

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RAPIDLY$\mathrm{c}\dot{\mathrm{o}}\mathrm{N}\mathrm{V}\mathrm{E}\mathrm{R}\mathrm{G}\mathrm{E}\mathrm{N}\mathrm{T}$

SERIESATIONS FOR $\zeta(2n+1)$$\mathrm{R}\mathrm{E}\mathrm{P}\mathrm{R}\mathrm{E}\mathrm{S}\mathrm{E}\mathrm{N}\dot{\mathrm{T}}$We start the proof of Theorem 1 with the observation that(22)$(|x|\leq 1)$$I(x)=- \sum\frac{\zeta(2k)}{(\mathit{2}k+1)(\mathit{2}k+\mathit{2})\cdots(2k+\mathit{2}n)}xk=\infty 02k$

This can be shown by moving the pathof the integral in (2.1) to the right, andcollecting the residues of the poles at $s=2k(k=0,1,2, \ldots)$ , because the order of theintegrand is $O\{(K+|t|)-2n-1|x|^{K}\}$ , as
, on the line $\sigma=\mathit{2}K+1(K=1,2, \ldots)$ .
$(\sigma_{0})$$tarrow\pm\infty$

We next transform the integral in (2.1) by applying the functional equation(2.3)$\zeta(1-S)=\mathit{2}^{1-s}\pi^{-s}\cos(\frac{1}{\mathit{2}}\pi s\mathrm{I}^{\Gamma(s})\zeta(_{S})$(cf. [Ti, p.16, Chapter II, (2.1.8)]), where
denotes the gamma function. Using (2.3)
and the formula $\Gamma(s)\mathrm{r}(1-s)=\pi/\sin\pi s$ , we have$\Gamma(s)$(2.4)$\cot(\frac{1}{2}\pi s)\zeta(S)=2s10\pi^{S-}\mathrm{c}\mathrm{s}(\frac{1}{2}\pi s)\Gamma(1-s)\zeta(1-s)$.

Substituting this into the integral in (2.1) and changing the variable into l-s, we obtain
$s$

(2.5)

,

$I(x)= \frac{1}{2_{\dot{i}}}x\int_{(\sigma_{1})}\sin(\frac{1}{\mathit{2}}\pi S)F(s)\zeta(_{S)}(\mathit{2}\pi x)-sd_{S}$where$\sigma_{1}=1-\sigma_{0}$and$F(s)= \frac{\Gamma(s)}{(s-2)(s-3)\cdots(s-2n-1)}$

Note thatsatisfies 1 $<\sigma_{1}<3/2$ . Since, it follows from (2.5) that$\sigma_{1}$.
converges absolutely for$\zeta(s)=\sum_{l=1}^{\infty}l^{-s}$$\sigma=\sigma_{1}$

(2.6)

,

$I(x)= \frac{1}{\mathit{2}}\pi ix\sum_{1\iota=}^{\infty}\{f(\mathit{2}\pi\dot{i}l_{X)-f}(-2\pi\dot{i}l_{X)\}}$

where
(2.7)

$f(z)= \frac{1}{2\pi\dot{i}}\int_{(\sigma_{1})}F(s)z^{-S}ds$.

This integral converges absolutely for $|\arg z|\leq\pi/2$ , since the order of the integrand isas(cf. Ivic [Iv, p.492, Appendix, (A.34)]), and sothat the interchange of the order of summation and integration is justified by the fact thatfor
. The identity$O \{|t|^{\sigma_{1}}-\frac{1}{2}-2ne-(\frac{1}{2}\pi-|\arg z|)|t|\}$$f(\pm \mathit{2}\pi ilX)=o(l^{-\sigma}1)$$l=1,\mathit{2},$$tarrow\pm\infty$$\ldots$$\frac{1}{(s-2)(s-\mathrm{s})\cdots(s-2n-1)}=\frac{1}{(s-1)\cdots(S-\mathit{2}n)}+\frac{2n}{(s-1)\cdots(s-2n-1)}$

shows that
(2.8)

$F(s)=\Gamma(s-2n)+\mathit{2}n\Gamma(S-2n-1)$

To evaluate the integral in (2.7), we need

.

