数理解析研究所講究録 1341 巻 2003 年 52-76

52
Some Topics in N-Fractional Calculus
デカルト 出版・西本勝之 (Katsuyuki Nishimoto)Research Institute for Applied Mathematics
Descartes Press Co.
Abstract
In this article, the following three topics in$\mathrm{N}$(Nishimoto’s )-fractional calculus

are reported, that is,
Part I.

N- Fractional calculus operator

.

Part

$\mathbb{I}$Part$\mathrm{m}$$N^{v}$(the set of them$\langle W$

$\}$

is an

action group ),N- Fractional calculus of the function $\log(z-c)$ and Beta function,. Application of -fractional calculus to the homogeneous Gauss$\mathrm{N}$equation and Kummerfs 24 functions.

Part I.

$\mathrm{N}$-Fractional calculus operator$N^{\mathrm{V}}$\S 0. Introduction (Definition of N Fractional Calculus)

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

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

be acurve along the cutjoining two points
be acurve along the cutjoining two Points

be adomain surrounded by C-,(Beta $D$

$D_{+}$

contains the points over the curve

Moreover, let$f_{v}\approx$

. and

,

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

$z$

$C$

$v$

for

$v$

$z$

$>0$ ,

the function $f$ , if$C_{+}$.

).

(v$\not\in T)$ ,$\in C$,$(m\in \mathrm{Z}^{*})$$= \lim_{varrow-m}(J)_{v}$for$v\in R$ ,$C_{-}$$\Gamma$,and integrals of$|(f)_{\nu}|<\infty$

.

(1)

,

(2)

$0\leq\arg(\zeta-\mathrm{z})\leq 2\pi$for$C_{+}$$v$(derivatives of,;Gamma function,

them $(f)_{v}$ is the fractional differintegration of arbitrary order

order

,

be adomain surrounded by$(f)_{\mathrm{v}} \approx_{c}(f)_{\mathrm{v}}=\frac{\Gamma(v+1)}{2\pi i}\int_{C}\frac{f(\zeta)}{(\zeta-z)^{v+1}}d\zeta$$-\pi\leq\arg(\zeta-z)\leq\pi$$\zeta\neq z$

and $-\infty+i{\rm Im}(z)$ ,$f(z)$ be aregular function in $D(z\in D)$ ,$f\simeq$$(f)_{-,l},$

where

$z$

$\mathrm{o}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{r}-v$

for $v<$$0$), with respect to$\mathrm{z}$

, of

53
\S 1. The Set of -Fractional Calculus Operator$\mathrm{N}$Definition of N- fractional calculus operator[I]$N^{\mathrm{v}}$We define N- fractional calculus operator (Nishimoto’s Operator)$N^{v}$ $:=( \frac{\Gamma(v+1)}{2\pi i}\int_{C}\frac{d\zeta}{(\zeta-z)^{\mathrm{v}+1}})$with$(v\not\in \mathrm{z}^{-})$$(m \in Z^{+})$$N^{-m}= \lim N^{\backslash \prime}$,as$N^{\mathrm{v}}$[Rejer to (1)1(1)(2),$varrow-m$

artd define the binary operatiortas$\circ=\mathrm{x}$$N^{\beta}\circ N^{a}f=N^{\beta}\mathrm{x}N^{a}f=N^{\beta}N^{a}f=$ $N^{\beta}(N^{a}f)$$(\alpha, \beta\in R)$.

(3)

Then we have$(v\not\in \mathrm{Z}^{-})$$N^{\mathrm{v}}f=N^{\mathrm{v}}f(z)=( \frac{\Gamma(v+1)}{2\pi i}\int_{c^{\frac{d\zeta}{(\zeta-z)^{\nu+1}}}})(f(\zeta))$

,

(4)

(S)

$= \frac{\Gamma(v+1)}{2\pi i}\int_{c}^{\frac{f(\zeta)}{(\zeta-z)^{\mathrm{v}+1}}d\zeta}$

$=$

[II1

(6)

$f_{\mathrm{v}}(z)=$ $f_{v}$An Abelan product group

Following Theorem 1is reported in JFC Vol. 4, Nov. (1994) [3]. However, it isshown again here for our convenience.
Theorem 1. The set$\{N^{v}\}\simeq$(7)$\{N^{\mathrm{v}}|v\in R\}$is an Abelian product group (having continuous index $v\in R$ ,the function

f$\in F=\{f;0\neq|f_{\mathrm{v}}|<\infty$

where f $=f(z)$ and z $\in C$ .
(In the following $0\neq f\in F$ and v,$\alpha$,$\beta$,$\mathrm{v}\mathrm{i}\mathrm{z}$. $-\infty<v$

) for

(8)

, v $\in R\}$ ,$\gamma\in R$$<\infty$.)

Proof.

We have the following for the multiplication of
We have;(i)$N^{\nu}$.

(9)(Closure )(10)$(\gamma= \alpha+\beta)$$N^{\beta}N^{a}f-N^{\beta+a}f=\mathcal{N}f$

Therefore, we obtain$N^{\beta}N^{a}=N^{\beta+a}=$ $N^{\gamma}\in\langle N^{\mathrm{v}}$

}.

54
(ii)

(11)

$N^{\gamma}(N^{\beta}N^{a})f=(N^{\gamma}N^{\beta})N^{a}f=N^{\gamma+\beta+\alpha}f$

Therefore, we obtain(Associative law )$N^{\gamma}(N^{\beta}N^{a})=(N^{\gamma}N^{\beta})N^{\alpha}$(12)

1 $\cdot f=N^{0}f=f$$(\mathrm{i}\mathrm{i}\mathrm{i})$(13)

Therefore, we obtain(Existence of unit element )$1=N^{0}$

(14)

(15)

$N^{-\mathrm{v}}N^{\mathrm{v}}f=N^{\nu}N^{-\nu}f=N^{0}f-f$$(\mathrm{i}\mathrm{v})$Therefore, we obtain$N^{-v}N^{\mathrm{v}}\approx$$N^{\nu}N^{-v}\approx$(v)$N^{0}=$$1$.

(Existence of inverse element)

(16)

(17)

$N^{\beta}N^{a}f=N^{a}N^{\beta}f=N^{a+\beta}f$Therefore, we obtain$N^{\beta}N^{a}\approx$$N^{a}N^{\beta}$

Therefore, we have this Theorem 1by

Then we call the set

this by ”F.O.G.

$\{N^{\nu}\}$Note 1. We have$N^{\beta}N^{a}f=$$\{N^{\nu}\}$as”(Commutative law)$(\mathrm{i})\sim(\mathrm{v})$.

Fractional calculus operator group and denote
1’”.

$([1] \mathrm{V}\mathrm{o}\mathrm{I}.\mathrm{I}, [2])$$(f_{a})_{\beta}=$(19)$\frac{\Gamma(\beta+1)}{\underline{\circ}\pi i}\int_{c_{(\eta-z)^{\beta+1}}}^{f(\eta)}\ovalbox{\tt\small REJECT} d\eta$$= \frac{\Gamma(\beta+1)\Gamma(\alpha+1)}{(2\pi i)^{2}}\int_{\mathrm{C}}f(\zeta)d\zeta\int_{c}\frac{d\eta}{(\zeta-\eta)^{a+1}(\eta-z)^{\beta+1}}$(22)$=f_{a+\beta}\simeq$ $N^{a+\beta}f$

Notice that letting

(20)

(21)

$= \frac{\Gamma(\alpha+\beta+1)}{2\pi i}\int_{c}\frac{f(\zeta)}{(\zeta-z)^{a+\beta+1}}d\zeta$

Note 2.

(18)$(N^{v})^{-1}$

be the inverse to$(N^{\mathrm{v}})^{-1}N^{\gamma}=N^{\nu}(N^{v})^{-1}=1$.

$N^{\nu}$

we have(23)

Then we obtain$(N^{\mathrm{v}})^{-1}=$from (24)and (16).
Therefore, we can see that

$N^{-\mathrm{v}}$

,

(24)

55
(25)

$((N^{1’})^{-1})^{-1}=(N^{-v})^{-1}=N_{1}^{\backslash \prime}$from (24 ).

Action group[III1

We have the following definition for action group. ([17] pp. 40-42. &PP. 113-133. )
Definition 1. Let $G=\{g\}$ be agroup, and $A=$

from

$G\mathrm{x}A\infty$$\{(g, a)|g \in G, a\in A\}$(i)$g_{1}\circ(g_{2}a)=(g_{1}\circ g_{2})a$

we say”$G$$A\subset$$\{a|a\in A\}$for all$g_{1}$

be aset. When the mapsatisfies, $g_{2}\in G$ ,$a\in A$

,

(26)
(27)

for all $a\in A$$1\circ a=a$$(\mathrm{i}\mathrm{i})$to$\{a\}\neq\phi$is agroup acting on aset A”.

Then we callas ”action group .
Obeying this definition we have the following theorem.
$\mathrm{I}\mathfrak{l}$$G$

Theorem 2. The set$\{N^{\nu}\}$is ”art action group which has continuous index$v$

for the set $F’|$.
Proof. Letting(i)$G=\{N^{\mathrm{v}}\}$, $A=$ $F$ and $a=f\in F$ in the above definition, we have$N^{\beta}(N^{a}f)=(N^{\beta}N^{a})f$

for all$N^{\beta}$,$N^{a}\in\{N^{\mathrm{v}}\}$,$f\in F$(28)

and$(\mathrm{i}\mathrm{i})$$N^{0}f=1\cdot$$f=f$

for all $f\in F$(29)

Therefore, we can see that the set {Nv} of our fractional calculus operator$N^{\mathrm{v}}$is agroup acting on aset $F$

Theorem 3. The set {having continuous index$N^{\nu}\rangle$$v$is ”an Abelian product group acting on a set $F$ ,E7?

Proof. It is clear by Theorems 1. and 2.

Part II.

