Quest articol chi l'è scrivuu in Koiné occidentala .
Ul A.Hurwitz al a pusaa la quistiun si al füdess pussíbil che una séria da puteenz
h
(
ξ
)
=
∑
k
=
0
∞
a
k
(
ξ
−
ξ
0
)
k
,
{\displaystyle h(\xi )=\sum _{k=0}^{\infty }a_{k}(\xi -\xi _{0})^{k},}
representaant una funziun difereent da
ξ
↦
c
e
ξ
{\displaystyle \xi \mapsto ce^{\xi }}
,
l'ametess cuntinuazziun analítica luungh un camin saraa inturna a
ξ
0
{\displaystyle \xi _{0}}
e, a la fin da la cuntinuazziun, la tuless la furma:
∑
k
=
1
∞
k
a
k
(
ξ
−
ξ
0
)
k
−
1
=
h
′
(
ξ
)
,
{\displaystyle \sum _{k=1}^{\infty }ka_{k}(\xi -\xi _{0})^{k-1}=h^{\prime }(\xi ),}
vargott a dí: sa pöö-la cuntinuá analiticameent una funziun ulumorfa
veers la suva derivada ?
Ul H.Lewy al a respundüü afermativameent, e al a daa una sulüzziun dal prublema che presentemm chí int una furma ligerameent mudifegada(vidée A.Naftalevich :
On a differential-difference equation , The Michigan Mathematical Journal, 22 (1975)).
Sa cunsideri la funziun
h
(
z
)
=
∫
R
+
exp
[
−
z
t
−
(
log
t
)
2
/
4
π
e
]
d
t
;
{\displaystyle h(z)=\int _{\mathbb {R} ^{+}}\exp \left[-zt-(\log t)^{2}/4\pi e\right]\,dt;}
h
{\displaystyle h}
a l'è ulumorfa par
ℜ
(
z
)
>
0
{\displaystyle \Re (z)>0}
e la pöö vess cuntinuada analiticameent
aj semipian
ℜ
(
z
e
−
e
ϑ
)
>
0
(
ϑ
∈
R
+
)
{\displaystyle \Re (ze^{-e\vartheta })>0\ (\vartheta \in \mathbb {R} ^{+})}
,
da la manera segueent:
al síes
N
∈
N
{\displaystyle N\in \mathbb {N} }
taal che
0
<
ϑ
/
N
<
π
/
2
{\displaystyle 0<\vartheta /N<\pi /2}
e femm
η
:=
ϑ
/
N
{\displaystyle \eta :=\vartheta /N}
.
Scrivemm, par
z
∈
{
ℜ
(
z
e
−
e
η
)
>
0
}
⋃
{
ℜ
(
z
)
>
0
}
{\displaystyle z\in \{\Re (ze^{-e\eta })>0\}\bigcup \{\Re (z)>0\}}
,
h
(
z
)
=
∫
R
+
e
x
p
[
z
e
−
i
η
e
i
η
t
−
log
(
e
−
i
η
e
i
η
t
)
2
4
π
i
]
d
t
{\displaystyle h(z)=\int _{\mathbb {R} ^{+}}exp\left[ze^{-i\eta }e^{i\eta }t-{\frac {\log(e^{-i\eta }e^{i\eta }t)^{2}}{4\pi i}}\right]\,dt}
=
∫
e
i
η
R
+
e
x
p
[
−
z
e
−
i
η
u
−
(
log
(
u
)
−
i
η
)
2
4
π
i
]
e
−
i
η
d
u
{\displaystyle =\int _{e^{i\eta }\mathbb {R} ^{+}}exp\left[-ze^{-i\eta }u-\displaystyle {\frac {(\log(u)-i\eta )^{2}}{4\pi i}}\right]e^{-i\eta }\,du}
=
lim
R
→
∞
{
∫
0
R
e
x
p
[
−
z
e
−
i
η
u
−
(
log
(
u
)
−
i
η
)
2
4
π
i
]
e
−
i
η
d
u
+
{\displaystyle =\lim _{R\to \infty }\left\{\int _{0}^{R}exp\left[-ze^{-i\eta }u-\displaystyle {\frac {(\log(u)-i\eta )^{2}}{4\pi i}}\right]e^{-i\eta }\,du+\right.}
+
∫
γ
R
e
x
p
[
−
z
e
−
i
η
u
−
(
log
(
u
)
−
i
η
)
2
4
π
i
]
e
−
i
η
d
u
}
.
