text:
\section{Introduction}

\addtocounter{sec}{1

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Formulation of problem.}
Consider the space $L^2$ on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ with respect to a positive weight $w(s)\,ds$.
Consider a subspace
$H$ consisting of functions $f(s)$ holomorphic in the strip $-1<\Im s<1$ smooth up to the boundary
$\Im s=\pm 1$ and sufficiently rapidly decreasing in the strip as $|s|\to\infty$.
We consider difference operators in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, w(s)\,ds)$  of the form
$$
\mathcal L f(s)=p(s)f(s+i)+q(s) f(s)+r(s) f(s-i)
,$$
where $i$ is the imaginary unit; the domain of definiteness of $\mathcal L$  is the subspace $H$.
For such operators we discuss essential self-adjointness and the eigenvalue
problem
$$
\mathcal L f(s)=\lambda f(s)
.$$
Our main purpose is spectral decomposition.
In fact, several problems of this kind were solved (see the list below). All
solved problems had the following form. Denote
\begin{equation}
\mu(s)=e^{cs}\frac{\prod_{k=1}^m \Gamma(a_k+ is)}{\prod_{l=1}^n\Gamma(b_l+is)}
\label{eq:mu}
,\end{equation}
where $c\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, and
\begin{equation}
\nu(s)=\overline{\mu(\overline s)}=
e^{c s}\frac{\prod_{k=1}^m \Gamma(\overline a_k- is)}{\prod_{l=1}^n\Gamma(\overline b_l-is)}
\label{eq:nu}
.
\end{equation}
Denote
\begin{align}
A(s):=\frac{\nu(s+i)}{\nu(s)}= e^{-i c}\frac{\prod_{k=1}^m
(\overline a_k- is)}{\prod_{l=1}^n(\overline b_l-is)}
,
\label{eq:A(s)}
\\
B(s):=\frac{\mu(s-i)}{\mu(s)}= e^{ic}\frac{\prod_{k=1}^m
(a_k+ is)}{\prod_{l=1}^n(b_l+is)}
.
\label{eq:B(s)}
\end{align}
We consider the space $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, w(s)\,ds)$ with respect to the weight
$$
w(s)\,ds:= \frac 1{2\pi}\mu(s)\nu(s)
$$
and the difference operator
\begin{equation}
\mathcal L f(s)= A(s)f(s+i)-\bigl(A(s)+B(s)\bigr)f(s)+B(s)f(s-i)
\label{eq:difference-operator}
.
\end{equation}

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Neo-classical orthogonal polynomials}.
Now we enumerate solved problems of this kind.
We use the standard notation for hypergeometric functions
$$
_pF_q\left[\begin{matrix} a_1,\dots,a_q\\ b_1,\dots, b_p \end{matrix}\,;\,z\right]
:=\sum_{n=0}^\infty \frac{(a_1)_n \dots (a_p)_n\, z^n}{(b_1)_n\dots (b_q)_n\, n!}
,$$
where
$(a)_n:=a(a+1)\dots(a+n-1)$ is the Pochhammer symbol.

Recall that there are 3 types of classical hypergeometric orthogonal
 polynomials,
see \cite{AW}, \cite{Koe}, \cite{Koe2}.
Polynomials of the first  type are solutions
 of the usual Sturm--Liouville problems
for second order differential operators: Jacobi (including Gegenbauer, Legendre, Chebyshev),
Laguerre, Hermite systems (see \cite{HTF2}).

Polynomials of
the second type are solutions of
 difference Sturm--Liouville problemы on
lattices: Racah, (Chebyshev)--Hahn, dual Hahn,
Meixner, Krawtchouk, Charlier, see \cite{NUS},  \cite{Koe}, \cite{Koe2}.

Polynomials of the  third type
are solutions of Sturm--Liouville problems  of the form
(\ref{eq:mu})--(\ref{eq:difference-operator}): Wilson, continuous Hahn,
continuous dual Hahn, Meixner--Pollaczek systems,
 see \cite{Koe}, \cite{AAR}.
 Recall that all classical polynomial orthogonal
systems are degenerations of the Wilson polynomials,
see \cite{AW}, \cite{Koe}, \cite{Koe2}.

\smallskip

a) {\it The Meixner--Pollaczek system} or {\it Meixner
polynomials of the second kind}, see \cite{Mei}, \cite{Koe}, Section 1.7. We take
$$
\mu(s)=e^{(\varphi} \def\le{\leqslant} \def\ge{\geqslant-\pi/2)s}\Gamma(a+is)
,$$
where parameters $a$, $\varphi} \def\le{\leqslant} \def\ge{\geqslant$ satisfy $a> 0$, $0<\varphi} \def\le{\leqslant} \def\ge{\geqslant<\pi$.
Therefore
\begin{equation}
w(s)=\frac 1{2\pi}
e^{(2\varphi} \def\le{\leqslant} \def\ge{\geqslant-\pi)s}\Gamma(a+is)\Gamma(a-is)
.
\label{eq:measure-meixner}
\end{equation}
and the difference operator is
\begin{multline}
\mathcal L f(s)
=
ie^{-i \varphi} \def\le{\leqslant} \def\ge{\geqslant}(a-is)f(s+i) + 2(-s\cos\varphi} \def\le{\leqslant} \def\ge{\geqslant+\lambda \sin\varphi} \def\le{\leqslant} \def\ge{\geqslant)f(s)-
\\=
ie^{i \varphi} \def\le{\leqslant} \def\ge{\geqslant}(a+is)f(s-i)
\label{eq:L-meixner}
.
\end{multline}

 The eigenfunctions are polynomials
\begin{align*}
P_n(s)&=\frac{(2a)_n}{n!} e^{in\varphi} \def\le{\leqslant} \def\ge{\geqslant}
 {}_2F_1\left[\begin{matrix} -n,a+is\\ 2a \end{matrix}; 1-e^{-2i\varphi} \def\le{\leqslant} \def\ge{\geqslant} \right],
\\
  \mathcal L P_n(s)&= n\sin \varphi} \def\le{\leqslant} \def\ge{\geqslant\, P_n(s)
.
\end{align*}
Norms of Meixner--Pollaczek polynomials are given by
$$
\|w(s)\|^2:=\int_{-\infty}^{\infty} |p_n(s)|^2 w(s)\,ds=
\frac{\Gamma(n+2a)}{(2\sin\varphi} \def\le{\leqslant} \def\ge{\geqslant)\,n!}
.
$$

Recall (see \cite{HTF1}, formula 1.18(6))  that
\begin{equation}
|\Gamma(a+is)|\sim \sqrt{2\pi} |s|^{a-1/2} e^{-\pi s/2},\qquad s\to\infty
\label{eq:gamma-is}
.
\end{equation}
Therefore the weight $w(s)$  exponentially decreases and the space
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$ contains all polynomials. The operator $\mathcal L$ send
 a polynomial to a polynomial of the same degree, therefore
 our Sturm--Liouville problem is pure algebraic. The same remarks
 hold for 3 polynomial systems discussed below.

\smallskip

b) {\it The continuous Hahn system,} see \cite{Car}, \cite{AW0}, \cite{NUS}, \cite{Koe}.
In this case,
$$
\mu(s)=\Gamma(a+is)\Gamma(b+is)
,$$
where the parameters $a$, $b$ satisfy $\Re a>0$, $\Re b>0$. The eigenfunctions
are polynomials
\begin{align*}
p_n(s)&:=i^n \frac{(a+\overline a)_n(a+\overline b)_n}{n!}\,
{}_3F_2\left[\begin{matrix} -n, n+a+b+\overline a+\overline b,a+is\\
a+\overline a, a+\overline b \end{matrix} ; 1 \right],
\\
\mathcal L p_n&=n(n+a+\overline a+b+\overline b) p_n
.
\end{align*}

\smallskip

c) {\it The continuous dual Hahn system,} see \cite{Wil}, \cite{Koe}, \cite{Koe2}.
In this case
$$
\mu(s)=\frac{\Gamma(a+is)\Gamma(b+is)\Gamma(c+is)}
{\Gamma(2is)}
,$$
where the parameters $a$, $b$, $c$ satisfy
$a>0$, $b>0$, $c>0$ or $a>0$, $\Re b>0$, $c=\overline b$.
We consider {\it even} orthogonal polynomials $p_n(s^2)$:
\begin{align*}
p_n(s^2)&:=(a+b)_n(a+c)_n{}\,\,{}_3F_2\left[\begin{matrix}-n,a+is,a-is\\ a+b,a+c \end{matrix};\,1  \right],
\\
\mathcal L p_n&=np_n
.\end{align*}

{\it d) Wilson system,} see \cite{Wil}, \cite{AAR},
\cite{Koe}, \cite{Ner-wilson}.
   In this case,
$$
\mu(s)=\frac{\Gamma(a+is)\Gamma(b+is)\Gamma(c+is)\Gamma(d+is)}{\Gamma(2is)}
,$$
where
$\Re a$, $\Re b$, $\Re c$, $\Re d>0$
and all parameters are real, or $a$, $b$ are real, $d=\overline c$,
or $b=\overline a$, $d=\overline c$.
Wilson polynomials are {\it even} polynomials given by
\begin{multline*}
P_n(a,b,c,d;s^2)=\\=(a+b)_n (a+c)_n (a+d)_n
\,\,
{}_4F_3 \left[\begin{matrix}
-n,n+a+b+c+d-1, a+is,a-is\\
a+b,a+c,a+d\end{matrix}; 1\right]
.
\end{multline*}
They satisfy to the difference equation
$$
\mathcal L P_n = n(a+b+c+d-1)P_n
.
$$

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Sturm--Liouville problems with continuous spectra.}
I know two solved problems.

\smallskip

a) We consider {\it even} functions $f(s)$ on the line,
and
$$
\mu(s)=\frac{\Gamma(a+is)\Gamma(b+is)}{\Gamma(2is)}
,$$
where $a$, $b>0$. Let $\mathcal L$ be the same as above.