N. Fractional calculus of the function$\log(z-c)$ and Beta function

Chapter 1. N- Fractional Calculus of the Power Functions andLogarithmic ones

56
\S 1. N. Fractional Calculus of Power Functions
Theorem 1. We have

where

$z$

$\in C1z\neq a$(1)$(| \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}|<\infty)$$((a-z)^{\beta})_{a}- \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(a-z)^{\beta-a}$

and a $(\in R)$ , $\beta(\in R)$ and $a(\in C)$ are constants.

Proof. Obeying our definition of N.fractional calculus, we have(2)$((a-z)^{q-1})_{-p}- \frac{\Gamma(1-p)}{2\pi i}\int_{C_{-}}\frac{(a-\zeta)^{q- 1}}{(\zeta-z)^{-p\star 1}}d\zeta$

set
$\{$

$- \frac{\Gamma(1-p)}{2\pi i}\int_{-\infty+l1\mathrm{m}(\mathrm{z})}^{(f*)}\frac{a^{q-1}\{1-(\zeta/a)\}^{r- 1}}{(\zeta-z)^{-,+1}}d\zeta$$- \frac{\Gamma(1-p)}{2\pi i}a^{\rho+q-1}\int_{-\sim}^{(0+)}\eta^{\rho-1}(1-t-\eta)^{q-1}d\eta$$- \frac{\Gamma(1-p)}{2\pi i}(a-z)^{\mu q-1}$,$a\xi$,$a-\delta e^{t|}j\delta,\phi\in R\mathfrak{l}\arg a\mathrm{I}-1\phi \mathrm{I}<\pi/2’)$$(\begin{array}{ll}z/a -t\xi-t- \eta\end{array})$$- \frac{\Gamma(1-p)}{2\pi i}a^{p+q-1}\int_{-\infty+l\mathrm{h}\mathrm{n}(\iota)}^{(t+)}\frac{(1-\xi)^{q- 1}}{(\xi-t)^{-\rho+1}}d\xi$$- \frac{\Gamma(1-p)}{2\pi i}(as)^{p*q-1}\int_{-\Phi}^{(0+)}u^{\rho-1}(1--u)^{q- 1}$$\zeta\simeq$$(\begin{array}{lll}\mathrm{l}- t- s\eta-s u\end{array})$I

du$\psi \mathrm{I}<\pi/2\psi-\arg_{S}.\mathrm{I}$$2i\sin\pi p\cdot B(1-p-q, p)$([21, p.20),$- \frac{\Gamma(1-q-p)}{\Gamma(1-q)}(a-z)^{p+q- 1}$(3)$\mathrm{t}4$

(5)

(6)

$(7)$

$\mathrm{C}8)$

slnce we have
(9)$\Gamma(p)\Gamma(1-p)-\frac{\pi}{s\mathrm{i}\mathrm{n}\pi p}$

Therefore, setting

we have (1), under the
Corollary 1.

.

$\mathrm{c}o\mathrm{n}\mathrm{d}\mathrm{l}\mathrm{t}\mathrm{l}\mathrm{o}\mathrm{n}\mathrm{s}q-\mathrm{l}-\beta$

and$-p-\alpha$We have$((z-a)^{\beta})_{a}$-e$\frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(z-a)^{\beta-a}$$\{$$<\infty)$(10)

Proof. It is clear by Theorem 1, since we have$e^{l\kappa\beta}((z-a)^{\beta})_{a}-e^{l\pi(\beta-a)} \frac{\Gamma(\alpha-\beta)}{\Gamma(-\beta)}(Z-a)^{\beta-a}$from (1).

$(| \frac{-\alpha-\beta)}{\mathrm{I}\mathrm{X}-\beta)}|<\infty)$

$)$

(11)

57
Corollary 2.

We have
(1 )

(I

$=0$

$\Gamma(\alpha)$1$<\infty$)(12)

Proof. We have

(1 )

(I

$-(z^{0})_{a}-e^{-\prime\pi a_{\frac{\Gamma(\alpha-0)}{\Gamma(-0)}}}z^{0-a}-0$),$<\infty$$\Gamma(\alpha)1$(13)

from (10).

\S 2. N- Fractional Calculus of Logarithmic Functions
Theorem 2. We have(1)$(\log$(a $-z))_{a}--\Gamma(\alpha)(a-z)^{-a}$ –e(14)$\Gamma(\alpha)(z-a)^{-a}$

and
(ii)

where 1$(\log$(z$-a))_{a}-(\log(a-z))_{\alpha}$$\mathrm{T}(\mathrm{a})1<$$\infty$, z $\in C$ , Z

$\#$

(15 I

a, and $\alpha(\in R)$ and a $(\in C$ ) are constants.

Proof of (i). We have$(\log(a-z))_{1}--(a-z)^{-1}$

Operate$N^{a-1}$(16)$(z\# a)$to the both sides of (16), we have then$((\log(a-z))_{1})_{a-1}--((a-z)^{-1})_{a- 1}$

(17)

hence
$(\log(a-z))_{a}--\Gamma(\alpha)(a-z)^{-a}$by our index law and by Theorem 1.

Proof of (il). We have(18)$(\log(a-z))_{a}-(\log e^{l\pi}(z-a))_{a}\mathrm{r}$ $(\log(z-a)+i\pi)_{a}$(19)- $(\log(z-a))_{a}$since$(i\pi)_{a}-0$

for

I

(20)

$\Gamma(\alpha)1<\infty$

Therefore,we have this theorem under the conditions.
Corollary 3.

We have$(\log az)_{a}-(\log z)_{a}\Leftrightarrow-e^{-i\mathrm{n}a}\Gamma(\alpha)z^{-a}$(I$\Gamma(\alpha)1$$<\infty$

)

(21)

where
.
Proof 1. Since we have$a\neq$ $0$$(\log az)_{a}-(\log \mathrm{Z} +\log a)_{a}-(\log z)_{a}+(\log a)_{a}$(22)

it is clear.

Proof 2.

Using the relationship$\log az$ $- \int_{0}\int_{1}^{a\iota}e^{-tt}dtd\mathrm{s}\approx\int_{0}\frac{e^{-s}-e^{-au}}{s}ds\infty\infty$

(23)

58
we obtain (21)by our definition of N- fractional calculus. $([11$VoI.I, pp$28- 30. )$( [2] pp. 50-51.)

Theorem 3. We have(i)$((a-z)^{-a})_{-a}=$(24)$- \frac{1}{\Gamma(\alpha)}\log(a-z)$

andwhere I$((z-a)^{-a})_{-a}\approx$$\mathrm{i}\mathrm{i})$$\mathrm{t}$$\Gamma(\alpha)1<\infty$

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

, and $z\neq a$.

We have(I$(\log(a-z))_{a}=-\Gamma(\alpha)(a-z)^{-a}$Operate(25)$-e^{i\pi a} \frac{1}{\Gamma(\alpha)}\log(z-a)$$\Gamma(\alpha)\mathrm{I}<$$\infty$)(26)

to the both sides of (26), we have then$N^{-a}$$((\log(a-z))_{a})_{-a}=$ $-\Gamma(\alpha)((a-z)^{-a})_{-a}$(27)hence we have$\log(a-z)\approx$ $-\Gamma(\alpha)((a-z)^{-a})_{-a}$$\mathrm{t}$$28)$

by our index law.Therefore, we have (24)from (28)clearly, under the conditions.

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

We have(29)$((a-z)^{-a})_{-a}\approx$ $e^{-i\pi a}((z-a)^{-a})_{-a}$Therefore, we have (25)from (24), (29)and (15 ), under the conditions.

Theorem 4. We have$( \log(z-a))_{-n}-\frac{(z-a)^{n}}{n!}\{.\log(z-a)-\sum_{k\cdot 1}^{n}\frac{1}{k}\}$where$z\neq a$

and

$n$

130)

$\in Z^{+}$

Proof. We have$(\log(z-a))_{-1}-\log(z-a)\cdot(z-a)-(z-a)$Then operating$N^{-m}(m\in \mathrm{Z}^{*})$

.

(31)

to the both sides of (31)we have$(\log(z-a))_{-1-n\iota}-(\log(z-a)\cdot(z-a))_{-m}-(z-a)_{-m}$$\mathrm{C}32)$

Now we have$( \log (z -a)\cdot(z-a))_{-m}-\geq_{-0}^{1}\frac{\Gamma(1-m)}{k!\Gamma(1-m-k)}(\log(z-a))_{-m-k}(z -a)_{k}$$-(\log(z-a))_{-m}(z-a)-m(\log(z -a))_{-m-1}$

(33)
(34)

since we have$(u \cdot v)_{a}-\geq_{0}.\frac{\mathrm{I}\mathrm{Y}1+\alpha)}{k!\Gamma(1+\alpha-k)}\infty u_{a-k}v_{k}$

([1

$]$

Vol.l )

(35)

59
(z$-a)_{-m}-e^{irtm} \frac{\Gamma(-1-m)}{\Gamma(-1)}(z -a)^{1\star m}=\frac{1}{(m+1)!}(z -a)^{1+m}$

Therefore, substituting (36) and (34) into (32 )$(\log$$\sqrt{\mathrm{e}}$(36)obtain(z $-a))_{-(1+m)} arrow\frac{1}{1+m}(\log(\mathrm{z}-a))_{-m}(z-a)-\frac{1}{(1+m)!(1+m)}(\mathrm{z} -a)^{1+m}$(37)

We have then, setting $m-0$ ,(31)$(\log(z-a))_{-1}-\log(z-a)\cdot(\mathrm{z}-a)-(z -a)1$

from (37), setting $marrow 1$ in (37)and using (31)we obtain$(\log(z-a))_{-2}$$-_{2}^{\mathrm{J}}$(38)(z$-a)^{2}\{\log(z -a)-(1+_{2}^{\mathrm{J}})\}$

Next setting $m-2$ in (37),and using (38)we obtain$( \log(z-a))_{-33!\overline{\overline{2}}}-\lrcorner(\mathrm{Z}-a)^{3}\{\log(z-a)-(1++\frac{1}{3}1)\rangle$

and so on.
Therefore, we have this theorem from (39)for$z\neq a$

and

(39)

$n\in Z^{*}$

Note 1. S.-T. Tu and D-.K. Chyan derived$(z^{\beta}\cdot\log z)_{a}-(z^{\beta})_{a}\mathfrak{p}o\mathrm{g}z$$+\psi$$(1+\beta)-\psi(1+\beta-\alpha)\rangle$$(z\neq 0)$

(42)

where

1
and

$\psi$(43)${\rm Re}(1+\beta)>0$$\Gamma(\beta-\alpha)/\Gamma(-\alpha)1$ $<\infty*$

is the Psi function.