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MASANORI KATSURADA

Lemma. Letwith$\sigma_{1}$be a constant with $1<\sigma_{1}<3/2$ . For any positive integer
we have$k\geq \mathit{2}$and any$|\arg z|\leq\pi/\mathit{2}_{f}$

$z$

(2.9)

$\frac{1}{\mathit{2}\pi\dot{i}}\int_{(\sigma_{1})}\mathrm{r}(s-k)_{Z}-sdS=z^{-k}\{e^{-z}-\sum_{h=0}\frac{(-z)^{h}}{h!}k-2\}$. Then changing the variable
equal
to
(2.9)
of
is
side
that the left-hand

Proof. Suppose first that$|\arg_{\mathcal{Z}}|<\pi/\mathit{2}$$\frac{1}{2\pi\dot{i}}z^{-k}\int_{(\sigma-k)}1S\mathrm{r}(s)_{Z^{-s}d}$.

$s$into $s+k$ , we see.

of this integral to the left, with noting $1-k<\sigma_{1}-k<3/2-k(<$
We move the path$2-k)$ . Collecting the residues of tlle poles at $s=-h(h=k-1, k, k+1, \ldots)$ , we find. This provesthat the left-hand side of (2.9) is further modified as
Lemma for $|\arg z|<\pi/2$ . The remaining case follows from the continuity of the integral
for $|\arg z|\leq\pi/2$ ,in (2.9), since the order of the integrand iswhen $tarrow\pm\infty$ .
$(\sigma_{1}-k)$$z^{-k_{\sum_{h}^{\infty}(}}=k-1-z)^{h}/h!$$O \{|t|^{\sigma_{1}}-k-\frac{1}{2}e-(\frac{1}{2}\pi-|\arg z|)|t|\}$$\square$

It follows from (2.7), (2.8) and Lemma that$f(2\pi\dot{i}lx)-f(-2\pi\dot{i}lx)$$=-4n(2\pi\dot{i}lx)-2n-1+4n(2\pi ilX)^{-}2n-1\mathrm{o}\mathrm{c}\mathrm{s}(\mathit{2}\pi l_{X})$.

$-2 \dot{i}(2\pi\dot{i}lx)^{-}2n\sin(2\pi lX)-4\sum_{k=}^{n}-11\frac{k}{(\mathit{2}n-\mathit{2}k)!}(\mathit{2}\pi\dot{i}lx)^{-2k}-1$

Substituting this into (2.6), we obtain$I(x)=-n(2 \pi\dot{i}x)^{-}2n\zeta(2n+1)+n(\mathit{2}\pi iX)^{-2n}\sum_{\mathrm{t}}^{\infty}\mathrm{t}\frac{\cos(\mathit{2}\pi l_{X})}{l^{2n+1}}=1$$+ \pi x(\mathit{2}\pi\dot{i}x)^{-}2n\sum_{l=1}^{\infty}\frac{\sin(\mathit{2}\pi lx)}{l^{2n}}$$- \sum_{=k1}^{1}\frac{k\zeta(\mathit{2}k+1)}{(2n-\mathit{2}k)!}(2\pi iX)n--2k$,which with (2.2) completes the poof of Theorem 1.

$\square$

3. PROOF OF THEOREM 2
In this section we prove Theorem 2. The skeleton of the proof is the same as that ofTheorem 1, so the details will be omitted. Throughout the following sections the constant.
andare fixed respectively
and$\sigma_{0}$$\sigma_{1}$$\mathrm{w}\mathrm{i}\mathrm{t}\mathrm{h}-1/2<\sigma_{0}<0$$1<\sigma_{1}<3/\mathit{2}$195

RAPIDLY CONVERGENT SERIES REPRESENTATIONS FOR $\zeta(2n+1)$

We begin the proof with the integral(3.1)$(|x|\leq 1)$$J(x)= \frac{1}{4_{\dot{i}}}I_{(\sigma)}0\cot(\frac{1}{2}\pi s)\zeta(S)\frac{X^{S}}{s(_{S+}1)\cdots(_{S+}\mathit{2}n+1)}d_{S}$Noting the facts$\zeta(0)=-1/\mathit{2},$$\zeta’(0)=-(1/2)\log \mathit{2}\pi$

.

, and

$\frac{1}{s(_{S+1})\cdots(S+\mathit{2}n+1)}=\frac{\Gamma(s)}{\Gamma(s+\mathit{2}n+\mathit{2})}=\frac{s^{-1}}{\Gamma(2n+2)}$.

$\frac{1+\psi(1)S+O(s^{2})}{1+\psi(2n+1)_{S}+O(s^{2})}$

with $\psi(s)=(\Gamma’/\Gamma)(s)$ , we see that the residue of the pole at $s=0$ of the integrand in(3.1) is$- \frac{1}{\mathit{2}(\mathit{2}n+1)!}(^{2n}\sum_{m=1}^{1}\frac{1}{m}-\log 2\pi x)+$Then moving the path of integration in (3.1) to thethe poles at $s=\mathit{2}k(k=0,1,2, \ldots)$ , we get(3.2)$\mathrm{r}\mathrm{i}\mathrm{g}\}_{1}\mathrm{t}$.

