数理解析研究所講究録 1414 巻 2005 年 15-25

15Some Doubly Infinite and Mixed Infinite Sums derived
from The N- Fractional Calculus of ALogarithmic

Function (with Some Examinations)Katsuyuki Nishimoto , Susana S. de Romero

and Ana L Prieto

Abstract
In this article theorems for some doubly infinite and mixed infinite sums derived

from the N-fractional calculus of alogarithmic function are reported.

Moreover some numerical examinations for the theorems are reported too.
\S 0. Introduction (Definition of Fractional Calculus)

Definition. (by K. Nishimoto)([1 ] Vol. 1)(I)

Let $D=\{D_{-}, D_{+}\}$ , $C=\{C_{-}, C_{+}\}$ ,$C_{-}$$C_{+}$$D_{-}$

be acurve along the cutjoining two points
be acurve along the cutjoining two points
be adomain surrounded by C-,$D_{+}$$z$

and $-\infty+i{\rm Im}(z)$ ,$z$

and $\infty+i{\rm Im}(z)$ ,be adomain surrounded by$C_{+}$(Here Z) contains the points over the curve $C$ ).

Moreover, let $f=f(z)$ be aregular function in $D(z\in D)$ ,$f_{\mathrm{v}}(z)=(f)_{v}=_{C}(f)_{v}= \frac{\Gamma(v+1)}{\underline{9}\pi i}\int_{C}\frac{f(\zeta)}{(\zeta-z)^{\mathrm{v}+1}}d\zeta$$-\pi\leq\arg(\zeta-z)$ $\leq\pi$$(f)_{\mathrm{v}}$$C_{-}$$z$$\in C-$$v\in R_{\backslash }$$0\leq\arg(\zeta-z)$ $\leq 2\pi$for$C_{+}$$\Gamma$order
for $v>0$ , and integrals of order-v forthe function $f$ . if.
$v$$|(f)_{\mathrm{v}}|<\infty$

(II)

(1)

(2)

; Gamma function,is the fractional differintegration of arbitrary order$\zeta\neq z$

then

for

,

$(m\in F)$ ,$(f)_{-m}= \lim_{\mathrm{v}arrow-m}(f^{\backslash })_{v}$where$(’\mathrm{v}\not\in T )$On the fractional calculus operator $N^{v}[3]$

$v$

$<0$

$v$

(derivatives of), with respect to

$z$

. of

16

Theorem A. Let fractional calcu$ll\mathit{4}S$operator (Nishimoto’s Operator)[Refer to $(1)$ ],$(v \not\in T)$$N^{v}=( \frac{\Gamma(v+1)}{\underline{?}\pi i}\int_{c}\frac{d\zeta}{(\zeta-z)^{v+1}})$$N^{v}$

be
(3)

with
$(m\in Z^{+})$$N^{-\prime n}= \lim_{\mathrm{v}arrow-m}N^{v}$

and define the binary operation$\circ$

,

(4)

as
(

$N^{\beta}\circ N^{\alpha}f=N^{\beta}N^{\alpha}f=N^{\beta}(N^{\alpha}f)$$\alpha$,$\beta$

ER},

( 5)

then $Che$ ser(6)$\{N^{\mathrm{v}}\}=\{N^{v}|v\in R\}$is an Abelian product group ( having $contim\iota ous$ index

transform operator
function such or$(N^{\mathrm{V}})^{-1}=N^{-\mathrm{v}}$$f$$\mathrm{f}$which has the inverseto the fractional calculus operator
for th)$v$$N^{v}$$f\in F=\{f:0\neq|f_{\mathrm{v}}|<\infty$, $v\in R\}$ , where $f=f(z)$

and

$e$

$z$

$\in C$.

(vis. $-\infty<v<\infty$ ).
(For our convenience, we call

Theorem B.

$\mathrm{I}$

F.O.G.