From (42)they got (30)having $a-0$ . [61Chapter 2. N. Fractional Calculus of the Function $\log$ (z -c)and Beta Functions
g1. Some Theorems Associated with The Beta Functions
for N.Fractional Calculus of Function $\log$ (z-c)
In the following$\alpha$,$\beta$,$\gamma$,$\delta\in R$,z$\neq c$and B()is Beta function.

Theorem 1. We have(1)m,$n\in Z_{0}^{+}$, $(m+\alpha)$ , $(m+\beta)$ , (n$+\alpha+\beta)\not\in \mathrm{Z}_{0}^{-}$

and

60
$[ \lambda]_{n}=\frac{\Gamma(\lambda+n)}{\Gamma(\lambda)}=\lambda$$(\lambda+1)\cdots(\lambda+n$-1) with$[\lambda]_{0}=1$.

Proof. We have$(\log$(z$-c))_{m+a}=-e^{-\mathrm{i}\pi(m+\alpha)}\Gamma(\dagger n+\alpha)(z-c)^{-(m+a)}$’$(\log$(z$-c))_{1’+\beta},=$ $-e^{-i\pi(m+\beta)}\Gamma(rn+\beta)(z-c)^{-(\prime\prime+\beta)}$$(|\Gamma(m+\alpha)|<\infty)$

,

(2)

(

,

(3)

$|\Gamma(m+\beta)|<\infty)$

and$(\log (z -c))_{\mathfrak{s}+a+\beta},=-e^{-i\pi(\prime 1+a+\beta)}\Gamma(ll+\alpha+\beta)(\overline{‘}-c)^{-(n+a+\beta)}$

Therefore, making (2)LHS of (1)$=$$\simeq$$\mathrm{x}(3)/(4)_{1}$$(|\Gamma(\prime l+\alpha+\beta)|<\infty)$,(4)

we have$-e^{l\pi(n-2m)_{\frac{[\alpha],\prime|[\beta]_{m}}{[\alpha+\beta]_{n}}\cdot\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}}}(z -c)^{n-2m}$$-e^{i\pi(n-2m)} \frac{[\alpha]_{m}[\beta]_{m}}{[\alpha+\beta]_{n}}B(\alpha, \beta)$(z $-c)^{n-2m}$

(5)

(1)

under the conditions, since$\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\Leftarrow B(\alpha, \beta)$.

(6)Corollary 1. We have(i)

(Theorem A)

(7)

where
$\alpha$,$\beta$,$(\alpha+\beta)\not\in Z_{0}^{-}$

(ii)

C8)

where
$(1+\alpha)$, $(1+\beta)$ , $(1+\alpha+\emptyset$$(\mathrm{i}\mathrm{i}\mathrm{i})$$\not\in Z_{0}^{-}$$\frac{\alpha\beta}{(\alpha+\beta)(\alpha+\beta+1)}B(\alpha, \beta)$

where$(1+\alpha)$, $(1+\beta)$ ,$(2+\alpha+\beta)\not\in \mathrm{Z}_{0}^{-}$

(9)

61
(iv)

(10)

$\frac{\lceil(1\mathrm{o}\mathrm{g}(z-c))_{\iota+\alpha},\rceil^{2}}{(1\mathrm{o}\mathrm{g}(z-c)),,\mathrm{I}+2\alpha},=(-1)^{\prime 1+1}’\frac{([\alpha],n)^{2}}{[2\alpha],\prime l}B(\alpha, \beta)(z-c)^{-m}$

where

m
Proof of (i).
Proof of
Proof of

Proof of$(\mathrm{i}\mathrm{i})$Set $m=n=$.

$+\alpha)$,$(m+\underline{9}\alpha)\not\in Z_{0}^{-}$in (1 ).

$0$

Set $m=n=$ in (1).
Set $m=$ , $n=$ in (1).
, $m=n$ in (1).
Set$1$$(\mathrm{i}\mathrm{i}\mathrm{i})$$(\mathrm{i}\mathrm{v})$(m.

$\in Z_{0}^{*}$

.

$2$

$1$

.

$\alpha=\beta$

Theorem 2. We have(11)

$B(\alpha, \beta)$

where$\gamma$,$\alpha$,$\beta$,$(\alpha+\beta)\not\in Z_{0}^{-}$

Proof. We have$(\log (z -c))_{\gamma}\approx$$-e^{-l\mathrm{n}\gamma}\Gamma(\gamma)(z-c)^{-\gamma}$$(|\Gamma(\gamma)|<\infty)$$(\log(z-c)),+1=$ $-e^{-l\pi(\gamma+1)}\Gamma(\gamma+1)(z-c)^{-(\gamma+1)}$,$(|\Gamma(\gamma+1)|< \infty)$

(12)

,

(13)

hence we have(14)$\frac{(1\mathrm{o}\mathrm{g}(z-c))+1}{(\log(z-c))_{\gamma}}$from (12)and (13).
Then operating$N^{-1}$to the both sides$\mathrm{f}\mathrm{o}(14)$

we otain$( \frac{(1\mathrm{o}\mathrm{g}(z-c))+1}{(\log(z-c))_{\gamma}})_{-1}=-\gamma((z-c)^{-1})_{-1}$(15)namely$\log(\log(z-c))_{\gamma}=-\gamma\log(z-c)$

Next operate$N^{a}$.

$\mathrm{t}$

$16$

)

to the both sides fo(16), we have then$(\log(\log(z-c))_{\gamma})_{a}=-\gamma(\log(z-c))_{a}$$\simeq$$\gamma e^{-i\pi\alpha}\Gamma(\alpha)(z-c)^{-a}$( $|\Gamma(\alpha)1 <\infty)$ .

(17)

(18)

62
In the same vvay we obtain$(|\Gamma(\beta)|<\infty)$$(\log(\log(z-c))_{\gamma})_{\beta}=\gamma e^{-i\pi\beta}\Gamma(\beta)(z -c)^{-\beta}$

.

(19)

and
$(|\Gamma(\alpha+\beta)|<\infty)$$(\log(\log(z-c))_{\gamma})_{a+\beta}=\gamma e^{-i\pi(\alpha+\rho\gamma}\Gamma(\alpha+\beta)(z-c)^{-(\alpha+\beta)}$

.

(20)

Therefore, making (18)$\mathrm{x}(19)/(\underline{?}0)$

LHS of (11), we obtain$= \gamma\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}=\gamma\cdot B(\alpha,\beta)$

,

(11)

under the conditions.
Corollary 2.

We have(21)$\frac{\lceil(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{a}]^{2}}{\overline{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{2a}}}=\gamma\cdot B(\alpha, \alpha)$where$\gamma$

Proof.

Set$\alpha-\beta$,$\alpha$,$\underline{9}\alpha\not\in \mathrm{z}_{0}^{-}$in (11 ).

Theorem 3. Let$S\simeq$.

$S(z)$ $=\log(\log(\log(z-c))_{\gamma})_{\delta}$(22)

We have then$\frac{S_{a}S_{\beta}}{S_{a+\beta}}=\delta$$\cdot B$(23)$(\alpha, \beta)$

where$\delta$,$\gamma$’$\alpha,\beta$,$(\alpha+\beta)\not\in Z_{0}^{-}$

Proof. We have$(\log(\log(z-c))_{\gamma})_{\delta}=-\gamma(\log(z -c))_{\delta}$(I$\Gamma(\gamma)$$=\gamma e^{-i\pi\delta}\Gamma(\delta)(z-c)^{-\delta}$(1

I$\Gamma(\delta)$

(24)

)

$<\infty$I$<\infty$

(25)

).

and
$(\log(\log(z-c))_{\gamma})_{\delta+1}\Rightarrow\gamma e^{-i\pi(\delta+1)}\Gamma(\delta+1)(z-c)^{-\delta-1}$(I $\Gamma(\delta+1)1$$<\infty$

).

(26)

from (16 ), respectively.
Then, making (26)
, we obtain$/(^{\underline{9}}5)$

(27)

63
Then operating

to the both sides$N^{-1}$$\mathrm{f}\mathrm{o}(27)$

we otain(28)$( \frac{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{\delta+1}}{(\log(\log(z-c))_{\gamma})_{\delta}})_{-1}=-\delta\cdot((z-c)^{-1})_{-1}$that is,

.

(29)

).

(30)

.

(31)

$S=\log(\log(\log(z-c))_{\gamma})_{\delta}=$ $-\delta\cdot\log(z-c)$

Next operate

to the both sides of (29), we have then$N^{\alpha}$$S_{\alpha}=\delta\cdot e^{-l\pi a}$

I ) $(z-c)^{-a}$$\alpha$( I$\Gamma(\alpha)$I$<\infty$

In the same way we obtain$S_{\rho}=\delta\cdot e^{-\mathit{1}\pi\beta}\Gamma(\beta)(z-c)^{-\beta}$$(1\Gamma(\beta)1 <\infty)$

and$S_{a+\beta}=$$\delta\cdot e^{-i\pi(\alpha+\beta)}\Gamma(\alpha+\beta)(z-c)^{-a-\beta}$$(1 \Gamma(\alpha+\beta)\mathfrak{l}<\infty)$.