We consider the operator (it is called the inverse Olevsky transform, \cite{Ole},
or the inverse Jacobi transform, \cite{Koo}):
$$
L^2\left({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+, \frac 1\pi
\left|\frac{\Gamma(a+is)\Gamma(b+is)}{\Gamma(2is)}\right|^2 ds\ \right)
\to
L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,x^{a+b-1}(1+x)^{a-b})
$$
defined by
$$
Jf(x)=\frac1{\pi\Gamma(a+b)}\int_0^\infty
{}_2F_1\left[\begin{matrix}a+is,a-is\\a+b \end{matrix}; x\right]
f(x)
\left|\frac{\Gamma(a+is)\Gamma(b+is)}{\Gamma(2is)}\right|^2 ds
.
$$
The $J$ send the difference operator $\mathcal L$ to the operator
$$
Mf(x)=xf(x)
.
$$
See  \cite{Ner-ber}, Theorem 2.1, but this is very special case
of Cherednik, \cite{Che1}.

\smallskip

b) Let
$$
\mu(s)=\frac{\Gamma(a+is)\Gamma(b+is)\Gamma(c+is)}{\Gamma(d+is)\Gamma(2is)}
.
$$
In this case the spectral decomposition was done by an integral operator, whose
kernel is a $_4F_3$-function, see Groenevelt \cite{Gro},
the discrete part of the spectrum was  found in
\cite{Ner-wilson}.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Partially solved problems. Romanovski-type systems of orthogonal polynomials.}
 Romanovski \cite{Rom} constructed orthogonal polynomials
on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ with respect to the weight $(1+ix)^{-a}(1-ix)^{-\overline a}$ on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$
and  with respect the weight $x^{a-1}(1+x)^{-b}$ on $(0,\infty)$.
Since the weights have polynomial decreasing, these
orthogonal  systems are finite.
However, Romanovski polynomials correspond to discrete part
of spectra of certain  Sturm--Liouville problems
(see  \cite{Dunford}, XIII.8, \cite{Koo}, \cite{Ner-double}).

Lesky (see, e.g., \cite{Les1}, \cite{Les2}) constructed numerous Romanovski type polynomial
systems related to difference Sturm--Liouville problems, his list contains
several difference problems in imaginary direction%
\footnote{More generally, Lesky's papers indicate numerous unsolved but
(certainly) solvable Sturm-Liouville
problems.}.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Multidimensional analogs.} See \cite{Che1}, \cite{Che2}.

\smallskip

{\bf \refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Results of the paper.} In Section 2, we show that the operators
 (\ref{eq:mu})--(\ref{eq:difference-operator}) are formally symmetric.
 Next,
we found spectral decomposition
for several operators $\mathcal L$.
In Sections 3, 4 we consider
$$
\mu(s)=\frac 1{\Gamma(is)}
\qquad\text{and}\qquad
\mu(s)=\frac{\Gamma(a+is)}{\Gamma(2is)}
$$
respectively.
In both cases the spectrum is the half-line $\lambda>0$.
The spectral decomposition is given respectively by the
inverse Kontorovich--Lebedev
transform and the inverse Wimp transform with Whittaker kernel.
Note that in a certain sense these problems (involving Bessel functions
$_0F_1$ and Kummer functions $_1F_1$) are simpler than neo-classical
polynomial problems
(involving  the Gauss function $_2F_1$ and higher hypergeometric functions
 $_3F_2(1)$, $_4F_3(1)$).

Next (Section 5), we consider $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ with respect to the measure
\begin{equation}
e^{\pi s}
| \Gamma(\alpha/2+is)|^2\,ds
\label{eq:measure-vil}
\end{equation}
and the difference operator
\begin{equation}
\mathcal L f(s)=i(\alpha/2+is) f(s-i)+ 2 \cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant \,s h(s)- i(\alpha/2-is) f(s+i)
.
\label{eq:L-vil}
\end{equation}
The form of this
 operator slightly differs from  (\ref{eq:mu})--(\ref{eq:difference-operator}).

 In Section 6 we discuss an example of a symmetric
non self-adjoint operator  and its essentially self-adjoint extensions.

\smallskip

In all cases essential self-adjointness is derived from the
explicit spectral decomposition. It is an interesting question
to find a priory proofs.

\smallskip

We also note that the problem (\ref{eq:measure-vil})--(\ref{eq:L-vil})
is an analytic continuation of the Meixner--Pollaszek problem
 (\ref{eq:measure-meixner})--(\ref{eq:L-meixner}).
 The objects of Section 6 also are "analytic continuations
 from integer
 points\footnote{i.e., construction of analytic continuation
  involves the Carlson theorem, see, e.g.,
 \cite{AAR}, Theorem 2.8.1}" of the Meixner--Pollaszek
 polynomials.

\section{Preliminaries}

\addtocounter{sec}{1

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Imaginary shift in $L^2$.}
We say that a function is holomorphic in a closed strip
$|\Im s|\le \alpha$ if it is holomorphic in a larger strip
$|\Im s|<\alpha+\delta.$

\begin{lemma}
\label{l:l-2.1}
Let $H\subset L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ be the subspace in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ consisting of functions $f(s)$ admitting
holomorphic continuation to the strip
$|\Im s|\le 1$ and satisfying the condition
$|f(s)|=O(s^{-1/2-\epsilon})$ in this strip.
The operators
$$
T_+f(s)=f(s+i), \qquad T_-f(s)=f(s-i)
$$
defined on $H$
are symmetric in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$.
\end{lemma}

{\sc Proof.}
$$
\int_{-\infty}^\infty f(s+i)\,\overline{g(\overline s)}\,ds=
\int_{i-\infty}^{i+\infty} f(t) \overline{g(\overline t+i)}\,dt=
\int_{-\infty}^{\infty} f(t) \overline{g(\overline t+i)}\,dt
.$$

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Lemma on symmetry.}
Now let $\mu(s)$, $\nu(s)$ be the same as above, see
 (\ref{eq:mu})--(\ref{eq:nu}). Therefore the weight
$w(s)$ is
$$
w(s)=\mu(s)\nu(s)=\frac 1{2\pi}
e^{2cs}
\frac{\prod_{k=1}^m \Gamma(a_k+ is)\Gamma(\overline a_k+- is)}
{\prod_{l=1}^n\Gamma(b_l+is)\Gamma(\overline b_l-is)}
.
$$
For real $s$ we can represent $w(s)$ in the form
$$
w(s)=\frac 1{2\pi}
e^{2cs}
\left|\frac{\prod_{k=1}^m \Gamma(a_k+ is)}{\prod_{l=1}^n\Gamma(b_l+is)}
\right|^2
.
$$

Let $A(s)$, $B(s)$ be as above
\begin{align*}
A(s)&:=\frac{\nu(s+i)}{\nu(s)}= e^{-i c}\frac{\prod_{k=1}^m
(\overline a_k- is)}{\prod_{l=1}^n(\overline b_l-is)}
\\
B(s)&:=\frac{\mu(s-i)}{\mu(s)}= e^{ic}\frac{\prod_{k=1}^m
(a_k+ is)}{\prod_{l=1}^n(b_l+is)}
.
\end{align*}

By (\ref{eq:gamma-is}), we have the following asymptotics of $w(s)$
 in any strip $|\Im s|<\alpha $
\begin{equation}
w(s)\sim \Psi(s):=\mathrm{const}\cdot |s|^{\sum(2 \Re a_k-1)-\sum(2 \Re b_l-1)}
 \exp\bigl(2cs+(n-m)\pi s\bigr),
\end{equation}
as
$s\to\infty$.
We say that a function $f$ is {\it $w$-decreasing} in
 a strip $|\Im s|\le\alpha$
 if
 $$
 f(s)=O\left(\Psi(s)^{-1/2}
s^{-m-1/2-\epsilon}\right), \qquad s\to\infty
. $$

 This condition provides
 $$
 f(s+i\beta),\,\, A(s)f(s+i\beta),\,\, B(s)f(s+i\beta)\in L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, w(s)\,ds)
 $$
  for all
 $\beta$ satisfying $|\beta|\le \alpha$.
 Denote by $\mathcal H[w]$ the space of all functions holomorphic in the strip
 $|\Im s|\le 1$ and $w$-decreasing in this strip.
 \hfill $\square$

\begin{lemma}
\label{l:l-2.2}
Let
$$
\Re a_j>0
$$
for all $j$.
The operator
$$
Rf(s)=A(s) f(s+i)
$$
defined on the domain $\mathcal H[w]$
is symmetric in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$.
\end{lemma}

{\sc Proof.}
We verify the identity
$\langle Rf,g\rangle=\langle f, Rg\rangle$ for $f$, $g\in \mathcal H[w]$:
\begin{multline*}
\int_{-\infty}^\infty
\frac{\nu(s+i)}{\nu(s)}f(s+i)\,\overline{g(\overline s)} \mu(s)\nu(s)\,ds
=
\int_{-\infty}^\infty
f(s+i)\,\overline{g(\overline s)} \mu(s)\nu(s+i)\,ds
=\\=
\int_{i-\infty}^{i+\infty}
f(s)\,\overline{g(\overline s+i)} \mu(s-i)\nu(s)\,ds=
\int_{-\infty}^\infty
f(s)\,\overline{g(\overline s+i)} \mu(s-i)\nu(s)\,ds=
\\=
\int_{-\infty}^\infty
f(s)\,\frac{\mu(s-i)}{\mu(s)} \overline{g(\overline s+i)}  \mu(s)\nu(s)\,ds
=
\int_{-\infty}^\infty
f(s)\,\overline{\frac{\nu(\overline s+i)}{\nu(\overline s)} g(\overline s+i)}  \mu(s)\nu(s)\,ds
.
\end{multline*}
The condition $\Re a_j>0$ provides absence of poles of
$\nu(s+i)\mu(s)$ in the strip $0<\Im s<1$.

\begin{corollary}
Under the same conditions the operator
$$
\mathcal L f(s)=A(s)f(s+i)-(A(s)+B(s)) f(s)+ B(s) f(s-i)
$$
is symmetric on the subspace $\mathcal H[w]\subset L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$.
\end{corollary}

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Change of a weight.}
Let $w_2(s)=\tau(s)\overline{\tau(\overline s))} w_1(s)$.
Then the operator
$$
Hf(s)=\tau(s) f(s)
$$
is a unitary operator $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w_2(s))\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w_1(s))$.
Evidently,
$$
H^{-1} T_+ H f(s)=
\frac {\tau(s+i)}{\tau(s)} T_+,\qquad
H^{-1} T_- H f(s)=
\frac {\tau(s-i)}{\tau(s)} T_+
.$$

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Operators in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$.}

\begin{lemma}
Let an operator
$$
R f(s)=L(s)f(s+i)
$$
be formally symmetric in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, ds)$. Then
\begin{equation}
L(s)=\overline{L(\overline s-i)}.
\label{eq:funct-eq}
\end{equation}
\end{lemma}

This is straightforward.