’$N^{\beta}\circ N^{\alpha}$$\{N^{v}\}\mathrm{T}l$index v\prime ’ for the ser of F

is art$\mathrm{r}$’as product of$N^{\beta}$

and$N^{a}1$)

Action product group which has continuous( F.O.G. ’. Fractional calculus operator group)

Theorem C. Let$S:=\{\pm N^{\mathrm{v}}\}\cup\{0\}=\{N^{\mathrm{v}}\}\cup\{-N^{\mathrm{v}}\}\cup\{0\}$Then$\mathrm{f}he$

ser

$S$

$(v \in R)$

.

( 7)

is a commutative ring for the function $f\in F$ , when the identity$N^{\alpha}+N^{\beta}=N^{\gamma}$(8)$(N_{3}^{\alpha}N^{\beta}, N^{\gamma}\in S)$

holds. [ 5]$(\mathrm{I}1\mathrm{I})$

Lemma. We have [ 1]$(\mathrm{i})$$((z -c)^{\beta})_{\alpha}=e^{-i\pi\alpha} \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(z -c)^{\beta-\alpha}$$(\log (z -c))_{\alpha}=-e^{-i\pi\alpha}\Gamma(\alpha)(z -c)^{-\alpha}$$(\mathrm{i}\mathrm{i})$$(\mathrm{i}\mathrm{i}\mathrm{i})$where$\mathrm{z}$$(\mathrm{i}\mathrm{v})$$((z-c)^{-\alpha})_{-\alpha}=-e^{iJ\mathrm{r}\alpha} \frac{1}{\Gamma(\alpha)}\log(z -c)$$-c\neq 0$in (i), and$z$$-c\neq 0$ ,1 in$(\mathrm{i}\mathrm{i})$and$(u \cdot v)_{\alpha}=\sum_{k=0}^{\infty}\frac{\Gamma(\alpha+1)}{k!\Gamma(\alpha+1-k)}u_{\alpha-k}v_{k}$$(| \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}|<\infty \mathrm{I}$,$(|\Gamma(\alpha)|<\infty)\mathrm{I}$(I$\Gamma(\alpha)1$$(\mathrm{i}\mathrm{i}\mathrm{i})$.

$($$\Gamma$$<\infty$)$|$; Gamma$(\begin{array}{l}u=u(z)v=v(z)\end{array})$$\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n})_{\mathrm{I}}$17\S 1. Doubly Infinite Sum and Mixed One
In the following$\alpha$, $\beta\in R$

Theorem 1. Let$M(\alpha, \beta ; k, m)$

(i)

When

$\beta\not\in Z_{0}^{+}$$:= \frac{\Gamma(\alpha+k)\Gamma(k+m)\Gamma(\beta+1)\Gamma(\beta-\alpha-m)}{k!\cdot m!\Gamma(\alpha)\Gamma(k)\Gamma(\beta+1-m)\Gamma(-\alpha)}$. we have the following doubly infinite sums

.

(1)

$j$

$\alpha-\beta$$\sum_{k\Rightarrow 0}^{\infty}\sum_{m=0}^{\infty}M(\alpha, \beta ; k, m)(\frac{z-c}{z})(m$(2)$\frac{c}{z})^{k}\subseteq\frac{\Gamma(\beta-\alpha)}{\Gamma(-\alpha)}($$\frac{z}{Z-\mathrm{C}})$

where
$z$

$-c\neq 0,1$ ,$z\neq 0,1$ ,$|$$(\mathrm{Z} -c)/z|<1$,$|c/z$

I $<1$ ,andI
The identity ( notation$(\mathrm{i}\mathrm{i})$When$s\in Z^{+}$$\Gamma(\alpha)|$$=$,$| \frac{\Gamma(\beta-\alpha-m)}{\Gamma(-\alpha)}|<\infty$) holds for$(\alpha-\beta)\in Z$

we have the following mixed infinite sums ;$\sum_{k=0}^{\infty}\sum_{m=0}^{s}M(\alpha, s ; k, m)(\frac{z-c}{z})^{m}($ $\frac{c}{z})^{k}\subseteq\frac{\Gamma(s-\alpha)}{\Gamma(-\alpha)}($(3)$\frac{z}{z-c})\alpha-s$

where
$z$

$-c\neq 0,1$ ,$z\neq 0,1$ ,$|c/\mathrm{z}$$|<1$ ,$|$$(\mathrm{Z} -c)/z1<\infty$,

and$|\Gamma(\alpha)|<\infty$

The identity ( notation
Proof of (i).