{\displaystyle \ \qquad \left.+\int _{\gamma _{R}}exp\left[-ze^{-i\eta }u-\displaystyle {\frac {(\log(u)-i\eta )^{2}}{4\pi i}}\right]e^{-i\eta }\,du\right\}.}
Chesta darera integrala,
che numinemnm
e
2
{\displaystyle e_{2}}
,
la gh'a da vess calcülada sura la cürva
γ
R
:
[
0
,
1
]
→
C
{\displaystyle \gamma _{R}:[0,1]\rightarrow \mathbb {C} }
definida par
γ
(
t
)
:=
R
e
i
t
η
{\displaystyle \gamma (t):=Re^{it\eta }}
.
A emm
e
2
≤
C
1
R
α
e
−
C
2
R
{\displaystyle e_{2}\leq C_{1}R^{\alpha }e^{-C_{2}R}}
par di custaant reaal pusitiiv
C
1
{\displaystyle C_{1}}
,
C
2
{\displaystyle {C_{2}}}
e
α
{\displaystyle {\alpha }}
, dunca
e
2
{\displaystyle e_{2}}
al teent a
0
{\displaystyle 0}
quan
R
→
∞
{\displaystyle R\to \infty }
.
Inscí, par
z
∈
{
ℜ
(
z
e
−
i
η
)
>
0
}
⋂
{
ℜ
(
z
)
>
0
}
{\displaystyle z\in \{\Re (ze^{-i\eta })>0\}\bigcap \{\Re (z)>0\}}
hom ha
h
(
z
)
=
∫
R
+
e
x
p
[
−
z
e
−
i
η
u
−
(
log
(
u
)
−
i
η
)
2
4
π
i
]
e
−
i
η
d
u
;
{\displaystyle h(z)=\int _{\mathbb {R} ^{+}}exp\left[-ze^{-i\eta }u-{\frac {(\log(u)-i\eta )^{2}}{4\pi i}}\right]e^{-i\eta }\,du;}
però
chesta darera integrala la cunveerg in
ℜ
(
z
e
−
e
η
)
>
0
{\displaystyle \Re (ze^{-e\eta })>0}
e dunca la ga definiss una cuntinuazziun analítica da
h
{\displaystyle h}
.
Repetemm ul prucedimeent
N
{\displaystyle N}
vöölt: vargott al na dà finalameent una cuntinuazziun
analítica da
h
{\displaystyle h}
al semiplan
ℜ
(
z
e
−
e
ϑ
)
>
0
{\displaystyle \Re (ze^{-e\vartheta })>0}
; dunca
h
{\displaystyle h}
la pöö vess cuntinuada analiticameent a tücc i puunt
p
∈
C
∖
{
0
}
{\displaystyle p\in \mathbb {C} \setminus \{0\}}
.
Finalmeent, si femm la cuntinuazziun analítica luungh ul camin
|
z
|
=
1
,
0
≤
arg
(
z
)
≤
2
π
{\displaystyle \vert z\vert =1,0\leq \arg(z)\leq 2\pi }
, utegnemm, designaant
h
^
{\displaystyle {\widehat {h}}}
l'element da funziun ulumorfa
utegnüü (int un intuurn da
z
=
1
{\displaystyle z=1}
) dapress una girada cumpleta,
h
^
(
z
)
=
∫
R
+
exp
[
−
e
2
π
e
z
t
−
(
log
t
+
2
π
e
)
2
/
4
π
e
]
d
t
=
{\displaystyle {\widehat {h}}(z)=\int _{\mathbb {R} ^{+}}\exp \left[-e^{2\pi e}zt-(\log t+2\pi e)^{2}/4\pi e\right]\,dt=}
=
∫
R
+
exp
[
−
z
t
−
(
log
t
)
2
−
4
π
2
+
4
π
e
log
t
4
π
e
]
d
t
=
{\displaystyle =\int _{\mathbb {R} ^{+}}\exp \left[-zt-\displaystyle {\frac {(\log t)^{2}-4\pi ^{2}+4\pi e\log t}{4\pi e}}\right]\,dt=}
=
∫
R
+
exp
[
−
z
t
−
e
2
π
e
t
−
(
log
t
)
2
/
4
π
e
−
π
e
+
log
t
]
d
t
=
{\displaystyle =\int _{\mathbb {R} ^{+}}\exp \left[\displaystyle -zt-e^{2\pi e}t-(\log t)^{2}/4\pi e-\pi e+\log t\right]\,dt=}
=
∫
R
+
(
−
t
)
exp
[
−
z
t
−
(
log
t
)
2
/
4
π
e
]
d
t
=
h
′
(
z
)
.
{\displaystyle =\int _{\mathbb {R} ^{+}}(-t)\exp \left[-zt-(\log t)^{2}/4\pi e\right]\,dt=h^{\prime }(z).}
Vargott al finiss la presentazziun da la sulüzziun da cheest prublema.