Note that if $L(s)$ satisfy (\ref{eq:funct-eq}), then
$L(s)^{-1}$  satisfy the same condition. Also, if $L_1(s)$, $L_2(s)$
satisfy (\ref{eq:funct-eq}), then $L_1(s)L_2(s)$ satisfy (\ref{eq:funct-eq}).

Obvious solutions are
\begin{align*}
L(s)&=i/2+s,
\\
L(s)&=(i/2+ia+s)(i/2-ia+s),
\\
L(s)&=h(e^{2\pi s})
.
\end{align*}

\section{The Kontorovich--Lebedev transform}

\addtocounter{sec}{1

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Difference operator.}
Now $\mu(s)=\Gamma(is)$, $w(s)=|\Gamma(is)|^{-2}$.
 We consider the space of {\it even} functions,
$f(s)=f(-s)$,
the inner product is given by
$$
\langle f,g\rangle=\frac 2 {\pi}\int_{-\infty}^\infty
f(s)\overline{g(\overline s)} \frac {ds}{|\Gamma(is)|^2}
=
\frac2{\pi^2}
\int_{-\infty}^\infty
f(s)\overline{g(\overline s)} s\sinh (\pi s)\,ds
.
$$
We consider a difference operator
$\mathcal L$ given by
\begin{equation}
\mathcal L f(s)= \frac 1{is}\bigl(f(s+i)-f(s-i)\bigr)
\label{eq:cL-KL}
\end{equation}
defined on the subspace $\mathcal H[w]\subset L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,|\Gamma(is)|^{-2}ds)$

\begin{lemma}
\label{l:sa-KL}
 The operator $\mathcal L$ is essentially self-adjoint.
\end{lemma}

The spectral decomposition is given by the inverse Kontorovich--Lebedev transform, see the next
subsection.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The Kontorovich--Lebedev transform. Preliminaries.}
The {\it Macdonald functions} $K_\nu(z)$ are solutions of the  modified
Bessel differential equation  (see \cite{HTF2},7.2(11)), i.e. the equation
$$
\left(z\frac d{dz}\right)^2 g(z)-z^2 g(z)=-\nu^2 g(z)
.
$$
They are defined by (see \cite{HTF2}, 7.2(13)),
$$
K_\nu(z)=
\frac\pi{\sin(\nu \pi)}
(I_{-\nu}(z)-I_{\nu}(z))
,$$
where $I_\nu(z)$ are the modified Bessel functions,
$$
I_\nu(z)=e^{-i\nu\pi/2}
J_\nu(ze^{i\pi/2})=
\sum_{m=0}^\infty \frac{(z/2)^{2m+\nu}}{m!\,\Gamma(m+\nu+1)}
.
$$
For each $z\ne 0$ the function $h_z(\nu):=K_\nu(z)$ is an entire function of the variable
$\nu$,
$$
K_\nu(z)=K_{-\nu}(z)
.
$$
For positive $z\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ and $\nu\in i{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$  values of $K_\nu(z)$ are real.

Below we use two identities (see \cite{Wat}, (3.71.1)--(3.71.2))
\begin{align}
K_{\nu-1}(z)-K_{\nu+1}(z)=-\frac{2\nu}z K_\nu(z)
\label{eq:W1}
,
\\
K_{\nu-1}(z)+K_{\nu+1}(z)=-\frac d{dz}K_\nu(z)
\label{eq:W2}
.
\end{align}

The Kontorovich--Lebedev transform \cite{KoLe}, \cite{Leb}, Section 6.5,
\cite{YaLu} is given by%
\footnote{Here and below we understand  integral operators
in the sense of the kernel theorem, see, e.g., \cite{Her}, Section 5.2.
However, for the Kontorovich--Lebedev transform and the Wimp transform discussed
below conditions of literal validness of formulas are well investigated.}
the formula
\begin{equation}
\mathfrak K g(s)=\int_0^\infty K_{is} (x) g(x)\frac{dx}x
.
\end{equation}

The inverse transform is
\begin{equation}
\mathfrak K^{-1} f(x)=\frac 2\pi \int_0^\infty
f(s) K_{is}(x) \frac{ds}{|\Gamma(is)|^2}
.
\end{equation}
The Kontorovich--Lebedev transform is a unitary operator
$$
L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,x^{-1}dx)\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+, 2\pi^{-1}|\Gamma(is)|^{-2}ds)
.
$$

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The statement.}

\begin{theorem}
The Kontorovich-Lebedev transform provides a unitary equivalence
between the operator
$$
Pg(x)=\frac 2x g(x)
$$
in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,x^{-1}dx)$ and the operator
$\mathcal L$ given by {\rm(\ref{eq:cL-KL})}.
\end{theorem}

{\sc Proof.} We use (\ref{eq:W1}),
\begin{multline*}
\int_0^\infty \frac 2x g(x)\cdot K_{is}(x)\frac {dx}x=
\int_0^\infty  g(x)\cdot \frac{2 K_{is}(x)}{x}\frac {dx}x
=\\=
\int_0^\infty g(x) \frac 1{is} \bigl(K_{i(s+i)}(x)-K_{i(s-i)}(x)\bigr)\frac {dx} x
=\frac 1{is}\bigl(\mathfrak K g(s+i)-\mathfrak K g(s-i)\bigr)
.
\end{multline*}
This proves the statement.

\smallskip

{\sc Remark.} However,  Lemma \ref{l:sa-KL} in this moment is not proved,
it a special case of Lemma \ref{l:sa-W} proved below.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } An additional remark.} Applying
(\ref{eq:W2}), we get the following statement

\begin{proposition}
The Kontorovich--Lebedev transform send
the operator
$$
Qg(x)=\left(\frac {d}{dx}-\frac 1x\right) g(x)
$$
to the operator
$$
\mathcal M f(s)=\frac 12\bigl(f(s+i)-f(s-i)\bigr)
.
$$
\end{proposition}

Therefore, we can evaluate the image of any operator
$x^{-m}\frac {d^n}{dx^n}$ under the Kontorovich--Lebedev transform.

\section{The Wimp transform}

\addtocounter{sec}{1

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Difference problem.}
Now
$
\mu(s)=\frac{\Gamma(1/2-\rho+is)}{\Gamma(2is)}
$,
we consider the space of {\it even} functions on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ with inner product
$$
\langle f,g\rangle =
\frac 1{4\pi}
\int_{-\infty}^{\infty} f(s)\overline{g(\overline{s})}
\left|\frac{\Gamma(1/2-\rho+is)}{\Gamma(2is)}\right|^2
ds
.
$$
We consider the following difference operator
\begin{multline}
\mathcal L f(s)=\frac{1-\rho-is}{(-2is)(1-2is)}f(s+i)-
\\-
\left( \frac{1-\rho-is}{(-2is)(1-2is)}+ \frac{1-\rho+is}{(+2is)(1+2is)}\right) f(s)
+ \frac{1-\rho+is}{(+2is)(1+2is)}f(s-i)
.
\label{eq:cL-wimp}
\end{multline}
As above, this operator is defined on the subspace $\mathcal H[w]\subset L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)ds)$.

\begin{lemma}
\label{l:sa-W}
Let $\rho<1/2$. Then
the operator $\mathcal L$ is essentially self-adjoint.
\end{lemma}

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Whittaker functions and the Wimp transform. Preliminaries.}
{\it Whittaker functions} $W_{\rho,\sigma}(z)$ are versions of confluent hypergeometric
functions. They are solutions of the Whittaker equation (see \cite{HTF1}, 6.1 (4))
\begin{equation}
\left(x^2\frac {d^2}{dx^2} -\frac{x^2}4+\rho x \right) f(x)
=(\sigma^2-1/4) f(x)
.
\label{eq:whitt}
\end{equation}
The explicit expression is
\begin{multline}
W_{\rho,\sigma} (x)=
e^{-x/2}
\Bigl(
\frac{\Gamma(-2\sigma) x^{1/2+\sigma}}{\Gamma(1/2-\rho-\sigma)}\,{}_1F_1\left[
\begin{matrix}1/2-\rho+\sigma\\ 1+ 2\sigma \end{matrix}; x\right]
+\\+
\frac{\Gamma(2\sigma) x^{1/2-\sigma}}{\Gamma(1/2-\rho+\sigma)}\,{}_1F_1\left[
\begin{matrix}1/2-\rho-\sigma\\ 1- 2\sigma \end{matrix}; x\right]
\Bigr)
\label{eq:W-F11}
,\end{multline}
see
\cite{Sla}, (1.9.10).
There are the following integral representations
(see \cite{HTF1}, 6.11(18), \cite{PBM1}, 2.3.6.9),
\begin{equation}
W_{\rho,\sigma}(x)=
\frac{e^{-x/2}x^{\rho}}{\Gamma(1/2-\rho+\sigma)}
\int_0^\infty
e^{-xt} t^{-1/2-\rho+\sigma}(1+t)^{-1/2+\rho+\sigma}dt
.
\label{eq:kummer}
\end{equation}
 and the Barnes representation (see \cite{PBM3}, 8.4.44.3, \cite{Sla},(3.5.16)),
\begin{multline}
W_{\rho,\sigma} (z)=
\frac
{ e^{-x/2}}
{2\pi\Gamma(1/2-\rho-\sigma) \Gamma(1/2-\rho+\sigma)}
\times\\ \times
\int_{-\infty}^\infty
\Gamma(it+1/2+\sigma) \Gamma(it+1/2-\sigma)\Gamma(-\rho-it)
x^{-it}dt
.
\label{eq:W-barnes}
\end{multline}