$\mathrm{C}32$)

Therefore, making (30) $\mathrm{x}(31)/(32)$ , we obtain$\frac{S_{a}S_{\beta}}{\overline{S_{a+\beta}}}=$(23)$\delta\cdot\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}=\delta\cdot B(\alpha, \beta)$

under the conditions.
Corollary 3.

We have(33)$\frac{(S_{a})^{2}}{S_{2a}}=\delta\cdot B(\alpha, \alpha)$

where
6,
Proof.

Set$\gamma$,$\alpha,\underline{\mathit{0}}_{\alpha\not\in \mathrm{Z}_{0}^{-}}$in (23 ).

$\alpha=\beta$

Theorem 4. Let$T=T(z)$$\mathrm{t}$$34\mathfrak{l}$$=(1\mathrm{o}\mathrm{g}(z-c))_{\mathrm{y}+1}\overline{\overline{(\log(}z-c))_{\gamma}}$

We have then$T_{a}T_{\beta}\overline{\overline{T_{a+\beta}}}<$(35)$- \frac{\alpha\beta\gamma}{\alpha+\beta}\cdot B(\alpha, \beta)(z -c)^{-1}$where$\gamma$

Proof. Operating$T_{a}=$$N^{a}$,$\alpha$,$\beta$,$(\alpha+\beta)\not\in \mathrm{Z}_{0}^{-}$

to the both sides of (34) we have$( \frac{(1\mathrm{o}\mathrm{g}(z-c))_{\gamma+1}}{(1\mathrm{o}\mathrm{g}(z-c))_{\gamma}})_{a}=$$-\gamma((z-c)^{-1})_{a}$($|\Gamma(\gamma)$

I$<\infty$

).

(36)

64
$=-\gamma\alpha e^{-\mathrm{i}\pi a}\Gamma(\alpha)(z -c)^{-1-\alpha}$$(|\Gamma(\alpha)|<\infty)$

.

(37)

),

(38)

In the same way, we obtain$T_{\beta}=-\gamma\beta e^{-i\pi\beta}\Gamma(\beta)(z -c)^{-1-\beta}$( I$\Gamma(\beta)$

I$<\infty$

and$T_{a+\beta}=-\gamma(\alpha+\beta)e^{-i\pi(a+\beta)}\Gamma(\alpha+\beta)(z-c)^{-1-a-\beta}$(I$\Gamma(\alpha+\beta)$

I$<\infty$)(39)

from (34).
Therefore, we have$T_{a}T_{\beta}\overline{\overline{T_{a+\beta}}}=$$- \frac{\alpha\beta\gamma}{\alpha+\beta}\cdot\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}$$(z -c)^{-1}$(40)from (37), (38)ancl (39).
We have then (35)from (40)clearly, under the conditions.
Corollary 4. We have$\frac{(T_{a})^{2}}{T_{2a}}=-\frac{\alpha\gamma}{2}\cdot B(\alpha, \alpha)(z -c)^{-1}$

(41)

where
$\gamma$

Proof.

(1)

Set$\alpha\approx$$\beta$,$\alpha$,$2\alpha\not\in \mathrm{Z}_{0}^{-}$in (35 ).

\S 2. Some examples of the Theorems
Example of Theorem 1

(i)

(ii)

(iii)

$\frac{(\log(z-c))_{1l2}(1\mathrm{o}\mathrm{g}(z-c))_{3l2}}{(1\mathrm{o}\mathrm{g}(z-c))_{(\mathrm{l}2)+(3/2)}},\approx$$\approx$$\frac{1\cdot 2}{3}B(1,$2)(z$-c)^{-1}= \frac{1}{3}(z-c)^{-1}$(2)$\frac{(\log(z-c))_{1+(1l2)}(1\mathrm{o}\mathrm{g}(z-c))_{1+(1\mathit{1}3)}}{(1\mathrm{o}\mathrm{g}(z-c))_{1+(1l1)+(1l3)}}=\frac{(1/2)(1/3)}{((1/2)+(1/3))}B(1/2,3/2)(z -c)^{-1}$$= \frac{1}{5}B(1/2,1/3)(\mathrm{z}-c)^{-1}$

(iv)

(1)

$-B(1/2,3/2)=- \frac{\pi}{2}$

(3)

(4)

65
Example of Theorem 2

(I1)

(i)

$\frac{(\log(\log(z-c))_{\gamma})_{\mathrm{l}/2}(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{3/2}}{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{\iota}}=\gamma\cdot B(1/2,3/2)=\frac{\gamma\pi}{\underline{?}}$

(III)

(i)

$\mathrm{C}6)$$\frac{[(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{1/2}]^{2}}{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{1}}=\gamma\cdot B(1/2,1/2)=$ $\gamma\cdot\pi$$(\mathrm{i}\mathrm{i})$Example of Theorem 3$\frac{S_{1\prime 2}S_{3/2}}{\mathrm{g}}=^{\frac{(\log(\log(\log(z-c))_{\gamma})_{\delta})_{1/2}(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{\delta})_{3/2}}{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{\delta})_{2}}}$(7)$- \delta\cdot B(1/2,3/2)=\frac{\delta\pi}{2}$$(\mathrm{i}\mathrm{i})$

(IV)

(i)

$\delta\cdot B(1/2,1/2)\Leftarrow\delta\cdot\pi$$\frac{(S_{1/2})^{2}}{S_{1}}-\frac{[(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(z-c))_{\gamma})_{4})_{1/2}]^{2}}{(1\mathrm{o}\mathrm{g}(1\mathrm{o}\mathrm{g}(\log(z-c))_{\gamma})_{\delta})_{1}}=$(8)

Example of Theorem 4$\frac{T_{1/2}T_{3/2}}{T_{2}}=(\frac{(1\mathrm{o}\mathrm{g}(z-c))_{\gamma+1}}{(\log(z-c))_{\gamma}-})_{1/2}\cdot(_{\overline{\overline{(\log(z-c))_{\gamma}}}}^{(1\mathrm{o}\mathrm{g}(z-c))_{\gamma+1)_{3/2}/}}(\frac{(1\mathrm{o}\mathrm{g}(z-c))_{\gamma+1}}{\overline{(\log(}z-c))_{\gamma}})_{2}$$=- \frac{3}{8}\gamma\cdot$$(\mathrm{i}1)$

(5)

19)

$B(1/2,3/2)(z-c)^{-1}\approx$ $- \frac{3}{16}\gamma\pi$ $(z-c)^{-1}$$2/(_{(\log(z-c))_{\gamma}}^{(1\mathrm{o}\mathrm{g}(-c))_{+1}}\underline{\underline{z}_{-\infty}}11$$\frac{(T_{1/2})^{\iota}}{T_{1}}-[(\frac{(1\mathrm{o}\mathrm{g}(z-c))_{r+1}}{\frac{}{(\log(z-c))_{\gamma}}})_{1/2}\mathrm{J}$$=- \frac{\gamma}{4}\cdot B(1/2,1/2)(z-c)^{-1}\simeq$

Part III. Application of(10)$- \frac{1}{4}\gamma\pi\cdot(z -c)^{-1}$-fractional calculus tothe homogeneous Gauss equation
and Kummer’s 24 functions.

Chapter 1.

$\mathrm{N}$-fractional calculus operator
homogeneous Gauss equation$\mathrm{N}$$.N^{\nu}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{d}$to66method to the homogeneous Gauss equation\S 1.
By our fractional calculus operator
method we obtain the following solutions$N^{\nu}$$N^{\vee}$

which contain the
Theorem 1. Let
equation,-fractional calculus.

$\mathrm{N}$, flten the homogeneous Gauss$\varphi\in p^{\mathrm{o}}=\{\varphi|\mathrm{U}\neq|\varphi_{\nu}| <\infty, V \in R\}$

$L1^{\varphi,Zj}\alpha$$\approx$$\beta$,$\gamma]$$\varphi_{2}\cdot(z^{2}-z)$$+\varphi_{1}\cdot\{z(\alpha+\beta+1)-\gamma\}+\varphi$$\cdot\alpha\beta\simeq$$(z \neq 0,1)$

$0$

(0)

has solutions of the form
(Group

$t$

)

$i$

$\varphi-K$($z^{a-\gamma}\cdot$$(z-1)^{\gamma-\beta-1)_{a- 1}}\Xi\varphi_{(1\}}$,denote)(1)$\varphi-K(z^{\beta-\prime}\cdot(z-1)^{\gamma-a-1)_{\beta-1}\sim\varphi_{(2)}}.$(2)$K((z-1)’-\beta-1.a-\prime z)_{a- 1}\Xi\varphi_{\{3)’}$(3)$\varphi=$ $K((z-1)^{\prime-a-1}\cdot z^{\beta-\prime})_{\beta- 1}=\varphi_{(t)\prime}$(4)$\varphi\approx$$(G\iota\cdot\sigma u\mu Il)_{i}$$\varphi-Kz^{1-\prime}$(z$Kz^{1-r(z^{\beta-1}\cdot(Z}$$\varphi\approx$a$\cdot(z.-1)^{-\beta})_{aarrow\prime}$$\varphi_{\{7)}$

Kz

(5)

,

(6)

,

(7)

,

(8)

$-1)^{-a})_{\beta-\prime}\overline{arrow}\varphi_{(\mathrm{f})}$$\varphi-KZ^{1-r((z-1)^{-\beta}\cdot z^{a- 1}})_{a-\prime}\cong$$\varphi\sim$,$\varphi_{(s\gamma}$.
$((z-1)^{-\mathrm{n}}\cdot z^{\beta-1})_{\beta-\prime}\equiv$$\varphi_{(\mathrm{d})}$(Group 111);$\varphi$,,(9)$u-1),-A-1\varphi_{(10)\prime}\Xi$(10)$-K(z-1)^{\prime-a-l(z^{-a}\cdot(z-1)^{\beta-1)}},-a-1\Rightarrow\varphi_{()}$$\varphi-K(z-1)^{\gamma-a-l(z}$

-l.