$=$) hoIds for$(\alpha-s)\in Z$

We have$\log\frac{z-c}{z}=\log(1-\frac{c}{z})=-\sum_{k\Leftrightarrow 1}^{\infty}\frac{c^{k}}{k}z^{-k}$Operate N- fractional calculus operator$N^{\alpha}$$(| \frac{-c}{z}|<1)$(4)

to the both sides of ( 4), we obtain$z^{-\alpha}-(z-c)- \alpha=-\sum_{k- 1}^{\infty}\frac{c^{k}}{k!}.\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}z^{-k-\alpha}$

sinc

.

$(|\Gamma(\alpha)|<\infty)$

,

(5)

18

$(z -c\neq 0,1.

$N^{\alpha}( \log\frac{z-c}{z})=(\log(z-c)-\log z)_{\alpha}$$=e^{-\iota\pi\alpha}\Gamma(\alpha)\{z^{-\alpha}-(z-c)^{-\alpha}\}$z\neq 0,1.)$$(|\Gamma(\alpha)|<\infty)$

(6)

$(7)$

and

$(8)$

$N^{\alpha}(z^{-k})=(z^{-k})_{\alpha}=e^{-i\pi\alpha} \frac{\Gamma(k+\alpha)}{\Gamma(k)}z^{-k-\alpha}$

by Lemmas$(\mathrm{i}\mathrm{i})$and (i).

Therefore, we have$(z -c)^{\alpha}-z$$\alpha=$$(9)$$- \sum_{k\cdot 1}^{\infty}\frac{c^{k}}{k!}\cdot\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}(z-c)^{\alpha}z^{-k}$

hence
$z$

from

$(\overline{5})$$(10)$$\alpha=\sum_{karrow 0}^{\infty}\frac{c^{k}}{k!}\cdot\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}(z-c)^{\alpha}z^{-k}$.

Next operate$\mathrm{N}arrow \mathrm{f}\mathrm{r}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\mathrm{a}\mathrm{l}$calculus operator$N^{\beta}$

to the both sides of ( 10), we

obtain$(z)_{\beta} \alpha=\sum_{k\approx 0}^{\infty}\frac{c^{k}}{k!}\cdot\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}((z-c)^{\alpha}z^{-k})_{\beta}$,$(11)$$\alpha-\beta$.

$(z^{\alpha})_{\beta}=e^{-iJ\mathrm{r}\beta} \frac{\Gamma(\beta-\alpha)}{\Gamma(-\alpha)}z$($(z-c)^{\alpha}z$$-k)_{\beta}= \sum_{m=0}^{\infty}\frac{\Gamma(\beta+1)}{m!\Gamma(\beta+1-m)}((z-c)^{a})_{\beta-m}(z^{-k})_{m^{1}}$$(12)$(13)

Now we have$((z-c)^{\alpha})_{\beta- m}=e^{-\mathrm{i}\pi(\beta-m)_{\frac{\Gamma(\beta-\alpha-m)}{\Gamma(-\alpha)}}}(z -c)^{\alpha-\beta+m}$

$\{$

(14)

and
$(z^{-k})_{m}=e^{-\mathrm{i}\pi\prime n} \frac{\Gamma(k+m)}{\Gamma(k)}$by Lemma$(\mathrm{i}\mathrm{v})$and$(\mathrm{i})_{1}$resp ctively.
$\mathrm{e}$

z-k-m

(15)

19

Therefore, we obtain$\frac{\Gamma(\beta-\alpha)}{\Gamma(-\alpha)}z^{\alpha-\beta}$$= \sum_{k\approx 0}^{\infty}\frac{\Gamma(\alpha+k)}{k!\Gamma(\alpha\grave{)}}\sum_{m=0}^{\infty}\frac{\Gamma(k+m)\Gamma(\beta+1)\Gamma(\beta-\alpha-m)}{m!\Gamma(k)\Gamma(\beta+1-m)\Gamma(-\alpha)}(\frac{z-c}{z})^{m}($$\frac{c}{z})(z-c)^{\alpha-\beta}k$$(16)$

from $(10)\sim( 15 )$ .
Therefore, we have$\sum_{k=0}^{\infty}\sum_{m\approx 0}^{\infty}M(\alpha, \beta ; k, m)(\frac{z-c}{z})^{m}($$\frac{c}{z})^{k}=\frac{\Gamma(\beta-\alpha)}{\Gamma(-\alpha)}($ $\frac{z}{z-c})\mathrm{m}\beta$(17)

from ( 16 ).