{\sc Remark.} If $\rho\in {\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$,  $\sigma\in i{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, $x>0$, then
$W_{\rho,\sigma}(x)$ is real. This follows from (\ref{eq:W-F11}).
\hfill $\square$

\smallskip

Fix real $\rho<1/2$. The {\it Wimp transform} $\frW_\rho$ is
the integral operator given by
$$
\frW_\rho g(s)=\int_0^\infty g(x) W_{\rho,is}(x) \frac{dx}{x^2}
$$
(see \cite{Wim}, \cite{YaLu}).
The inverse transform is
$$
\frW_\rho^{-1} f(x)=\frac 1{2\pi} \int_0^\infty f(x) W_{\rho,is}(x)
\left|\frac{\Gamma(1/2-\rho+is)}{\Gamma(2is)}\right|^2
ds
.
$$
The Wimp transform is a unitary operator
$$
L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,x^{-2}dx)\to L^2\left({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,
\frac1{2\pi}\left|\frac{\Gamma(1/2-\rho+is)}{\Gamma(2is)}\right|^2
ds  \right)
.
$$

{\sc Remark.} This theorem can be obtained by writing of explicit spectral decomposition
of the differential operator (\ref{eq:whitt}) as it
is explained in \cite{Dunford}, Chapter XIII.
\hfill $\square$

\smallskip

The Macdonald function $K_\nu$ admits the following expression in the terms
of Whittaker functions:
$$
K_\nu(x)=\sqrt{\frac\pi{2x}} W_{0,\nu}(x)
.
$$
Therefore the Kontorovich--Lebedev transform is a special case of
Wimp transforms.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The statement.}

\begin{theorem}
The Wimp transform send the operator
\begin{equation}
Rg(x)=x^{-1} g(x)
\label{eq:}
\end{equation}
to the difference  operator $\mathcal L$ defined by
{\rm(\ref{eq:cL-wimp})}.
\end{theorem}

Theorem is a corollary of the following lemma.

\begin{lemma}
The Whittaker functions satisfy the difference equation
\begin{multline}
\frac{1-\rho-\sigma}{(-2\sigma)(1-2\sigma) }( W_{\rho,\sigma-1}(x)
-W_{\rho,\sigma})+
\frac{1-\rho+\sigma}{(2\sigma)(1+2\sigma) }( W_{\rho,\sigma+1}(x)
-W_{\rho,\sigma})=
\\=
\frac 1x W_{\rho,\sigma}(x)
\label{eq:diff-eq-W}
.
\end{multline}
\end{lemma}

{\sc Proof.} We use the Barnes integral (\ref{eq:W-barnes}).
We multiply both sides of (\ref{eq:diff-eq-W}) by $e^{x/2}$
and pass to Mellin transforms
(see  below (\ref{eq:mellin})--(\ref{eq:mellin-inverse})).
Denote by $h(t)$ the Mellin transform of $e^{x/2}W_{\rho,\sigma}(x)$, i.e.,
$$
h(t)=
\frac
{\Gamma(it+1/2+\sigma) \Gamma(it+1/2-\sigma)\Gamma(-\rho-it)}
{\Gamma(1/2-\rho-\sigma) \Gamma(1/2-\rho+\sigma)}
.
$$
The Mellin transforms of $e^{x/2}W_{\rho,\sigma\pm1 }(x)$ are
$\gamma_{\pm}(t)h(t)$, where
$$
\gamma_\pm(t)=\frac{(-1/2-\rho\mp\sigma)(t+1/2\pm\sigma)}
{(1/2-\rho\pm\sigma)(t-1/2\mp\sigma)}
.
$$
In the left-hand side we get
 \begin{multline*}
h(s)\cdot\left\{ \frac{1-\rho-\sigma}{(-2\sigma)(1-2\sigma) }(\gamma_-(t)-1)
+
\frac{1-\rho+\sigma}{(2\sigma)(1+2\sigma) }(\gamma_+(t)-1)\right\}
=\\
=h(s)\frac{-t-\rho}{(t-1/2-\sigma)(t-1/2+\sigma)}
=\\=
\frac
{\Gamma(it-1/2+\sigma) \Gamma(it-1/2-\sigma)\Gamma(-\rho-it+1)}
{\Gamma(1/2-\rho-\sigma) \Gamma(1/2-\rho+\sigma)}=
h(s+i)
.
 \end{multline*}
 Shift of a Mellin transform by $i$ is equivalent to multiplication of the original by
 $1/x$.
 \hfill $\square$

 \smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Proof of self-adjointness.}
The space $C^{\infty}_c({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+)$ of smooth functions
with compact support on
 $(0,\infty)$ is a domain of essential self-adjointness
of the operator (\ref{eq:}). It is sufficient to prove
the following lemma.

\begin{lemma}
\label{l:W-image}
$\frW_\rho\bigl(C^{\infty}_c({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+)\bigr)\subset\mathcal H[w]$.
\end{lemma}

\begin{lemma}
\label{l:compact}
Fix $\rho<1/2$.
For $(\sigma,x)$ ranging in a domain
\begin{equation}
 |\Re\sigma|\le 1 , \qquad 0<c\le x\le C< \infty
\label{eq:compact}
\end{equation}
the following uniform estimate holds
\begin{equation}
|W_{\rho,\sigma}(x)|=O\left(  e^{\pi |\Im\sigma|/2}
|\Im\sigma|^{\rho+1}\right)
\label{eq:W-estimate}
.
\end{equation}
\end{lemma}

{\sc Proof of Lemma \ref{l:compact}.}
The integral  formula (\ref{eq:kummer})  converges
if $\Re\sigma>\rho-1/2$ and admits the holomorphic
to the whole plane $\sigma\in\C$.

The statement is very simple if $\rho<-1/2$ (the integral in
(\ref{eq:kummer}) is bounded and the desired estimate is obtained from
an estimate of a pre-integral factor. But we wish to cover also
the interval $-1/2<\rho<1/2$.

Fix $A>B>1$.
Represent $1$ as $1=\varphi} \def\le{\leqslant} \def\ge{\geqslant(t)+\psi(t)$, where $\varphi} \def\le{\leqslant} \def\ge{\geqslant$, $\psi(t)\ge 0$
are smooth nonnegative on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+$, $\psi(t)=0$ for $t<A$, and
$\varphi} \def\le{\leqslant} \def\ge{\geqslant=0$ for  $t>B$.
We write the integral in (\ref{eq:kummer}) as
\begin{multline}
\int_0^\infty=
\int_0^A
e^{-xt} t^{-1/2-\rho+\sigma}(1+t)^{-1/2+\rho+\sigma}\varphi} \def\le{\leqslant} \def\ge{\geqslant(t)\,dt
+\\+
\int_B^\infty
e^{-xt} t^{-1/2-\rho+\sigma}(1+t)^{-1/2+\rho+\sigma}\psi(t)\,dt
.
\label{eq:two-summands}
\end{multline}
The second summand is uniformly bounded in our domain
(\ref{eq:compact}),  the integrand is dominated  by
$$
e^{-ct} t^{1/2-\rho}(1+t)^{1/2+\rho}
$$

Next, we represent the first summand of (\ref{eq:two-summands}) as
$$
\int_0^A=
\int_0^A
 t^{-1/2-\rho+\sigma}
\left(e^{-xt} (1+t)^{-1/2+\rho+\sigma}\varphi} \def\le{\leqslant} \def\ge{\geqslant(t)-1\right)\,dt
+
\int_0^A  t^{-1/2-\rho+\sigma} \,dt
$$
Denote by $Q(t,x,\sigma)$ the first integrand.
Then $|Q(t,x,\sigma)|$ depend on $t$, $x$, $\Re\sigma$,
 these variables
range in a compact set, the function $Q$ is continuous on this set.
Therefore
first summand is uniformly bounded in (\ref{eq:compact}),
 the second
summand is uniformly bounded in (\ref{eq:compact})
outside a neighborhood of
$\sigma=\rho-1/2$.

Thus $\int_0^\infty$ is uniformly bounded in
in (\ref{eq:compact})
outside a neighborhood of
$\sigma=\rho-1/2$.