(11)

$\varphi-K(z-1)^{\prime-\mathrm{n}-l((z-1)^{l-1}\cdot z^{-\mathrm{B}),-\varphi_{(11)\prime}}}-a-1$$\varphi-K(z-1)’-a-\beta((z-1)^{a-1}\cdot z^{-t)_{\gamma-\beta-1}\sim}$ $\varphi_{\{12)}$

whereconstant,$\varphi_{k}-d^{k}\varphi/d\mathrm{z}^{k}$$\alpha$,$\beta$and$(k-0,1, 2)$ ,given constan$\varphi_{0}=\varphi=$$\gamma a’.e$$fs$

Proof of Group I
Operate fractional calculus operator
have then$\varphi(z)$

.

,

$z$

$\in C$

, and

,

(12)

$K$

is art arbitrary$j$$\mathrm{N}$$N^{\vee}$

directly to the both sides of (0), we$N^{\vee}\{L[\varphi.z ; \alpha, \beta, \gamma]\}$$\infty$$\varphi_{2\mathrm{s}\nu}\cdot(z^{2}-z)$$+\varphi_{1\mathrm{s}\nu}\cdot\{z(2v+\alpha+\beta+1)-\mathrm{v}-\gamma\}$$+\varphi_{\vee}\cdot\{v^{2}+v(\alpha+\beta)+$a$\beta\}arrow 0$$(z\sim 0,1)$

(13)

since
$N^{\nu}(\varphi_{m}\cdot z^{n})\propto$ $( \varphi_{m}\cdot z^{n})_{v}-z_{-0}^{l\mathrm{I}}\frac{\Gamma(v+1)}{\Gamma(v+1-k)\Gamma(k+1)}(\varphi_{m})_{\nu-k}(z^{n})_{k}$

where

$n$

$\in \mathrm{Z}_{0}^{*}(-Z^{*}\cup\{0\})$Choose such that
we have then
$v$

(14)

.
$v^{2}+v(a \cdot\vdash/J)+\alpha/f\infty 0$$v\approx$$-\alpha$and$-[i.$

,

(15)

(1)

67
Substitute $v=$ $-1X$ into (13), yield$,p_{2-\prime\prime}.(z^{2}-z)$$cdot\prime p_{1-\mathrm{t}\mathrm{I}}\cdot\langle \mathrm{z}$( $-\alpha+$ $(l +1)+rx$

selling$.1^{\cdot}\mathrm{I}1\mathrm{C}^{\backslash }\mathrm{I}^{\cdot}\mathrm{G}\mathrm{f}\mathrm{c}".1_{-J}^{\backslash }$$\varphi_{l- a}=n$$\mathrm{r}\mathrm{v}\iota^{\iota}$$=u(z )$

(18)$(\varphi =u_{\mathfrak{n}-1})$

have$u_{1}+\iota r$

fro

(17)

U.

$-\gamma\}\simeq$$\mathrm{m}$(17).

$’\iota\cdot 11\mathrm{e}$solution of$\mathrm{t}\mathrm{l}\dot{\mathrm{u}}\mathrm{s}$(19)$(z\# 0,1)$$\cdot\frac{z(-\alpha+\beta+1)+\alpha-\underline{\gamma}}{\overline{z^{l}}-z}=0$differential equation is given by(20)$u-[(z^{n-\prime}(z-1)^{\prime-fl- 1}.$Thus$\iota \mathrm{v}\mathrm{t}^{1}\mathrm{t}’ b\mathrm{t}\mathrm{n}\mathrm{i}\mathrm{l}\mathrm{l}$$\varphi\infty$$K(z^{} \cdot(z-1)’-\beta-\mathrm{I})_{a- 1}=$(dellOte)$\varphi_{[1)}$$(z\# 0,1)$$\mathrm{C}1)$

from (20) and (18). Where $K$ is an arbitrary conslant.
Inversely, tlte funcliongiven by (20) satisfies(19) clearly. Hence (1)satisfies equation(17). l’herefore, the function (1)satisfies equation (0).
and [ in (1 ), because, in the same((or merely by the change of
For $v=$and ) we obtain lher solution
tlte equation (0) is symmetry for$-\beta$$\iota \mathrm{v}\mathfrak{n}\mathrm{y}$$/J$$\alpha$$\varphi=$$K(z\cdot(\rho_{-\prime}z-1)’- a-1)_{\beta- \mathrm{I}}$which is different from (.1), if
Moreover, changing tlte orderVol. 1&[7|)$\alpha\neq$$z^{a-\gamma}$$\beta$.

and$(z-1)^{\prime-\beta- 1}$.

$\varphi_{(2)}$in (1)$\varphiarrow K((z-1)^{\prime-fl- 1}\cdot z^{\alpha-\prime})_{a-1}=\varphi_{(3)}$different from (1) when(2)$(z\#\mathrm{t}\mathrm{I} , 1)$have other solution $([6|$(3)$(z\# 0, 1)$. In the same way we have other solution$(\alpha-1)\not\in \mathrm{Z}_{0}^{*}$$\varphi\approx$$J$$\alpha$$\mathrm{o}$$K((z-1)^{\prime-oe-1}\cdot z^{t-\prime})_{\beta- 1}\cong$$\varphi_{(\mathrm{t})}$from (2), which is different from (2) when $(\sqrt-1)\not\in.4^{*}$(4)$(z\mathrm{r} 0,1)$.

Proof of Group II;

Set$arrow z^{\lambda}\phi$$\varphi$(Hence,$\phi\approx$$\phi(z)$$(z\# 0,1)$

and
Substitute (21) into (0), we have then$\lambda z^{\lambda-1}\phi$$+z^{1}\phi_{1}$$\varphi_{1}-$$\varphi_{2}-\lambda(\lambda-1)z^{\lambda-2}\phi$ $+2\lambda z^{\lambda-1}\phi_{1}+z^{\lambda}\phi_{2}$

(21)

).

$\phi_{2}.(z^{t}-z)$ $+\phi_{1}\cdot\{z(\alpha+\beta+1+2\lambda)-2\lambda-\gamma\}$(22)$+\phi\cdot\langle\lambda(\lambda-1)+\lambda(\alpha+\beta+ 1)+\alpha\beta-z^{-1}\lambda(\lambda-1+\gamma)\}-0$

.

where
$(k-0,1, 2)$ andHere, we choose Asuch tltat$\phi_{k}\approx$$d^{k}\phi/d\mathrm{z}^{k}$$\phi_{0}=\phi$.
(23)$\lambda(\lambda-1+\gamma)\approx$ $0$

that is,$\lambda=$ $0$,$1-\gamma$

.

(24)

case A-11
tlUs case xve have the same results as Group 1.
(ii)
tlte case A $-1-\gamma$
Substituting A $-1-\gamma$ into (22) we have11}$\ln 11_{1}\mathrm{c}$

$\ln$

$\ln$

$\phi_{2}.(z^{2}-z)$ $+\phi_{1}\cdot\{z(\alpha+\beta-2\gamma+3)+\gamma-2\}+\phi\cdot\{(1-\gamma)+\alpha\}\{(1-\gamma)+\beta\}=$Next, operate$N^{\vee}$

to tlte both sides of (25),$\backslash \mathrm{v}\mathrm{e}$$0$. (25)

have then$\phi_{2*\nu}\cdot(z^{2}-z)+\phi_{1\mathrm{s}\nu}\cdot\{z(\alpha+ /J-2\gamma+3+2v)+\gamma-2-v\}$$+\phi_{\nu}\cdot\{v^{2}+v(\alpha+\beta-2\gamma+2)+(1-\gamma)(\alpha+\beta+1-\gamma)+\alpha\beta\}=0$

.

(25)

68
Here we dtoose

such that$v$$v^{2}+v(\alpha+\beta-2\gamma+2)+(1-\gamma)(\alpha+\beta+1-\gamma)+\alpha\beta=\mathrm{t}\mathrm{I}$$\mathrm{t}\mathrm{h}\mathrm{a}\mathrm{t}$

,

(27)

is,
(28),

$v=$ $\gamma-\alpha$ $-1$

1} For tlxe case of (28)$\mathrm{n}$

1td

(29)

$v-\gamma-\beta-1$$\cdot$$j$

Substituting (28) into (26), we have$\phi_{1+\gamma-a}\cdot(z^{2}-z)+\phi,-a.\{z(\beta-\alpha+1)+a-1\}-0$

Set{$\emptyset_{\gamma-a}-u-u(z)$we have then$\phi$$arrow u_{a-\gamma})$

.

(30)

,

(31)
(32)

$u_{1}\cdot(z^{2}-z)+u\cdot\{z(\beta-a+1)+\alpha-1\}\approx$ $0$

from (30). The solution to this differential equation is given by$(z* 0, 1)$$u=Kz^{\tau*-1}(z-1)^{-\beta}$

Here

$K$

(33)

is an arbitrary constant.

Therefore, we obtain$(z\# 0,1)$$\phi-K(z^{a-1}\cdot(z-1)^{-\beta})_{a-\gamma}$

from (33) and (31), hence we have$\varphi$$-Kz^{1-}$ ’($z^{a- 1}$.

$(z-1)^{-\beta}$

)

(3.4)

$(z\sim.0, 1)$$\mathrm{r}$$\varphi_{[5)}$(5)from (34) and (21).
Inversely, (33) satisfies (32), then (34) satisfies (30) dearly. Therefore, (5)satisfies(0), since we have (21).
2) For tlxe case of (29);
111 the same way as 1) (or merely by the change and in (5))we obtain$\beta$$\alpha$$\varphi\simeq$$Kz^{1-\gamma}(z^{\beta-1}\cdot(z-1)^{-a)_{\beta-\prime}}\vee$‘$\varphi_{()}$$(z\# 0,1)$(6).
as the solutions to tlte equation (0), which is different from (5) if
in (5) we have other solution
Moreover, changing the orderand$\alpha$$\varphi-Kz^{1-\prime}((z-1)^{-\beta}\cdot z^{a-1)_{a-\prime}}\cdot\varphi_{(7\}}$$(\alpha-\gamma)\not\in \mathrm{Z}_{0}^{\mathrm{g}}$$\varphi-Kz^{1-\prime}((z.