However the LHS (left hand side ) of ( 17) is always one valued function, on thecontrary the RHS ( right hand side ) of ( 17 ) is many valued function for$(\alpha-\beta)\not\in Z$

and one valued one for$(\alpha-\beta,\mathrm{I}\in Z$

Hence we must calculate as$( \frac{z}{z-c})^{\alpha-\beta}=($ $e^{i2n\pi} \frac{z}{z-c})\alpha-\beta$$($$(\alpha-\beta)\not\in Zn\in \mathrm{z}_{\mathrm{I}}$(18)because we are now being in the field of complex analysis.
Moreover, when $(\alpha-\beta)\in Z$

both of the LHS and RHS of ( 17 ) are one valued

functions respectively. In this case we have ( 17) strictly.
Therefore, we obtain ( 2) from ( 17), considering ( 18 ) finally.

Proof of$(\mathrm{i}\mathrm{i})$comditions.

Set$\beta=s\in Z^{+}$in ( 2 ), we have then ( 3) clearly, under the20\S 2. Some Numerical Examinations for Theorem 1
Examination of Theorem 1. ( 2)

[I]

Set
$z=10$ ,$c=1$ .

$\alpha=1/4$

and$\beta=1/2$

in Theorem 1. ( 2), we obtain$\sum_{k-0}^{\infty}\sum_{m\Rightarrow 0}^{\infty}M($

1/4, 1/2;$k$, $m)( \frac{1}{10})^{k}($$\frac{9}{10})^{m}$$(n \in Z)$$\subseteq-\frac{1}{4}\cdot\frac{\Gamma(1/4)}{\Gamma(3/4)}(e^{i2n\pi}\frac{9}{10})^{1/4}$(1)$=\{\begin{array}{l}-0.720\triangleleft 4\cdots(\mathrm{f}\mathrm{o}\mathrm{r}n=0)(2)-i0.7^{\underline{\gamma}}0\triangleleft 4\cdots(\mathrm{f}\mathrm{o}\mathrm{r}n=1)(3)0.72044\cdots(\mathrm{f}\mathrm{o}\mathrm{r}n=2)(4)i0.720\triangleleft 4\cdots(\mathrm{f}\mathrm{o}\mathrm{r}n=3)(5)\end{array}$When$\alpha$,$\beta$,$c$, $z\in R$ , the left hand side ( LHS ) of ( 3) is real, then we mustchoose ( 2) and ( 4) from the set

$\{$

( 2

$)_{1}$( 3 ), ( 4), ( 5)

$\}$

.

Now we have$M($ 1/41/2-. 0 , 0$\sqrt$$\frac{1}{10})^{0}(\frac{9}{10})^{0}$$\{$$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{L}\mathrm{H}\mathrm{S}\mathrm{o}\mathrm{f}(\mathrm{l})\mathrm{f}\mathrm{i}\mathrm{r}\mathrm{s}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{o}\mathrm{f})$$= \frac{\Gamma(1/4)}{\Gamma(-1/4)}=-\frac{1}{4}\cdot\frac{\Gamma(1/4)}{\Gamma(3/4)}<0$(6)

Then choosing ( 2) from the set {(2 ), (4)}, since the sign of the double iffiimite

sum of LHS of ( 1) is decided by the sign of its first tem (with $k=m=0$ ), when$>|M_{k+1,m+1}( \frac{1}{10})(k+1$ $\frac{9}{10})^{m+1}|$$l$$M_{k.m}=M(\alpha,\beta;k,m)$ ,we have then$\sum_{k\Rightarrow 0}^{\infty},\sum_{n\Rightarrow 0}^{\infty}M($

1/4, 1/2 ;from ( 2 ), considering (6 ).