Next, we  multiply the integral  (\ref{eq:kummer})
 by the pre-integral factor
$\frac{e^{-x/2}x^{\rho}}{\Gamma(1/2-\rho+\sigma)}$.
Since $\Re(1/2-\rho+\sigma)\in (-1/2-\rho,3/2-\rho)$, we have
$$
\Gamma(1/2-\rho+\sigma)^{-1}
= O(e^{\pi |\Im\sigma|/2} |\Im \sigma|^{\rho+1}), \qquad
|\Im \sigma|\to\infty
$$
and we get (\ref{eq:W-estimate}).
\hfill$\square$

\smallskip

{\sc Proof of Lemma \ref{l:W-image}.}
By Lemma \ref{l:compact},
 for a function $f\in C^\infty_c({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+)$ with compact support,
we have
\begin{equation}
|\frW_\rho f(s)| \le C\cdot e^{\pi |\Re s|/2} |\Re s|^{\rho+1/2}
.
\label{eq:estimate}
\end{equation}
Next, we use (\ref{eq:whitt}),
\begin{multline*}
-(1/4+s^2) \frW_\rho f(s)=\int_0^{\infty} (-1/4-s^2) W_{\rho,is}(x))\cdot f(x)\frac{dx}x=
\\=
\int_0^{\infty}\left(x^2\frac {d^2}{dx^2} -\frac{x^2}4+\rho x \right)
W_{\rho,is}(x)\cdot f(x)\frac{dx}x
=\\=\int_0^\infty W_{\rho,is}(x) \cdot
\left[ x\frac{d^2}{dx^2}(x f(x))-\frac14 x^2 f(x)+\rho x f(x)\right]
\frac{dx}x
.
\end{multline*}
We apply (\ref{eq:estimate}) for the function
 in square brackets and get
$$
|\frW_\rho f(s)| \le C\cdot
   (s^2+1/4)^{-1}\cdot e^{\pi |\Re s|/2} |\Re s|^{\rho+1}
$$
and $\frW_\rho f(s)\in \mathcal H[w]$.
\hfill $\square$

\section{The Vilenkin transform}

\addtocounter{sec}{1

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Difference problem.} Fix $\alpha>0$, $\varphi} \def\le{\leqslant} \def\ge{\geqslant>0$.
We consider the weight
$$
w(t)=\frac 1{2\pi}|\Gamma(\alpha/2+it)|^2 e^{\pi t}
,$$
 the corresponding space $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(t)\,dt)$, and the difference operator
$$
\mathcal L f(t)=i(\alpha/2+it) f(t-i)+ 2 \cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant \,t h(t)- i(\alpha/2-it) f(t+i)
.
$$
This operator differs from (\ref{eq:mu})-(\ref{eq:difference-operator}),
 but
it is symmetric (proof is the same as in  Lemma \ref{l:l-2.2}).

\begin{theorem}
The operator $\mathcal L$ is essentially self-adjoint on the space $\mathcal H[w]$.
\label{th:vilenkin-1}
\end{theorem}

\smallskip

{\bf \refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The Vilenkin transform.}

\begin{theorem}
\label{th:vilenkin-2}
 The Vilenkin transform
$$
\frV_\alpha g(t)
= (1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant})^{\alpha/2} e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant it}
\int\limits_{-\infty}^\infty
g(s)\,\,
{}_2F_1 \left[
\begin{matrix} \alpha/2-is,\alpha/2+it\\ \alpha\end{matrix};1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant}  \right]
w(s)\,ds
.
$$
is a unitary operator $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$.
\end{theorem}

This  is  a minor  modification
of Vilenkin \cite{Vil}, \S 7.4, see also \cite{VK}, 7.7.7.

Since the operator $\frV$ is unitary, the inversion formula is
$$
\frV_\alpha^{-1} f(s)
= (1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant})^{\alpha/2}
\int\limits_{-\infty}^\infty
f(t)\,\,
{}_2F_1 \left[
\begin{matrix} \alpha/2+is,\alpha/2-it\\ \alpha\end{matrix};1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant}  \right]
e^{\varphi} \def\le{\leqslant} \def\ge{\geqslant it}
w(t)\,dt
.
$$

\begin{theorem}
\label{th:vilenkin-3}
The inverse Vilenkin transform $\frV_\alpha^{-1}$ send the operator $\mathcal L$
to the operator
$$
\mathcal N f(s)=2s\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant f(s)
.
$$
\end{theorem}

To prove these statements,
 we decompose the Vilenkin transform as a product
of 3 simple transformations, see below formula (\ref{eq:JTJ}).

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Highest weight representations of $\mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$.}
The group $\mathrm {SL}(2,{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ is the group of $2\times 2$ real matrices
$g=\begin{pmatrix}a&b\\c&d  \end{pmatrix}$ with $\det g=1$.
Denote by $\Pi$ the half-plane $\Im z>0$.

Fix $\alpha>0$. Consider the Hilbert space $H_\alpha$ of holomorphic functions
on $\Pi$ determined by reproducing kernel
(see, e.g., \cite{Ner-gauss}, Section 7.1)
$$
K(z,u)=\left(\frac{z-\overline u}{2i}\right)^{-\alpha}
.$$
In other words, denote $\Psi_a(z):=K(z,a)$. Then for any $F\in H_\alpha$
we have
\begin{equation}
\langle F,  \Psi_a\rangle =F(a)
\label{eq:reproducing}
.
\end{equation}
For $\alpha>1$ the inner product in $H_\alpha$ admits
the following
integral representation
$$
\langle F, G\rangle=\mathop {\mathrm {const}}\nolimits(\alpha)
\int_{\Pi} F(z)\overline{G(z)} (\Im z)^{\alpha-2} dz\,d\overline z
.
$$
Consider the following operators
in $H_\alpha$
$$
T_\alpha\begin{pmatrix}a&b\\c&d\end{pmatrix}
F(z)=F\left(\frac{b+zd}{a+zc} \right)(a+zc)^{-\alpha}
.
$$
The function $(a+zc)^{-\alpha}$ is multi-valued. We choose arbitrary branch of
this function on $\Pi$.
Then operators $T_\alpha(g)$ are unitary and satisfy the condition
$$
T_\alpha(g_1)T_\alpha(g_2)=\lambda(g_1,g_2)
T_\alpha(g_1 g_2)
,$$
where $\lambda(g_1,g_2)\in \C$. Thus we get a projective unitary representation
of $\mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$, such representations are called {\it highest weight representations}.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The Mellin transform. Preliminaries.} See, e.g., \cite{Tit}.
 For a function $f$ on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+$ we define a Mellin transform
$\mathfrak M f(s)$ as
\begin{equation}
\mathfrak M f(s)=\int_0^\infty f(x) x^{is-1} dx
.
\label{eq:mellin}
\end{equation}
The inverse transform is given by
\begin{equation}
\mathfrak M^{-1} g(x)=\frac 1{2\pi} \int_{-\infty}^\infty g(s) x^{-is} ds
.
\label{eq:mellin-inverse}
\end{equation}
The Mellin transform is a unitary operator
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+,x^{-1}dx)\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\frac 1{2\pi} ds)$.

Notice, that changing variable $x=e^t$ in (\ref{eq:mellin}), we come to the usual Fourier
transform.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Spectral decomposition of dilatation operators.}
Consider a one-parametric subgroup $A\simeq {\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}_+^\times$
 in $\mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ consisting of matrices
of the form
$D(a)=\begin{pmatrix} a^{1/2}&0\{\ov 0}&a^{-1/2}
\end{pmatrix}$, where $a>0$.
The subgroup $A$ acts in the space $H_\alpha$ by the transformations
\begin{equation}
T_\alpha(D(a))=f(z)
=f(a^{-1}z)a^{-\alpha/2}
\label{eq:D1}
.
\end{equation}
Next, consider the measure $d\mu(s)$ on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ given by
$$
\mu(s)\,ds:=\frac 1{2\pi\Gamma(\alpha)} |\Gamma(\alpha/2+is)|^2\,ds
$$
and the action of the same group in the space
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)$ given by the formula
\begin{equation}
\tau_\alpha(D(a))f(s)=f(s)a^{is}
\label{eq:D2}
.
\end{equation}
Consider the operator
 $J:L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)\to H_\alpha$
given by
\begin{equation}
F(z)=J_\alpha f(z)=\frac{2^{\alpha}}{2\pi\Gamma(\alpha)}
\int_{-\infty}^\infty f(s) \left(\frac zi\right)^{-\alpha/2-is} |\Gamma(\alpha/2+is)|^2ds
\label{eq:mellin-twist}
,
\end{equation}
we choose a branch of $(z/i)^{-\alpha/2-is}=e^{-(\alpha/2+is)\ln(z/i)}$
such  that $\ln z/i$ is real for $z=ip$, $p>0$.

Therefore $F(ip)(p)^{\alpha/2}$ is the inverse Mellin transform of
$\frac {2^\alpha}{\Gamma(\alpha)}f(s)|\Gamma(\alpha/2+is)|^2$.

Applying the direct Mellin transform, we get
\begin{equation}
f(s)\cdot \frac{2^{\alpha}}{\Gamma(\alpha)}|\Gamma(\alpha/2+is)|^2
=\int_0^\infty F(ip)\,p^{\alpha/2+is-1} dp
.
\end{equation}

\begin{proposition}
The transform $J_\alpha$ is a unitary operator
$$J_\alpha:L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)\to H_\alpha$$
intertwining actions {\rm (\ref{eq:D1})} and {\rm (\ref{eq:D2})}.
\end{proposition}

{\sc Proof.} A verification of
$$
J\circ \tau_\alpha(D(a))=T_\alpha (D(a))\circ J
$$
is straightforward.
 Next,
$$
T_\alpha(D(a)) \Psi_i=a^{\alpha/2} \Psi_{a i}
.
$$
By (\ref{eq:reproducing}), the system of vectors $\Psi_{a i}$, where $a>0$
 is total
in the Hilbert space $H_\alpha$.