$\beta$$(z-1)^{-\beta}$$z^{a- 1}$

different from (5) when$\mathrm{r}$$(z\sim 0, 1)$(7). In the same way we have other solution-1)^{-aa}\cdot z^{l-\downarrow)_{\beta-\gamma}}\mathrm{z}$from (6), which is different from (6) when$\varphi_{[l)}$$(\beta-\gamma)\not\in Z_{0}^{*}$$(z\mathrm{r} 0,1)$(8).

Proof of Group III;

,
Set
(0),
we have thenand substitute (35) into$\varphi\simeq$$\phiarrow\phi(z)$$(z-1)^{\lambda}\phi$$(z\neq 11:$\phi_{2}.(z^{2}-z)+\phi_{1}\cdot\{z(\alpha +\beta+1+2\lambda)-\gamma$ $\}$$+\phi$$\cdot\{(z-1)^{-1}(\lambda^{l}+\lambda a +\lambda\beta-\lambda\gamma)+(\lambda+\alpha)(\lambda_{-}+\beta)\}-0$

Here, we choose Asuch that$\lambda(\lambda+\alpha +\beta-\gamma)-0$

that is,
(i)
$1n$

A

$-1\mathrm{I}$,$\gamma-\alpha-\beta$

111 tlue case A
lbis case we have the snmu results as Group 1.
$-\mathfrak{l}\mathrm{I}$.

,$(3\mathrm{S})$

1)$

.

(36)
(37)

(38)

69
(ii)

tbe case$\ln$

In tliis case,$\lambda=\gamma-\alpha-\beta$$\mathrm{s}\mathrm{u}\mathrm{b}\mathrm{s}\mathrm{t}_{1}|\mathrm{t}\mathrm{u}\mathrm{t}\mathrm{i}_{\mathfrak{l}\mathrm{l}}\mathrm{g}$$\lambda=$$\gamma-\alpha-\beta$

into (36) we have$\phi_{2}\cdot(z^{2}-z)+\phi_{1}\cdot\{z(2\gamma -\alpha-\beta+1)-\gamma\}+\phi$

Next operate$N^{v}$$\cdot(\gamma-\alpha)(\gamma-\beta)\approx 11$

.

(36)

to the both sides of (39), we ahve then$\cdot$$\phi_{2+\nu}\cdot(z^{2}-z)+\phi_{1\mathrm{k}\nu}\cdot\{z(2v+2\gamma-\alpha-\beta+1)-v-\gamma\}$$+\phi_{\nu}\cdot\{v^{2}+v(2\gamma-\alpha-\beta)+(\gamma-\alpha)(\gamma-\beta)\}-0$

Here, we choose

$V$

tbat

(40)

such that$v^{2}+v(2\gamma -\alpha -\beta)+(\gamma -\alpha)(\gamma-\beta)-0$$1\mathrm{s}$

.

,

(41)

,

alld

(42),

$v-\alpha-\gamma$$v-\beta$ $-\gamma$

(43)

.

(44)

For tlte case of (42);Substituting (42) into (40), we have1)$\phi_{2\iota a-\prime}\cdot(z^{2}-z)+\phi_{1\iota*-\gamma}\cdot\{z(\alpha-\beta+1)-\alpha\}-0$

Next set$\phi_{1\mathrm{s}u-\prime}\approx$$(\phi-w, -a-1)$$w-\dagger\nu(z)$

,

(45)

we have then(46)$w_{1}\cdot(z^{2}-z)+w$ $\cdot\{z(\alpha-\beta+1)-\alpha\}-0$

from (44). The$\epsilon \mathrm{o}\mathrm{l}\mathrm{u}\mathrm{t}\mathrm{i}\mathrm{c}’ 11$

to this equation is given by

w-Kz
arbitrary constant.

where $K$ is
Hence we have$\mathrm{n}1\tau$$\phi$$(z-1)^{l- 1}$

(47)

),

(48)

(

$-K$ $z^{-a}(z -1)^{l- 1}$from (47) ated (45). Therefore we have$\varphi-K(z-1)^{\prime-\prime\cdot-l}$(z$\cdot${denote))$(z-1)^{fl-1),\Rightarrow\varphi_{(’\}}}-a-1$.

$(z\mathrm{r} 0,1)$(9)

from (48) and (35) wlticlt has
Inversely, (47) satisfies (46), then (48) satisfies (44) clearly. Therefore, (9) satisfies(0), since we have (35).
2) For tlte case of (43);111 tlte same svay (or merely by the change and in (9))we obtain$\lambda-\gamma-\alpha-\beta$$\beta$$\alpha$$(z\# 0,1)$$\varphi-K(z -1)^{\prime-a-l^{1}}(Z^{-n_{(z-1)^{a-1)_{r- l\mathrm{I}-1}\cong\varphi_{(10\}}}}}\cdot$

Moreover, changing the order$z^{-n}$and $(z-1)^{fl- 1}$ in (9) we have other solution$\varphi=K(z-1)’-’\cdot-\beta((z-1)^{fl-1}z^{-u)},-n- 1$

from (9) thenisthe same way we have other solution$\mathrm{w}\mathrm{l}\dot{\mathrm{u}}\mathrm{d}\mathrm{l}$$\ln$$(\gamma-\alpha-1)\not\in \mathrm{a}^{\mathrm{g}}$$\mathrm{d}\mathrm{i}\mathrm{f}\mathrm{f}\iota.\mathrm{r}\mathrm{e}\mathrm{l}\mathrm{t}\mathrm{t}$

(10)

a

$\varphi_{(11)}$$(z\# 1\mathrm{I}, 1)$(11)$(z\sim 0,1)$

(12)

.

$\varphi-K(z-1)^{\gamma-\pi-l(\mathrm{B}-1- l)_{\gamma-\beta-1}\mathrm{r}\varphi_{(12)}}(z-1)z$.
from (40), which is different from (10) when
, we have the following iden tities,
Theorem 2. when,,,,respectivel
Proof. It ls clear because they overlap each other when$(\gamma-\beta-1)\not\in 4^{*}$$\alpha\approx$$\varphi_{(1)}\approx$$\beta$$\varphi_{(2)}$$\varphi_{\{5)}\simeq$ $\varphi_{()}‘$$\varphi_{\{3)}-\varphi_{(4\}}$$a’\iota d$$\varphi_{(?)}\approx$$\varphi_{(\theta)}$$\varphi_{(’)}\mathrm{r}$$\varphi_{(10)}$

(49)

’

$\varphi_{(11)}-\varphi_{(12\}}$,$y$$\alpha-\beta$, respectively

(50)

70
TheOrem3. We have the following identities,

for
for
for

$\varphi_{[1\}}=\varphi_{(3)}$$\varphi_{(2)}\infty$$\varphi_{(4)}$$\varphi_{(S)}=\varphi_{(7)}$and$\varphi_{(6)}\simeq$$\varphi_{\{\mathrm{B})}$$\varphi_{[’)}=$$\varphi_{(11)}$$f_{\mathit{0}\mathfrak{l}}$$\varphi_{(10)}\approx\varphi_{\{12|}$

Proof. Let $u=$$u(z)\in p^{\mathrm{o}}$and$(u\cdot \mathrm{v})_{\nu}\neq(\mathrm{v}\cdot u)_{\nu}$Therefore, we have this theorem dearly. ((51)$(\beta-1)\in \mathrm{Z}_{0}^{\mathrm{k}}$

,

(52)

$(\beta-\gamma)\in \mathrm{Z}_{0}^{\mathrm{k}}$}”$\varphi_{(2\}}\approx\varphi_{\{3)}\approx$$\varphi_{\{5)}-\varphi_{(\mathrm{t})}\sim$

and

for
for

$\varphi_{(4)}$$\varphi_{\{7)}=\varphi_{(8)}$$\varphi_{(’)}-\varphi_{(10)}-\varphi_{(11)}-\varphi_{[12)}$

for

(53)

,

(54)

$(\gamma-\alpha-1)\in \mathrm{a}^{\mathrm{b}}$

,

(55)

$(\gamma-\beta-1)\in \mathrm{Z}_{0}^{*}$

,

(56)

$v\in \mathrm{Z}_{0}^{\star}$$u\sim \mathrm{v}$

,

,
and

VoLl)
Theorem 4. We have the following identities,$\varphi(1$,, we have then

for
for

$(u \cdot \mathrm{v})_{\nu}=(\mathrm{v}\cdot u)_{\nu}$and.

$\mathrm{v}arrow \mathrm{v}(z)\in p^{\mathrm{Q}}$,$(\alpha-\gamma)\in \mathrm{Z}_{0}^{*}$

for
for

respectively.

$(\alpha-1)\in Z_{0}^{+}$$v\not\in \mathrm{Z}_{0}^{*}$

(57)

.

(58)

$[1\mathrm{J}$,,$a-\beta$ ,$\alpha-\beta$$(\alpha-1)\in Z_{0}^{*}$

,

(59)

$\alpha=$$(\alpha-\gamma)\in Z_{\mathit{0}}^{*}$

,

(60)

$\beta$.

$(\gamma-\alpha-1)\in \mathrm{z}_{0}^{\mathrm{s}}$(61)Proof. lt is clear by Theorems 2. and 3.

Chapter 2. More familiar forms of the solutions obtained in
Chapter 1and Rummer’s twenty- four functions\S 1. More familiar forms of the solutions Group Iin Otap. 1, \S 1.