$k$, $m)( \frac{1}{10})^{k}($ $\frac{9}{10})^{m}=-0.720\triangleleft 4\cdots$

(7)

21
Indeed we have
LHS of$(7)= \sum_{k=0}^{\infty}\frac{\Gamma(_{4}^{1}+k)}{k!\Gamma(\frac{1}{4})}(\frac{1}{10})\sum_{m\approx 0}^{k\infty}\frac{\Gamma(k+m)_{\underline{7}}^{\lrcorner}\Gamma(_{2}^{\mathrm{J}})\Gamma(_{4}^{1}-m)}{m!\Gamma(k)\Gamma(\frac{1}{2}+1-m)\Gamma(-\frac{1}{4})}($$\frac{9}{10})m$(8)$= \frac{\Gamma(_{4}^{[perp]})}{\Gamma(-\frac{1}{4})}\sum_{k=0}^{\infty}\frac{\Gamma(_{4}^{1}+k)}{\Gamma(\frac{1}{4})}(\frac{1}{10})^{k}+\frac{\Gamma(-\frac{3}{4})}{2!\Gamma(-\frac{1}{4})}($$\frac{9}{10})\sum_{k-0}^{\infty}\frac{\Gamma(_{4}^{1}+k)\cdot k}{k!\Gamma(\frac{1}{4})}($ $\frac{1}{10})^{k}$$- \frac{1}{\underline{?}!2^{2}}\cdot\frac{\Gamma(-\frac{7}{4})}{\Gamma(-\frac{1}{4})}(\frac{9}{10})^{2}\sum_{k-0}^{\infty}\frac{\Gamma(_{4}^{1}+k)\cdot k(k+1)}{k!\Gamma(\frac{1}{4})}($$\frac{1}{10})^{k}$$\frac{1}{10})^{k}$$+ \frac{3}{3!2^{3}}\cdot\frac{\Gamma(-\frac{11}{4})}{\Gamma(-\frac{1}{4})}(\frac{9}{10})^{3}\sum_{k=0}^{\infty}\frac{\Gamma(_{4}^{[perp]}+k)\cdot k(k+1)(k+2)}{k!\Gamma(\frac{1}{4})}($$- \frac{3\cdot 5}{4!2^{3}}.\frac{\Gamma(-\frac{15}{4})}{\Gamma(-\frac{1}{4})}(\frac{9}{10})^{4}\sum_{k=0}^{\infty}\frac{\Gamma(_{4}^{\mathrm{J}}+k)\cdot k(k+1)(k+2)(k+}{k!\Gamma(\frac{1}{4})}(\underline{3)}$ $\frac{1}{10})^{k}$$+\cdots\ldots\ldots\ldots\ldots$$= \frac{1}{(-4)}\cdot\frac{\Gamma(\begin{array}{l}-\mathrm{l}4\end{array})}{\Gamma(\frac{3}{4})}\{1+\frac{1}{40}+\frac{5}{2!40^{2}}+\frac{5\cdot 9}{3!40^{3}}+\frac{5\cdot 9\cdot 13}{4!40^{4}}+\frac{5\cdot 9\cdot 13\cdot 17}{5!40^{S}}+\cdots\}$$+ \frac{1}{2\cdot 3}\cdot\frac{\Gamma(_{4}^{\mathrm{J}})}{\Gamma(\frac{3}{4})}(\frac{9}{10})\{$$0+ \frac{1}{40}+\frac{5}{40^{2}}+\frac{5\cdot 9}{2!40^{3}}+\frac{5\cdot 9\cdot 13}{3!40^{4}}+\frac{5\cdot 9\cdot 13\cdot 17}{4!40^{S}}+\cdots\}$$+ \frac{1}{\underline{?}!3\cdot 7}\cdot\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}(\frac{9}{10})^{\mathrm{Z}}\{0+\frac{2}{4\mathrm{O}}+\frac{3\cdot 5}{40^{2}}$$+ \frac{4(5\cdot 9)}{2!40^{3}}+\frac{5(5\cdot 9\cdot 13)}{3!40^{4}}+\frac{6(5\cdot 9\cdot 13\cdot 17)}{4!40^{5}}+\cdots\}$$+ \frac{1}{3\cdot 7\cdot 11}\cdot\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}(\frac{9}{10})^{3}\{$$0+ \frac{\underline{?}\cdot 3}{40}+\frac{3\cdot 4\cdot(5)}{40^{2}}$$+ \frac{4\cdot 5(5\cdot 9)}{2!40^{3}}+\frac{5\cdot 6(5\cdot 9\cdot 13)}{3!40^{4}}+\frac{6\cdot 7(5\cdot 9\cdot 13\cdot 17)}{4!40^{5}}+\cdots\}$$+ \frac{5}{2!3\cdot 7\cdot 11\cdot 15}.\frac{\Gamma(\frac{1}{4})}{\Gamma(\frac{3}{4})}(\frac{9}{10}.)\{4$ $0+ \frac{2\cdot 3\cdot 4}{40}+\frac{3\cdot 4\cdot 5(5)}{40^{2}}$$+ \frac{4\cdot 5\cdot 6(5\cdot 9)}{\underline{?}!40^{3}}+\frac{5\cdot 6\cdot 7(5\cdot 9\cdot 13)}{3!40^{4}}+\frac{6\cdot 7\cdot 8(5\cdot 9\cdot 13\cdot 17)}{4!40^{5}}+\cdots\}$