Next, we consider functions
$$\Phi_a:=a^{\alpha/2+is}$$
in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds$. Then
$$
\tau_\alpha(D(a)) \Phi_1=\Phi_a\cdot a^{\alpha/2}
.
$$
To prove unitarity, it is sufficient to show (see, e.g., \cite{Ner-gauss}, Theorem 7.1.4) that
\begin{align}
 J\Phi_a&=\Psi_{ia}
, \label{eq:JPhi}
 \\
\langle\Phi_a,\Phi_b \rangle_{L^2}&=
\langle \Psi_{ia},\Psi_{ib}\rangle_{V_{\alpha}}=\left(\frac{a+b}2\right)^{-\alpha}
.
\label{eq:PhiPhi}
 \end{align}
 First, note that
 $$
 \int_0^\infty (1+x)^{-\alpha} x^{-is}dx=\mathrm B(is,\alpha-is)=
 \frac{\Gamma(is)\Gamma(\alpha-is)}{\Gamma(\alpha)}
. $$
 Applying the inversion formula for the Mellin transform, we
 get (\cite{PBM3}, 8.5.2.5)
 $$
 \frac 1{2\pi}\int_{-\infty}^\infty \Gamma(is)\Gamma(\alpha-is) x^{is-1}ds=
   {\Gamma(\alpha)} (1+x)^{-\alpha}
 .
 $$
 Both formulas (\ref{eq:JPhi})--(\ref{eq:PhiPhi}) are reduced to the latter integral.
 \hfill $\square$

\smallskip

 {\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Calculation. Proof of Theorem \ref{th:vilenkin-2}}
 Set
 \begin{equation}
 r_\varphi} \def\le{\leqslant} \def\ge{\geqslant:=\frac 1{\sqrt{2\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant}}\begin{pmatrix}
 1&1\\e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}&e^\varphi} \def\le{\leqslant} \def\ge{\geqslant
 \end{pmatrix}\in \mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})
 .
 \label{eq:r-phi}
 \end{equation}

\begin{lemma}
The operator
\begin{multline}
J_\alpha^{-1}T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)J_\alpha f(t)
= (2\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant)^{\alpha/2} e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant(\alpha/2+it)}e^{\pi t/2}
\times\\ \times
\int_{-\infty}^\infty
f(s)\,
{}_2F_1 \left[\begin{matrix} \alpha/2-is,\alpha/2+it\\ \alpha\end{matrix};1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant}  \right]
e^{-\pi s/2}
\mu(s)\,ds
.
\label{eq:JTJ}
\end{multline}
is a unitary operator
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)$.
\end{lemma}

 {\sc Proof.}
 The operator
 $$
 J_\alpha^{-1} T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant) J_\alpha
 $$
 is unitary by definition as a product of 3 unitary operators
 $$
 L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds) \to H_\alpha\to H_\alpha \to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},\mu(s)\,ds)
 .
 $$
 We must find explicit formula for composition.
   Write $J_\alpha$ in the form
 \begin{equation}
 J_\alpha f(z)=e^{i\pi\alpha/4}2^{\alpha}
 \int_{-\infty}^\infty f(s) z^{-\alpha/2-is} e^{-\pi s/2} d\mu(s)
,
\label{eq:Jf}
\end{equation}
we use $(e^{-i\pi/2})^{-is}=e^{-\pi s/2}$.
In this formula we take the branch of
 $z^{-\alpha/2-is}$ given by
\begin{equation}
  z^{-\alpha/2-is}=e^{-(\alpha/2+is)\ln z}
  \label{eq:lnz}
,  \end{equation}
 where the logarithm is real on the semi-axis
$z>0$.
Then the inversion formula is
\begin{equation}
J_\alpha^{-1} F(t)
=
\frac{e^{-i\pi\alpha/4} 2^{-\alpha}\Gamma(\alpha)e^{\pi t/2}}
{\Gamma(\alpha/2+it) \Gamma(\alpha/2-it)}
\int_0^{\infty} F(z) z^{\alpha/2+it-1}ds
\label{eq:J-}
.
\end{equation}

Recall that $r_\varphi} \def\le{\leqslant} \def\ge{\geqslant$ is given by (\ref{eq:r-phi}),
\begin{multline*}
T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)J_\alpha f(z)=
 e^{i\pi\alpha/4}2^{\alpha}(2\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant)^{\alpha/2}
 \times\\ \times
\int_{-\infty}^\infty  \left(\frac {e^\varphi} \def\le{\leqslant} \def\ge{\geqslant z+1}{e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}z+1}\right)^{\alpha/2-is}
(e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}z+1)^{-\alpha} f(s)e^{-\pi s/2}\,d\mu(s)
=
\end{multline*}
\begin{multline}
=\frac{e^{i\pi\alpha/4}2^{\alpha}(2\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant)^{\alpha/2}}
{2\pi\Gamma(\alpha)}
 \times\\ \times
 \int_{-\infty}^\infty  (e^\varphi} \def\le{\leqslant} \def\ge{\geqslant z+1)^{-\alpha/2-is}
(e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}z+1)^{-\alpha/2+is} f(s)\,e^{-\pi s/2}|\Gamma(\alpha+is)|^2ds
\label{eq:TJ}
.
\end{multline}

Next, we apply the inverse transform $J_\alpha^{-1}$,
\begin{multline}
J_\alpha^{-1}T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)J_\alpha f(t)
= \frac{(2\sinh \varphi} \def\le{\leqslant} \def\ge{\geqslant)^{\alpha/2} e^{\pi t/2}}
{\Gamma(\alpha/2+it) \Gamma(\alpha/2-it)}
\times\\ \times
\int_0^\infty \int_{-\infty}^\infty
z^{\alpha/2+it-1} (e^\varphi} \def\le{\leqslant} \def\ge{\geqslant z+1)^{-\alpha/2-is}
(e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}z+1)^{-\alpha/2+is} f(s)
\times\\ \times
e^{-\pi s/2}
|\Gamma(\alpha+is)|^2ds\,dz
.
\label{eq:j-inverse}
\end{multline}
We must evaluate the integral in $z$,
\begin{multline*}
\int_0^\infty z^{\alpha/2+it-1} (e^\varphi} \def\le{\leqslant} \def\ge{\geqslant z+1)^{-\alpha/2-is}
(e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}z+1)^{-\alpha/2+is}\,dz=\\=
e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant(\alpha/2+it)}
\int_0^\infty
u^{\alpha/2+it-1} (1+u)^{-\alpha/2-is}
(1+e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant}u)^{-\alpha/2+is}\,du
=\\=
e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant(\alpha/2+it)}\cdot
\frac{\Gamma(\alpha/2+it)\Gamma(\alpha/2-it)}{\Gamma(\alpha)}
\,\,
{}_2F_1 \left[\begin{matrix} \alpha/2-is,\alpha/2+it\\ \alpha\end{matrix};1-e^{-2\varphi} \def\le{\leqslant} \def\ge{\geqslant}  \right]
,
\end{multline*}
here we applied an integral representation
of the Gauss hypergeometric function,
\begin{equation}
_2F_1
\left[\begin{matrix} a,b\\c\end{matrix}; 1-u \right]
=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}
\int_0^\infty y^{b-1}(1+y)^{a-c}(1+y u)^{-a}dy
\label{eq:F21}
,
\end{equation}
see \cite{HTF1}, 2.12(5); this is valid for $|\arg u|<\pi$.

Thus, we get (\ref{eq:JTJ}).
\hfill $\square$

\smallskip

{\sc Proof of Theorem \ref{th:vilenkin-2}.}
Finally, we change function by the rule
\begin{equation}
\label{eq::}
g(s)=e^{-\pi s/2} f(s)
.
\end{equation}
This is equivalent to passing to the space $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$,
where $w(s)=e^{\pi s} d\mu(s)$.

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  }  Calculations. The difference operator.}
Now we  evaluate the image of the operator
$$
f(s)\mapsto s f(s)

$$
under  $J_\alpha^{-1}T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)J_\alpha$.
Differentiating (\ref{eq:TJ}) by parameter $z$, we get
that  $J_\alpha^{-1}T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)$ send
the operator
$$
f(s)\mapsto -2is\sinh\varphi} \def\le{\leqslant} \def\ge{\geqslant f(s)
$$
to
$$
D:=
(z^2+2z\cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant+1) \frac  d{dz}+\alpha(z+\cosh\varphi} \def\le{\leqslant} \def\ge{\geqslant)
.$$

Next, we evaluate the corresponding operator in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(s)\,ds)$.
First, set
\begin{equation}
g(t)=\int_0^\infty F(z) z^{\alpha/2+it-1}dz, \qquad
h(t)=\frac {g(t)}{\Gamma(\alpha/2+is)\Gamma(\alpha/2-is)}
\label{eq:gh}
\end{equation}
and evaluate
\begin{multline}
\int_0^\infty DF(z) z^{\alpha/2+it-1}dz
=\\=
\int_0^\infty
F'(z)\left(z^{\alpha/2+it+1} + 2\cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant\, z^{\alpha/2+it} +z^{\alpha/2+it-1}
\right)\,dz+\\+ \alpha
\int_0^\infty F(z)\left(z^{\alpha/2+it} +
\cosh z^{\alpha/2+it-1}\right)\,dz
\label{eq:DF}
.
\end{multline}
Next we {\it formally} integrate by parts and come
to
$$
(\alpha/2-it-1)g(t-i)-2\cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant\,t g(t) +(-\alpha/2-it-1)g(t+i)
.
$$
For functions $h\in L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},w(t)\,dt)$ we get the transformation
$$
h(t)\mapsto
(\alpha/2+it) h(t-i)- 2i \cosh \varphi} \def\le{\leqslant} \def\ge{\geqslant \,t h(t)- (\alpha/2-it) h(t+i)
.$$

 {\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Self-adjointness. Proof of Theorem \ref{th:vilenkin-1}.}
  Denote by $W_R$ the space of functions $f(s)$ holomorphic in the strip
\begin{equation}
|\Im s|<R
\label{eq:strip}
\end{equation}
 satisfying the condition:
 for any $A>0$ there is $C$ such that
$$
|f(s)|< C\cdot \exp (-A|\Re s|)
.
$$
The operator $f\mapsto sf(s)$ in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},d\mu(s))$
is essentially self-adjoint
on $W_R$.