, and collecting the residues of$J(x)=- \frac{1}{2(\mathit{2}n+1)!}(^{2n}\sum_{1m=}^{1}\frac{1}{m}-\log \mathit{2}TX)+$$- \sum_{k=1}^{\infty}\frac{\zeta.(\mathit{2}.k)}{(2k)(2k+1)\cdot(2k+\mathit{2}n+1)}X2k$$s$

On the other side, we substitute (2.4) into the integral in (3.1), then change the variable
into l–s, and obtain(3.3)$J(x)= \frac{1}{2_{\dot{i}}}x\int_{(\sigma_{1})}\sin(\frac{1}{2}\pi s)c(S)\zeta(S)(2\pi x)-sds$

,

where
(3.4)

$c(_{S})= \frac{\Gamma(s)}{(s-1)(s-2)\cdots(s-2n-\mathit{2})}=\mathrm{r}(_{S\mathit{2}}-n-2)$.

Remark. In comparison with (2.8), the gamma factor (3.4) does not split in this case; theevaluation of $J(x)$ becomes simple.r than that of $I(x)$ in the preceding case.

Substituting the representation
into the integral in (3.3) and changing
the order of summation and integration, we obtain$\zeta(s)=\sum_{l=1}^{\infty}l^{-s}$(3.5)$J(x)= \frac{1}{2}\pi ix\sum_{=l1}^{\infty}\{g(2\pi\dot{i}x)-g(-\mathit{2}\pi\dot{i}X)\}$

where$g(z)= \frac{1}{\mathit{2}\pi\dot{i}}\int_{(\sigma_{1})}G(Z)z^{-S}ds$

,

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MASANORI KATSURADA

for $|\arg z|\leq\pi/2$ . Hence by Lemma and (3.5),$J(x)= \frac{1}{\mathit{2}}(2\pi\dot{i}x)^{-}2n\zeta(\mathit{2}n+1)+\pi x(2\pi i_{X)^{-}}2n-2\sum_{l=1}\frac{\sin(2\pi lx)}{l^{2n+2}}\infty$$+ \frac{1}{2}\sum_{k=1}^{1}\frac{\zeta(\mathit{2}k+1)}{(2n-\mathit{2}k+1)!}n-(\mathit{2}\pi\dot{i}X)-2k$.

This with (3.2) establishes Theorem 2.

$\square$

4. PROOF OF THEOREM 3
We first treat the even character case (i) of Theorem 3. In this case the functional
equation is of the form
(4.1)

$L(1-s, \overline{\chi})=\mathit{2}\mathcal{T}(x)-1(\frac{2\pi}{q})^{-s}\cos(\frac{1}{2}\pi s)\mathrm{r}(_{S})L(_{S}, \chi)$(cf. Washington [Wa, p.29. Chapter 4]). This suggests to adopt the integral(4.2)$s=\mathit{2}k(k=0,1,\mathit{2}, \ldots)$,

to the right, passing over the poles atof the integrand, and obtainas an initial setting. We first move the path
(43)$(|x|\leq 1)$$K(x)= \frac{1}{4_{\dot{i}}}\int_{(\sigma_{0})}\cot(\frac{1}{\mathit{2}}\pi S)L(_{S}, \overline{\chi})\frac{X^{S}}{(s+1)(S+\mathit{2})\cdots(_{S+}2n)}ds$$(\sigma_{0})$$K(x)=- \sum_{0k=}^{\infty}\frac{L(2k,\overline{x})}{(2k+1)(2k+2)\cdots(2k+2n)}x^{2}k$

Next changing the variable$s$$(|x|\leq 1)$

into l–s in (4.2), and then substituting (4.1), we get$K(x)= \frac{1}{2_{\dot{i}}}x\tau(\chi)-1\int_{(\sigma_{1})}\sin(\frac{1}{\mathit{2}}\pi s)F(S)L(s, x)(\frac{2\pi x}{q})-sds$

and hence noting that $L(s, \chi)=\sum_{l=1}^{\infty}x(l)l-s$ converges absolutely for,$\sigma=\sigma_{1}$$K(x)= \frac{1}{2}\pi\dot{i}X\tau(\chi)-1\sum x(ll=\infty 1)\{f(\frac{\mathit{2}\pi\dot{i}lx}{q})-f(-\frac{\mathit{2}\pi\dot{i}lx}{q})\}$, we obtain.