(9)

22

(10)

$+\cdots\cdots\cdots\cdots$$=-(0.75941\cdots)+(0.01265\cdots)+(0.00205\cdots)$( 11)$+(0.00182\cdots)+(0.00117\cdots)+\cdots\cdots$$=-0.74172\cdots$[I I1$(12)$

Examination of Theorem 1. ( 3) for $(\alpha-s)\not\in Z$

Set$c=1-$$z$.

$=3_{-}$

and$\alpha=1/2$$s=1$

in Theorem 1. ( 3), we obtain$\sum_{k=0}^{\infty}\sum_{tn=0}^{1}M(1/2, 1 ; k, m)(\frac{1}{3})^{k}($ $\frac{2}{3})m$$(n \in Z)$$)^{1/2}3\neg$$\subseteq-\frac{1}{2}(e^{i2n\pi}--,$$=\{$-0.408240–
0.408240$\cdots$(for $n=0$ )

(14)

(for

(10).

$n=1$ )

(13)

Now we have$M(’1/2, 1 ; 0,0)( \frac{1}{3})^{0}($$\frac{2}{3})^{\alpha}$$\{$$\mathrm{t}\mathrm{h}\mathrm{e}\mathrm{L}\mathrm{H}\mathrm{S}\mathrm{o}\mathrm{f}(13)\tau \mathrm{l}\mathrm{f}\mathrm{S}\mathrm{t}\mathrm{t}\mathrm{e}\mathrm{m}\mathrm{o}\mathrm{r}_{\mathrm{I}}$(16)$= \frac{\Gamma(1/2)}{\Gamma(-1/2)}=-\frac{1}{2}<0$Then choosing ( 14) from the set { ( 14 ), ( 15 ) }, since the sign of themixed sum of LHS of ( 13 ) is decided by the sign of its first tem (with$k$$\inf$

inite$=m=0$ ),when$|M_{k,m}( \frac{1}{3})^{k}($ $\frac{2}{3})^{m}|>|M_{k+1,m+1}($ $\frac{1}{3})^{k+1}($ $\frac{\underline{?}}{3})^{m+1}|$$M_{k.m}=M(1/2,1 ; k, m)$ ,we have then$\sum_{k\approx 0}^{\infty}\sum_{m- 0}^{1}M’(1/2,1 ; k , m)$$( \frac{1}{3})^{k}($$\frac{2}{3})^{m}=-0.408240\cdots$

(17)