Theorem \ref{th:vilenkin-1} is a corollary of the following lemma.

\begin{lemma}
If $R$ is sufficiently large, then for any $f\in W_R$
we have $\frV_\alpha f\in \mathcal H[w]$.
\end{lemma}

 {\sc Proof.}
 Since $f(z)$ super-exponentially decreases,
  $J_\alpha f(z)$, see (\ref{eq:Jf})--(\ref{eq:lnz}), is
a well-defined
 analytic function on the universal covering of $\C\setminus 0$.
In other words, we can assume in (\ref{eq:lnz}) that
 $-\infty<\arg z<+\infty$.
Since $f$ is analytic in the strip, the Fourier transform
of $f$  exponentially
decreases, therefore the Mellin transform decreases as $O(|z|^R)$
 as $|z|\to 0$
and as $O(|z|^{-R})$  as $z\to \infty$ (see \cite{Tit},  Theorem 31),
both $O(\cdot)$ are uniform in any  sector $|\arg z|<C$ with
 finite central angle.

 After the transform $T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)$ we get a function
 $F(z):=T_\alpha(r_\varphi} \def\le{\leqslant} \def\ge{\geqslant)J_\alpha f(z)$ on the universal covering over
 $$
 \C\setminus \{-e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant},-e^{\varphi} \def\le{\leqslant} \def\ge{\geqslant}\}
.
 $$
It has the following behavior near the ramification points:

\smallskip

1. Near $\infty$ the function $F(z)$
 has form $z^{-\alpha}\gamma(1/z)$, where $\gamma$ is
holomorphic near $0$.

\smallskip

2. Near $e^\varphi} \def\le{\leqslant} \def\ge{\geqslant$ we have $F(z)=O(z-e^\varphi} \def\le{\leqslant} \def\ge{\geqslant)^R$.

\smallskip

3. Near $e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant}$ we have  $F(z)=O(z-e^{-\varphi} \def\le{\leqslant} \def\ge{\geqslant})^{R-\alpha}$.

\smallskip

Dominants $O(\cdot)$ are uniform in all sectors
with finite central angles.

Next, we   examine the function $g(t)$ given by (\ref{eq:gh}).
The function $F(z)z^{\alpha/2}$ is holomorphic in the sector
$|\arg z|<\pi$ and admit estimates $O(|z|^{\alpha/2})$ at zero
and $O(|z|^{-\alpha/2})$
at $\infty$. Therefore (see \cite{Tit}, Theorem 31),
 its Mellin transform $g(t)$ is

--- holomorphic in the strip
$|\Im t|<\alpha/2$,

--- decreases as $O(e^{-(\pi-\epsilon) |\Re t|})$ as
$\Re t\to\pm\infty$.

Both consequences are not sufficient for our purposes%
\footnote{If $\alpha\le 2$, then the width of the strip
is not sufficient. }.
For this reason, we improve a behavior of $F(z)$ at zero and at infinity
(in the spirit of Watson's Lemma%
\footnote{See, e.g., \cite{AAR}, Theorem C.3.1}).

 Consider the functions
 \begin{align*}\tau_1(z)&=
 \exp(-z^{1/3})\left(1+z^{1/3}+\frac 1{2!}z^{2/3}\right),
 \\
 \tau_2(z)&=z^{-\alpha}
 \exp(-z^{-1/3})\left(1+z^{-1/3}+\frac 1{2!}z^{-2/3}\right)
 .
 \end{align*}

 \begin{lemma}
 The functions
$$ R(t):= \int_0^\infty \tau_1(z) z^{\alpha/2+it-1}dz,\qquad
Q(t):= \int_0^\infty \tau_2(z) z^{\alpha/2+it-1}dz
$$
are meromorphic in the strip
$$
-\alpha/2-1<\Im t<\alpha/2+1.
$$
A unique singularity of $R(t)$ in the strip is a simple pole at $t=i\alpha$.
A unique singularity of $Q(t)$ in the strip is a simple pole at $t=-i\alpha$.
Both functions admit the following estimate in the strip
\begin{equation}
 O( |t|^{3\alpha/2-1/2} e^{-3\pi |t|/2}), \qquad \Re t\to\pm\infty
.
\label{eq:corr}
\end{equation}
 \end{lemma}

{\sc Proof.}
$$
R(t)=3\left(\Gamma(3\alpha/2+3it)+ \Gamma(3\alpha/2+1+3it)+\frac 12\Gamma(3\alpha/2+2+3it)
\right)
. $$
 Poles of summands are $i\alpha/2$, $i\alpha/2+i/3$, $i\alpha/2+2i/3$, but the last
 two poles cancel.
 \hfill $\square$

\smallskip

Next, consider the function
$$
F^\circ(z):=F(z)-F(0)\tau_1(z)-\bigl(z^\alpha F(z)\bigr)\Bigl|_{z=\infty}
 \cdot \tau_2(z)
.
$$
Denote
$$
g(t)=\int_0^\infty z^{\alpha/2} F(z) z^{is-1}ds,
\qquad
g^\circ(t)=\int_0^\infty z^{\alpha/2} F^\circ(z) z^{is-1}ds,
$$

The  function $z^{\alpha/2}F^\circ(z)$  admits the following expansions
near $0$ and $\infty$
\begin{align}
z^{\alpha/2}F^\circ(z)&=
p_1z^{\alpha/2+1}+ p_2 z^{\alpha/2+2}+p_3 z^{\alpha/2+3} +\dots,
 \quad |z|\to 0,
\label{eq:add1}
\\
  z^{\alpha/2} F^\circ(z)&=
q_1z^{-\alpha/2-1}+ q_2 z^{-\alpha/2-2}+q_3 z^{-\alpha/2-3}
 +\dots,
,\quad  |z|\to\infty
\label{eq:add2}
\end{align}
in the sector $|\arg z|\le\pi$. It is  continuous up to the boundary
of the sector if $R>\alpha$.
The functions
$$
\gamma_\pm(x):=z^{\alpha/2}F^\circ(z)\Bigr|_{z=e^{x\pm i\pi}}
$$
 have $R-\alpha$ derivatives. Expansions
(\ref{eq:add1})--(\ref{eq:add2}) imply the following lemma.

 \begin{lemma}
 \label{l:decreasing}
All derivatives  $\frac{d^k}{dx^k}\gamma_\pm(x)$ tend to zero as  $x\to\pm\infty$.
\end{lemma}

Therefore (see \cite{Tit}, Theorem 31 and proof of Theorem 26),
 $g^\circ(t)$ is holomorphic in the strip
$|\Im t|<\alpha/2+1$ and satisfy the estimate
$$
|g^\circ(t)|=O(e^{-\pi |\Re t|} |\Re t|^{-(R-\alpha)}),
\qquad \Re t \to\pm\infty
.
$$

The function $g(t)$ satisfy the same estimate  (because $g(t)-g^\circ(t)$
is (\ref{eq:corr})) at infinity, but it is meromorphic in the strip with
simple poles at $t=\pm\alpha/2$.

Now it remains to divide%
\footnote{The formula (\ref{eq:j-inverse}) contains also a multiplication by
$e^{\pi t/2}$, but this factor cancels after (\ref{eq::}).}
 $g(t)$ by $\Gamma(\alpha/2+it)\Gamma(\alpha/2-it)$.
Poles at $t=\pm i\alpha/2$ disappear, we get a function holomorphic in the strip
$|\Im t|<1+\alpha/2$, and decreasing as $O(t^{-(R-2\alpha+1)})$.
It remains to choose a sufficiently wide strip (\ref{eq:strip}).
\hfill $\square$

\section{Example of self-adjoint extensions}

\addtocounter{sec}{1

This section  contains another construction in Vilenkin's style, see
 \cite{VK}, Section 7.7.11. A representation-theoretic standpoint
 of our considerations is explained at the end of the section.

 \smallskip

{\bf \refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The  difference operator.}
Consider the space $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ and the subspace
 $\mathcal V$ consisting of functions
holomorphic in the strip $|\Im s|\le 1$ and decreasing as
$$
|f(s)|=O(\Re s)^{-3/2-\epsilon},\qquad |\Re s|\to\infty
.
$$

Fix $\tau\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, $0<\varphi} \def\le{\leqslant} \def\ge{\geqslant<\pi$ and consider the operator
$$
\mathcal L f(s)=i(1/2-is) f(s+i)+ 2(s-\tau)\cos\varphi} \def\le{\leqslant} \def\ge{\geqslant f(s) -i(1/2+is-2it) f(s-i)
.
$$

It is symmetric, see Subsection 2.4.

\begin{proposition}
\label{l:1-1}
The  operator $\mathcal L$ is not self-adjoint. Its defect indices are $(1,1)$.
\end{proposition}

The author does not know are self-adjoint extensions of $\mathcal L$
 natural objects or not.
For this reason we consider another example.

Consider the operator $\mathcal L\oplus\mathcal L$ acting in the space
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},ds)\oplus L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},e^{2\pi s}ds)$.

 Consider the space $\mathcal H$ consisting of pair of functions
$(f_1,f_2)$ meromorphic  in the strip
$|\Im s|\le 1$ such that
\begin{align}
f_1(s)&= O(\Re s)^{-3/2-\epsilon},
\qquad |\Re s|\to\infty
\label{eq:f1f2-1}
\\
 e^{2\pi s} f_2(s)&= O(\Re s)^{-3/2-\epsilon},
\qquad |\Re s|\to\infty
\label{eq:f1f2-2}
.
\end{align}

Fix $\sigma\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$. Consider the space $\mathcal H_\sigma$ consisting of pair of functions
$(f_1,f_2)$ meromorphic  in the strip
$|\Im s|\le 1$ and satisfying (\ref{eq:}), with simple poles at points
$i/2$ and $-i/2+2\tau$. We also require
\begin{align}
\mathop {\mathrm {res}}\nolimits\limits_{s=i/2} f_1(s)&=
\mathop {\mathrm {res}}\nolimits\limits_{s=i/2} f_2(s),
\label{eq:res1}
\\
\mathop {\mathrm {res}}\nolimits\limits_{s=-i/2+2\tau} f_1(s)&=-e^{2\pi(\tau+i\sigma)}
 \mathop {\mathrm {res}}\nolimits\limits_{s=-i/2+2\tau} f_2(s)
\label{eq:res2}
.
\end{align}
The parameter $\sigma$ is present only in the last condition, it is a parameter
of a self-adjoint extension.

\begin{proposition}
{\rm a)} The operator $\mathcal L\oplus \mathcal L$  has defect indices $(2,2)$ on $\mathcal H$.