Theorem 1. By Me fractional calculus of products (using the generalized Leibniz$mle$ we have ([6] Vol. 1&[7|)$)$$\varphi_{(1)}\approx$$K(_{Z^{a-r}(z-1)}.\gamma-\beta-1)_{a-1}-z^{1-\gamma}(1-z)_{2}^{\gamma- l- 1}F_{1}(1+\beta-\gamma,$ $1- \alpha;2-\gamma|.\frac{z}{z-1})(1)$$\sim$

for$V_{(20)}$$0,1$ and $|z/(z-1)|<1$ , wher\^e,2
usual Hypergeometric functions of Gauss.
Note. For the notations$(k =1,2, \cdots, 24)$ refer to the list shown in
Proof. We have$|(z^{a-\gamma})_{a-1-n}|<\infty(n \in \mathrm{Z}^{*}\cup\{0\})$$z\neq$$\varphi_{(1)}\approx$$K \sum_{n-0}^{-}\frac{\Gamma(\alpha)}{\Gamma(\alpha-n)\Gamma(n+1)}(z^{a-t})_{a-1-n}((z-1)^{\gamma-\beta-1)_{n}}$$1$$\mathrm{g}2$$V_{(\mathrm{A})}$$(z\neq 0,1)$

is the

$F$

.
(2)
(3)
(4)

under
(5)

71
Therefore, choosing)( $M=$$K=1/M$(6)$e^{\dot{|}\pi(\gamma-a-\beta)}\Gamma(\gamma-1)/\Gamma(\gamma-\alpha)$

we have (1) from (4).
By the change of

$a$

and

in (1), we have$\beta$$z^{1-\prime}(1-z)_{2}^{\gamma-a-1}F_{1}(1+\alpha$$-\gamma$’$1- \beta j2-\gamma j\frac{z}{z-1})=$$V_{(1’)}$.

(7)

Theorem 2. By the fractional calculus of produc , we have$ts$$\varphi_{(3)}=$$K((z-1)’-\beta-1.z^{a-\prime})_{a- 1}$ $-z^{a-}$ ’ $(1-z)_{2}^{\gamma-a-\beta}F_{1}\{$ $1-\alpha$,$\gamma-\alpha;1-\alpha-\beta+\gamma$;$1- \frac{1}{z})$(8)$\approx$$V_{(23)}$$for|((z-1)^{\gamma-\beta-1})_{a-1-n}|<\infty(ll \in \mathrm{Z}^{*}\cup\{0\})$, z$\neq 0$,1

$a’\iota d$$|(z-1)/z|<$ 1.

Proof. We have
(9)
(10)
(11)

under the conditions. Therefore, choosing$K-1/M$

we have (8) from (11).
By the change of and$\alpha$$\beta$

(

),

(12)

$M$ $-e^{l\pi(r\beta-1)}"\Gamma(\alpha+\beta -\gamma )/\Gamma(1+\beta-\gamma)$in (8), we have$z^{\beta-\gamma}(1-z)_{2}^{\gamma-a-\beta}F_{1}(1-\beta,$$\gamma-\beta$; $1-\alpha-\beta+\gamma$ ;$1- \frac{1}{z})-V_{(u)}$.

(1)

Theorem 3. Without the $ue$ of generalized Leibniz rule, we have$\varphi_{(1)}\Leftrightarrow K(_{Z^{a-\prime}(z-1)}.r-\beta-1)_{a-1}=(1-z)^{-\iota_{2}}F_{1}(\gamma-\alpha,$$\beta;\beta-\alpha+1|.\frac{1}{1-z})$

(14)

(15)

(16)
(17)

(18)
(19)

72
$=$ $Ke^{\mathrm{I}\pi(a-\beta-1)_{\frac{\Gamma(\beta)}{\Gamma(\beta-a+1)}(1-z)_{2}^{-\beta}F_{1}}}$($V$$-\alpha$,$\beta$;$\beta-\alpha+1;\frac{1}{1-z}$

),

(20)

under the conditions.
Therefore choosing$K=1/M$

we have (14) from (20).
By the change and$.\alpha$$\beta$(21)$(M=e^{\mathrm{i}\mathrm{n}(a-\beta-1)}\Gamma(\beta)/\Gamma(\beta-\alpha+1))$

in (14), we have$(1-z)^{-\mathrm{n}_{2}}F_{1}(\gamma-\beta$,;$\alpha$$a- \beta+1;\frac{1}{1-z})\approx V_{(11)}$

.

(22)

Theorem 4. Without the use of generalized Leibniz rule, we have$\varphi_{(1)}=$$K(z^{a-\gamma}\cdot(z-1)^{\gamma-\beta-1)_{a-1}=}$ $(1-z)_{2}^{\prime-a-\beta}F_{1}(\gamma-\alpha’ \gamma-\beta;1-\alpha-\beta+\gamma|.1-z)(23)$$arrow V_{(21)}$for$|((1-z)^{k-\beta-1)_{a-1}1<\infty}\star’$$(k\in \mathrm{Z}^{*}\cup\{0\})$, $z\neq 0$ ,1 and$|1$$-z|<1$.

Proof. Using$(|1-z|<1)$(24)

we have
(25)
(26)

(27)
(28)

(29)

under the conditions. Therefore, choosing$K-1/M$$(M-e^{l\pi \mathrm{t}y-\beta-1)}\Gamma(a+\beta-\gamma)/\Gamma(1+\beta-\gamma))$(30)

we have (23) from (29).
in (23) we have (23) itself again.
Theorem 5. Wuhout he use of generalized Leibniz rule, we have
By the change$\alpha$and$\beta$$\varphi_{(1)}-K(z^{\mathrm{n}-\gamma}\cdot(z-1)^{\gamma-\beta-1)_{a- 1}}arrow(-z)_{2}^{-\beta}F_{1}(\beta-\gamma+1,$$\beta_{1}.\beta-a+1;\frac{1}{z})$(21)$-V_{(13)}$$for|(z^{a-\beta- 1-\mathit{1})_{a-1}1<\infty}$(k$\in \mathrm{Z}^{*}\cup\{0\}).$, z$\neq$

0,1 and |z $|>1$ .

Proof. Using the identity$(z -1)^{\lambda}=$$z^{\mathrm{A}}\{$$1- \frac{1}{z})^{\lambda}=$$z^{\mathrm{A}} \sum_{k-0}\cdot\frac{(-1)^{\mathit{1}}\Gamma(\lambda+1)}{\overline{\Gamma(k+1)\Gamma(\lambda+1-k)}}z^{-\mathit{1}}$$(|z|>1 )$

(32)

73
$K(z^{a-\gamma}(z-1)^{\lambda})_{a- 1}$$\varphi_{(1)}=$(33)$(z\neq 0, 1)$$(\lambda\approx \gamma-\beta-1)$$=K \sum_{k-0}^{\mathrm{r}}\overline{\overline{\Gamma(k+1)\Gamma(\gamma-\beta-k)}}((-1)^{k}\Gamma(\gamma-\beta)z^{a-\beta-1-k)_{a-1}}$(33)$\approx-Ke^{-i\pi a}z^{-\beta}\sum\frac{(-1)^{k}\Gamma(\gamma-\beta)\Gamma(\beta+k)}{\Gamma(k+1)\Gamma(\gamma-\beta-k)\Gamma(\beta+1+k-\alpha)}z^{-k}$(35)$-Ke^{j\pi(\beta-a)_{\frac{\Gamma(\beta)}{\Gamma(\beta-\alpha+1)}(-z)_{\iota}^{-\beta}F_{1}(\beta-\gamma+1}}$$\approx$,(35)$\beta;\beta-\alpha+1;\frac{1}{z})$under the conditions. Therfore, choosing$K\approx$

$1/M$

(37)

$(M=-e^{\mathfrak{l}\pi(\beta-a}\mathrm{T}(\beta)/\Gamma(\beta-\alpha+1))$

we have (31) from (36).
By the change of

and$\alpha$$\beta$in (31), we have$(-z)^{-a_{2}}F_{1} \alpha\backslash -\gamma+1(, \alpha_{1}\cdot\alpha-\beta+1;\frac{1}{z})=V_{(’)}$

Theorem 6.
$\varphi_{(1)}=$tlte use of generalized Leipniz rule, we have

$W\iota tl\iota out$$K$($z^{a-\gamma}$.

$|(z^{k+a-\gamma})_{a-1}|<\infty$

(38)

.

; )

(39)

$(|z|<1)$(40)$(z -1)^{t^{-\beta-1)_{a-1}-z_{2}^{1-\prime}F_{1}(\alpha-\gamma+1,\beta-\gamma+1}}.$ $2-\gamma$$\approx$for.

$V_{(17)}$(A EEZ’ $\cup\{0\}$ ), z$z$,$\neq 0,$

1 and $|z|<1$ .

Proof. Using the identity$(z-1)^{\lambda}\approx$$e^{i\mathrm{r}\lambda}(1-z)^{\lambda} \approx e^{\mathrm{i}\pi \mathrm{A}}\sum_{k\cdot 0}^{\sim}\overline{\overline{\Gamma(k+1})\Gamma(\lambda+1-k)}(-1)^{k}\Gamma(\lambda\underline{+1})_{-z^{k}}$we have(A $-\gamma-\beta-1$ )$K(z^{a-\gamma}\cdot(z-1)^{1})_{a-1}$$\varphi_{(1\}}\approx$(41)$(z\sim 0,1)$

(42)
(33)

; Z)

(44)

)

(45)

under the conditions. Therefore, choosing($K-1/M_{*}$

we have (39) from (44).
andBy the change$\alpha$$\beta$$M\propto$ $e^{\mathrm{f}\pi(\gamma-oe-\beta)}\Gamma(\gamma-1)/\Gamma(\gamma-\alpha)$in (39), we have (39) itself again.