23
from ( 14), considering ( 16).
Indeed we have
LHS of ( 17)(18)$=- \frac{1}{2}+\frac{1}{\underline{7}\cdot 3}\cdot\frac{1}{6}+\frac{1}{2!}\cdot\frac{1}{2^{\mathrm{z}}\cdot 3}.\frac{5}{6}+\frac{1}{3!}\cdot\frac{1}{2^{3}\cdot 3^{2}}.\frac{9}{6}$(19)$+ \frac{1}{4!}\cdot\frac{5\cdot 7}{2^{4}\cdot 3^{3}}.\frac{13}{6}+\frac{1}{5!}\cdot\frac{5\cdot 7\cdot 9}{2^{\mathrm{S}}\cdot 3^{4}}.\frac{17}{6}+\cdots\cdots$$=-0.5+(0.0\underline{\circ}777\cdots)+(0.034722\cdots)+(0.017361\cdots)$$+(0.007314\cdots)+(0.002869\cdots)+(0.0001083\cdots)+\cdots\cdots$$=-0.40887\cdots$$\mathrm{L}\lceil$

II ]$\mathrm{I}$

(20)
(21)

Examination of Theorem 1. ( 3) for $(\alpha-s)\in Z$

Set$c=1$ .

$z$$=3$ ,$\alpha=2$

and

$s=1$

in Theorem 1. ( 3), we obtain$\sum_{k-0}^{\infty}\sum_{\dagger n-0}^{1}M(2,1 ; k, m)(\frac{1}{3})(k$without the ad hoc shown in [I] and [I
mmy valued ones.
That is, in this case both sides of
tively, then we have the notation$=\mathrm{i}\mathrm{n}$$\frac{2}{3})m$$\mathrm{I}$$= \frac{\Gamma(-1)}{\Gamma(-\underline{?})}($$\frac{3}{9,\sim},)=-3$(22)] in which the RHS of ( 1) and ( 13 ) are\S 1. ( 3) are one valued functions respec-\S 1. (3)Indeed we have
LHS of ( 22)=$=- \underline{?}\sum_{k=0}^{\infty}$$\sum_{k-0}^{\infty}\frac{\Gamma(\underline{?}+k)}{k!\Gamma(k)}(\frac{1}{3})^{k}\sum_{m\Rightarrow 0}^{1}\frac{\Gamma(k+m)\Gamma(-1-m)}{m!\Gamma(^{\underline{\gamma}}-m)\Gamma(-2)}($$(k +1)( \frac{1}{3})+\frac{2}{3}\sum_{k-0}^{\infty}k(k+1)(k$$\frac{1}{3})^{k}$$\frac{2}{3})^{m}$(23)$(24)$

24$=-2(1+- \underline{)\prime}+\frac{3}{3^{2}}+\frac{4}{3^{3}}+\frac{5}{3^{4}}+\cdots)3$(25)$+ \frac{2}{3}(0+\frac{2}{3}+\frac{\underline{9}\cdot 3}{3^{\mathrm{z}}}+\frac{3\cdot 4}{3^{3}}+\frac{4\cdot 5}{3^{4}}+\frac{5\cdot 6}{3^{5}}+\cdots)$$\approx$(26)$-2(2.2478 \cdots)+\frac{\underline{?}}{3}(\underline{?}.\underline{?}313\cdots)$$=-3.00807\cdots$(27)\S 3. Commentary
1. Notice that the LHS of \S 1. ( 2 ) is always one valued function, on
the contrary its RHS is many valued function forand one$(\alpha-\beta)\not\in Z$

valued one for $(\alpha-\beta)\in Z$
And notice that when both of the LHS and the RHS of \S 1. ( 2)are one valued functions respectively, namely in the case of $(\alpha-\beta)\in Z$ .

we have the identity ( notation
2.

”

Fractional calculus

”

$=$

)

in \S 1. ( 2) always.

is essentially a problem in the field of complex

analysis. We should not forget that we are now being in the field of fracti0-

nal calculus, that is, in that of complex analysis.
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Zensho, (1957), Iwanmi, Japan.

Katsuyuki Nishimoto
Institute of Applied Mathematics

Descartes Press Co.
2 . 13 . 10 Kaguike, Koriyama

963-8833 Japan

Susana S. de Romero
and Ana I. Prieto
C.I.M.A., Universidad del Zulia
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Venezuel