Here $f(z)$ is given by (2.7). The evaluation of $f(2\pi\dot{i}lx/q)-f(-2\pi\dot{i}lx/q)$ is the same as in
the proof of Theorem 1, so that
$K(x)=-nq \tau(x)-1(\frac{2\pi\dot{i}X}{q})^{-2n}L(\mathit{2}n+1, \chi)$$+nq \tau(x)^{-}1(\frac{2\pi\dot{i}X}{q})2n\sum^{-}\frac{\chi(l)\cos(\mathit{2}\pi l_{X}/q)}{l^{2n+1}}l=\infty 1$$+ \pi x\tau(\chi)^{-}1(\frac{\mathit{2}\pi\dot{i}X}{q})2n\sum^{-}\frac{\chi(l)\sin(2\pi lx/q)}{l^{2n}}l=\infty 1$$-q \tau(\chi)^{-}1\sum_{1k=}^{n}-1\frac{kL(\mathit{2}k+1,\chi)}{(2n-2k)!}(\frac{2\pi\dot{i}X}{q})^{-2k}$

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RAPIDLY CONVERGENT SERIES REPRESENTATIONS FOR $\zeta(2n+1)$

This with (4.3) establishes the former half of Theorem 3.

$\square$

We roceed to treat the odd character case (ii) of Theorem 3. The functional equationin this case asserts that$..\mathrm{p}$(4.4)$L(1-s, \overline{\chi})=2\dot{i}T(x)^{-}1(\frac{2\pi}{q})-s\sin(\frac{1}{2}\pi s)\Gamma(S)L(S, \chi)$(cf. [Wa, p.29, Chapter 4]). This suggests to adopt the integral(4.5)$(|x|\leq 1)$$H(X)= \frac{1}{4_{\dot{i}}}\int_{(\sigma_{0)}}\tan(\frac{1}{2}\pi s)L(_{S},\overline{\chi})\frac{x^{s}}{s(_{S+}1)\cdots(S+2n-1)}ds$,

as a starting point. We first move the path of integration in (4.5) to the right, passing
over the poles at $s=2k+1(k=0,1,2, \ldots)$ of the integrand, and obtain(46)$H(x)=k \sum^{\infty}\frac{L(\mathit{2}k+1,.\overline{\chi})}{(\mathit{2}k+1)(\mathit{2}k+\mathit{2})\cdot\cdot(\mathit{2}k+2n)}=0X^{2k+}1$

Next changing the variable$s$into l–s in (4.5) and using (4.4), we get.

$H(x)=- \frac{1}{\mathit{2}}X\tau(x)-1\int_{(\sigma_{1})}\cos(\frac{1}{2}\pi s)\mathrm{r}(_{S}-2n)L(S, \chi)(\frac{2\pi x}{q})-Sds$

This yields that
$H(x)= \frac{1}{2}\pi\dot{i}X\mathcal{T}(\chi)^{-}1\sum_{=l1}^{\infty}x(l)\{h(\frac{2\pi\dot{i}lx}{q})+h(-\frac{2\pi ilx}{q})\}$

,

where
$h(z)= \frac{1}{2\pi\dot{i}}\int_{(\sigma_{1})}\mathrm{r}(s-2n)Z^{-}Sds$

for $|\arg_{Z}|\leq\pi/2$ . The evaluation of $h(2\pi\dot{i}lx/q)+h(-\mathit{2}\pi il_{X}/q)$ is performed by Lemma,
and it is seen that
$H(x)=- \frac{1}{\mathit{2}}q\tau(\chi)^{-}1(\frac{2\pi\dot{i}X}{q})-2n+1\mathit{2}L(n, x)$$+ \frac{1}{\mathit{2}}q\tau(\chi)^{-}1(\frac{\mathit{2}\pi ix}{q})-2n+1\sum_{\iota=1}\frac{\chi(l)\cos(2\pi lX/q)}{l^{2n}}\infty$$- \frac{1}{2}q\tau(\chi)^{-}1\sum_{k=1}^{n-1}\frac{L(\mathit{2}k,x)}{(2n-2k)!}(\frac{2\pi\dot{i}X}{q})-2k+1$

This with (4.6) establishes the latter half of Theorem 3. The proof of Theorem 3 is therefore
complete.
$\square$

198

MASANORI KATSURADAREFERENCES[Ay]

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