52. Commentaries, is equal an integer, each of,
Kummer)(duesatisfies the homogeneous Gauss
the following twenty-four functions
to
(0)
[8].
in
1.
equation \S 1,
Chap.
(I)

When none of the numbers$\gamma$,$\alpha-\beta$$\gamma-\alpha-\beta$$\mathrm{t}\dot{\mathrm{o}}$74
List of the twenty-four functions by Kitmmer$1_{(1)}’=_{2}F_{1}(\alpha, \beta;\gamma;z)$$V_{1}=F_{1}5|2(\alpha, \beta;\alpha+\beta+1 -\gamma;1 -z)$$V_{(2\}}=(1-z)_{\mathit{1}}^{\gamma-a-\beta}F_{1}(\gamma-\alpha, \gamma-\beta|.

$V_{13\mathrm{I}}=(1-z)^{-\alpha_{2}}F_{1}(\alpha,$\gamma;z)$$V_{\{\iota\downarrow}=z^{1-\nu_{1}}\Gamma_{1}.(\alpha+\mathrm{I}$$\gamma-\beta;\gamma|.\frac{z}{z-1})$$V_{11}‘=(1-z)_{l}^{-\beta}F_{1}(\gamma-\alpha,$$V_{\{}7\mathfrak{l}=z^{-\mathrm{r}}\Gamma 2|1(\alpha,$$\alpha+1$$-\gamma j\alpha+1$$V_{(10)}=(-z)^{l-\nu}(\mathrm{I}-z)_{2}^{\gamma-\alpha-\rho}\Gamma_{1}.(\mathrm{I}$$V_{\mathrm{I}11\}}=(1-\mathrm{z})^{-\mathrm{r}_{2}}F_{1}(\alpha,$,$\beta+\downarrow$$-\gamma;\alpha+\beta+\mathrm{I}$$\alpha+1-\gamma j\alpha+\beta+1-\gamma;1$.

$\gamma-\beta|$$-/ \mathit{1}_{j}\frac{\mathrm{I}}{z})$$-/l$ , $\gamma-\beta;\alpha+1-\beta j\frac{1}{z})$$\alpha+1-\beta j\frac{1}{1-z})$.

$V_{\mathrm{t}12\mathrm{I}}=(-z)^{1-\gamma}(1-z)^{\gamma-\mathrm{r}-1}2\Gamma_{1}$ $(\alpha+1-\gamma.$$V_{|131}=(-z)^{-\iota_{2}}F_{1}(\dot{\beta}+1-\gamma,$$\beta;\beta+\mathrm{I}$$1- \beta;\alpha+1-\beta;\frac{1}{1-z})$$- \alpha;\frac{\mathrm{I}}{z})$$r_{|14)}=(-z)^{*-f}(1-z \mathrm{I}’-a-’\Gamma 2^{\cdot}1(1-\alpha, \gamma-\alpha j\beta+1-\alpha j\frac{1}{z})$$\psi_{15)},=(1-z)-r_{\mathit{1}1}l^{\tau}.(/l,$$\gamma-\alpha j/l+\mathrm{I}-\alpha j\frac{1}{1-z})$$V_{1|},‘=(-Z)^{1-}{}^{t}(1-z \mathfrak{l}^{\gamma-\rho-1}\mathrm{z}^{F_{1}(\beta+1-\gamma}, 1-\alpha;\beta+1-\alpha j\frac{1}{1-z})$$V_{(17|}=z^{1-\nu_{2}}F_{1}(\alpha+1-\gamma, \beta+1-\gamma j2-\gamma jz)$

$V_{(1\epsilon \mathrm{I}}=z^{1-r}(1-z)^{\gamma-a-\iota_{2}}F_{1}(1-a, 1-\beta j2-\gamma jZ)$

$V_{(19)}=z^{1-\gamma}(1-\mathrm{z})_{2}^{\gamma-*-1}F_{1}(\alpha+\mathrm{I}$$-\gamma$$V_{\mathrm{t}20)}=z^{1-\nu}(1-z\mathfrak{l}^{r-l-1}2F1(\beta+1-\gamma$$V_{\mathrm{I}21|}=(1-z)’-\cdot-\iota_{2}F_{1}$($\gamma-\alpha,$, $1- \beta j2-\gamma j\frac{z}{z-1})$, I $- \alpha;2-\gamma;\frac{z}{z-1})$$\gamma-\beta j\gamma+1$ $-\alpha-\beta j$

I $-z$)$V_{l2\}},=z^{1-\nu}(1-z)_{2}^{\nu-a-\prime}F_{1}(\mathrm{I}-\alpha, 1-\beta;\gamma+1-\alpha-\beta_{j}1-z)$I$V_{(13)}=z^{a-\gamma}(1-z)^{\gamma-*-\iota_{l}}F_{1}(\gamma-\alpha,$$1-\alpha;\gamma+1-\alpha-\beta;$$V_{\{24\}}=z^{l^{-}\nu}(1-z)^{\gamma-\cdot-\iota_{2}}F_{1}(\gamma-\beta,$$1- \beta_{j}\gamma+1-\alpha-\beta_{j}\mathrm{I}-\frac{1}{z})$$- \frac{1}{z})$.

Moreover, we have the following six identities.
$(\mathrm{i})$(ii)$V_{(1)}arrow V_{(2|}-V_{\beta)}=$$V_{(5)}\sim$$V_{[)}‘\approx$$V_{(4)}$$V_{\{7)}-V_{(\mathfrak{h}}$,,(Hi) $V_{\mathrm{t}’ 1}-V_{(10)}-V_{(11)}-V_{(12)}$ ,$-\gamma i$$- \frac{\mathrm{I}}{z})$$V01=z^{-\prime}12F1(\beta+1-\gamma,$ $\beta;\alpha+\beta+1-\gamma i1-\frac{1}{z})$$\beta j\gamma j\frac{z}{z-\mathrm{I}})$$V_{(9\mathrm{I}}=[-z)^{-a_{2}}\Gamma_{1}.(\alpha,$$-\gamma$(iv)$(\mathrm{v})$(vi)$V_{(13)}\infty$$V_{(14)}=V_{(15)}=$$V_{(1?)}\simeq$$\mathrm{P}_{\acute{(}10\}}-V_{(1)},-V_{(20)}$$V_{(1)}‘$$V_{(21\}}-V_{(22)}-V_{(\mathrm{z}\iota)}-V_{(u)}$

’

,

.

I

75
By our -fractional calculus erator
method to the homogeneous Gauss equation we obtained the solutions shown in Chap. 1, which have the fractional differintegrated forms respectively.
The translations from our solutions Chap.1, Group Ito the more familiar forms which
wn Gauss Hypergeometric functions yield, as we see in Chap. 2, 51,
contain the
,
,
,
,
,
,
,
,
,
(II)

$N^{v}$$\mathrm{N}$$0$$\mathrm{w}\mathrm{e}\mathrm{U}$$\mathrm{k}\mathrm{n}\mathrm{o}$$V_{(20)}$$V_{(1’)}$$V_{(23)}$$V_{(11)}$$V_{(15)}$$V_{(u)}$$V_{(13)}$$V_{\{21)}$$V_{(17\}}$$V_{(’)}$(refer to the list described above).

In the same way as the procedure shown in Chapter 2. \S 1, the transrations from thesolutions of Group yield
,
,
,
,
,
,
,
,
,
’
and the transrations from the solutions of Group III yield
,
.
,
,
,
,
,
,
in the list
Therefore, we see that almost all functions of the group
described above can be derived directly from our solutions of the Groups , IJ and I1I in
Chap. 1, which have fractional differintegrated forms, except only two functions(For the calculations from the solutions of Group II and III, refer toVol.
and10, November 1996, pp 9-23.)$\mathrm{U}$$V_{(3)}$$V_{(1)},$$V_{(4)}$.

$V_{(\mathrm{P})}$$V_{(u\}}$$V_{(7)}$$V_{(\mathrm{d})}$.

$V_{(1)}‘$$V_{(12)}$$V_{(\mathrm{Z}3)}$$V_{\{11)}$$V_{(1S)}$$V_{(’)}$$V_{(22)}$$V_{(5)}$$V_{(10)}$$V_{(1)}$$V_{(13)}$$V_{(14)}$$V_{(18)}$$V_{(1)}\simarrow V_{(u)}$$\mathrm{I}$$V_{(2)}$$V_{()}‘$.

$|\mathrm{F}\mathrm{C}$However, we have the relationships (i) ind (ii) respectively.
from our solutions
Therefore, we can derive the Kummer’s twenty-four$\mathrm{f}\mathrm{u}\mathrm{n}\mathrm{c}\downarrow 1^{1}\mathrm{o}\mathrm{n}\mathrm{s}$in Gap. 1.
That is, the solutions obtained by ouroperator method cover the$N^{\nu}$$\mathrm{K}\mathrm{u}\mathrm{m}\mathrm{m}\mathrm{e}\mathrm{r}.\epsilon$$24$functions.
-ffactlonal calculus, N-transformation(II) AU mathematicians should compare ourto the ordinary and partial differential
and N-fractional calculus operator
calculus,
for example, Riemann-Liouville,
equations with that of other fractional
Weyl, Osier, Oldham and Spanier’s ones.
in(IV) Hitherto, only the solutions of the forms in Chap 1, Group Ihad been treatedthis
However,of fractional calculus to differential
the applicatio
1I1,
insufficient Namely, we must add tlte solutions guch as the foms of Croup Il anddifferentialothersolutionsthetowhidt are shown in Chap. 1, \S 1. For
equations the situations are same.
$\mathrm{N}$$N^{\nu}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{h}\mathrm{o}\mathrm{d}$$\cdot$$\mathrm{e}\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{H}\mathrm{o}\mathrm{n}\epsilon$$\mathrm{n}\mathrm{s}$.

$\mathrm{i}\epsilon$$\mathrm{t}11\epsilon$

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