\smallskip

{\rm b)} The operator $\mathcal L\oplus \mathcal L$ is essentially self-adjoint
 on the domain
$\mathcal H_\alpha$.
\label{l:ss-extension}
\end{proposition}

Next, consider the following elements of the space
$\mathcal H_\alpha$:
\begin{equation}
\left(\Psi_1^{(n)}(s),\Psi_2^{(n)}(s)\right),
\label{eq:Psi-Psi}
\end{equation}
 where both functions $\Psi_1^{(n)}$, $\Psi_2^{(n)}$ are given by the same formula
 $$
 B(1/2+is,1/2+2i\tau-is)
\,\,
 {}_2F_1\left[\begin{matrix}1/2+is,1/2-\sigma+i\tau-n\{\ov 1}+2i\tau \end{matrix}
; 1-e^{-2i\varphi} \def\le{\leqslant} \def\ge{\geqslant}
 \right]
. $$
The function $\Psi_1^{(n)}$ is obtained by analytic continuation of
$$
B(\dots)\,\,
 {}_2F_1\left[\begin{matrix}1/2+is,1/2-\sigma+i\tau-n\{\ov 1}+2i\tau \end{matrix}
; z
 \right]
$$
 from $z=0$ along the path $z=1-e^{-2i\theta}$ with $\theta\in [0,\varphi} \def\le{\leqslant} \def\ge{\geqslant]$;
$\Psi_2^{(n)}$  along the path $z =1-e^{2i\theta}$ with $\theta\in [0,\pi-\varphi} \def\le{\leqslant} \def\ge{\geqslant]$.

\begin{proposition}
\label{pr:basis}
{\rm a) (\cite{Ner-mei})} Elements
$\left(\Psi_1^{(n)}(s),\Psi_2^{(n)}(s)\right)$,
where $n$ ranges in $\Z$, form an orthogonal basis
in the space
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},ds)\oplus L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},e^{2\pi s}ds)$.

\smallskip

{\rm b)} They also are eigenfunctions of the operator
 $\mathcal L\oplus \mathcal L$ defined on $\mathcal H_\alpha$. The eigenvalues
 are $2\sin\varphi} \def\le{\leqslant} \def\ge{\geqslant(\sigma+n)$.
\end{proposition}

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } A family of orthogonal bases in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$.}
Fix $\tau\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, $\sigma\in\C$ and $\varphi} \def\le{\leqslant} \def\ge{\geqslant\in (0,\pi)$.
Define functions
$$\Delta_{\sigma}(x)=\Delta_{\sigma}(x;\tau,\varphi} \def\le{\leqslant} \def\ge{\geqslant)
=
(1+xe^{i\varphi} \def\le{\leqslant} \def\ge{\geqslant})^{-1/2-i\tau-\sigma}
(1+xe^{-i\varphi} \def\le{\leqslant} \def\ge{\geqslant})^{-1/2-i\tau+\sigma}
.
$$
We choose a branch of $\Delta_{\sigma}(x)$
 by the condition $\Delta_{\sigma}(0)=1$.

\begin{lemma}
For any $\tau$, $\sigma\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, the functions
 $\Delta_{\sigma+n}$,
where $n$ ranges in $\Z$, form an orthogonal basis in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$.
\end{lemma}

{\sc Proof.} We pass to a new variable $\theta\in [0,2\pi]$
defined by
$$
e^{i\theta}=\frac{1+e^{i\varphi} \def\le{\leqslant} \def\ge{\geqslant}x}{1+e^{-i\varphi} \def\le{\leqslant} \def\ge{\geqslant}x},
\qquad d\theta=
\frac{2\sin\varphi} \def\le{\leqslant} \def\ge{\geqslant\, dx}{(1+e^{i\varphi} \def\le{\leqslant} \def\ge{\geqslant}x)(1+e^{-i\varphi} \def\le{\leqslant} \def\ge{\geqslant}x)}
.$$
Then
$$
(2\sin\varphi} \def\le{\leqslant} \def\ge{\geqslant)^{1/2+i\tau}\cdot \Delta_{\sigma+n}=
e^{-i(\sigma+n)\theta}
 \theta'(x)^{1/2+i\tau}
$$
We consider the map from $L^2[0,2\pi]$ to
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ given by
$$
S f(x)=f(\theta(x))\theta'(x)^{1/2+i\tau}
$$
Evidently, it is unitary.
The system $\Delta_{\sigma+n}$ is the image of the
complete orthogonal
system $e^{-i(\sigma+n)\theta}$ under the map $S$.
\hfill $\square$

\smallskip

{\bf \refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } A differential operator.}
Fix $\tau\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$, $\varphi} \def\le{\leqslant} \def\ge{\geqslant\in(0,\pi)$. We consider
the following symmetric differential operator
\begin{equation}
D=D_{\tau,\varphi} \def\le{\leqslant} \def\ge{\geqslant}
=i(x^2+2\cos\varphi} \def\le{\leqslant} \def\ge{\geqslant x+1)\frac {d}{dx}+i(1+2i\tau)(x+\cos\varphi} \def\le{\leqslant} \def\ge{\geqslant)
\label{eq:Dtauphi}
\end{equation}
in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},dx/2\pi)$.

The functions $\Delta_{\sigma}$
  are formal eigenfunctions of the operator $D$,
\begin{equation}
D \Delta_{\sigma}(x)= (2\sin\varphi} \def\le{\leqslant} \def\ge{\geqslant)\,\sigma
\Delta_{\sigma}(x)
.
\end{equation}

\begin{lemma}
{\rm a)} Defect indices of the operator $D$ defined on the subspace
$C^\infty_c({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ are $(1,1)$.

\smallskip

{\rm b)} Defect indices of the operator $D$ defined on the subspace
$C^\infty_c\bigl((0,\infty) \bigr)$ are $(1,1)$.
\end{lemma}

{\sc Proof.} Indeed, functions $\Delta_{\sigma}$ are contained in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$
for all $\sigma\in\C$. Therefore, $\dim\ker(D^*\pm i)=1$.
\hfill $\square$

Fix $\sigma\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$.
Denote by $W_\alpha$ the space of $C^\infty$-functions on ${\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$ such that there is
a function $h(y)$ smooth near zero such that
\begin{equation}
f(x)=\begin{cases}
x^{-1-2i\tau} h(1/x),& \text{for sufficiently large positive $x$}
\\
e^{-2\pi i\sigma}(-x)^{-1-2i\tau} h(1/x),& \text{for sufficiently small negative $x$}
\end{cases}
\label{eq:cases-h}
\end{equation}

\begin{lemma}
The operator $D$ is essentially self-adjoint on the subspace $W_\sigma$
and $\Delta_{\sigma+n}$ are its eigenfunctions.
\end{lemma}

{\sc Proof.} Verification of symmetry of $D$ on $W_\sigma$
 is straightforward.
 The subspace $W_\sigma$ contains vectors $\Delta_{\sigma+n}$.
Other functions $\Delta_{\kappa}$ are not in the domain of definiteness
of $D^*$ and therefore defect indices are $(0,0)$.
\hfill$\square$

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The double  Mellin transform.}
Let $f\in L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$. Consider the pair of functions
\begin{align}
g_1(s)&=\int_0^\infty f(x) x^{is-1/2} dx
\label{eq:g1}
,
\\
g_2^\circ(s)&=\int_{-\infty}^0 f(x)(-x)^{is-1/2} dx
\label{eq:g2-circ}
.
\end{align}
Obviously,
$$
\int_{-\infty}^\infty |f(x)|^2 dx=
\frac 1{2\pi}
\left\{
\int_{-\infty}^\infty |g_1(s)|^2ds+
\int_{-\infty}^\infty |g_2^\circ(s)|^2ds
\right\}
.
$$

Thus we get a unitary operator
$L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},dx)\to L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, ds/2\pi)\oplus L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}, ds/2\pi)$.
Let modify this transform and set
\begin{equation}
g_2(s)=\int_{-\infty}^\infty f(x) x^{is-1/2}ds=-i
e^{-\pi s}\int_0^\infty f(x) (-x)^{is-1/2}ds
,
\label{eq:g2}
\end{equation}
here we take a branch of $x^{is-1/2}$ that is analytic in
 the upper half-plane
and real for $x>0$. Now we get
$$
\int_{-\infty}^\infty |f(x)|^2 dx=
\frac 1{2\pi}
\left\{
\int_{-\infty}^\infty |g_1(s)|^2ds+
\int_{-\infty}^\infty |g_2(s)|^2 e^{2\pi s}ds
\right\}
.
$$
We denote the operator $f\mapsto (g_1,g_2)$ by $\widetilde\mathfrak M$

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The difference operator.} We evaluate the $\widetilde\mathfrak M$-image
of $Df$ as in (\ref{eq:DF}) and get the formal difference  operator
$\mathcal L\oplus \mathcal L$ in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},ds)\oplus L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K},e^{2\pi s} ds)$.

 Propositions \ref{l:1-1},
 \ref{l:ss-extension}.a are corollaries of the following lemma.

\begin{lemma}
{\rm a)} The image of $C^\infty_{c}(0,\infty)$ under (\ref{eq:g1}) is contained in
$\mathcal V$.

\smallskip

{\rm b)} The image of $C^\infty_{c}(-\infty,0)+C^\infty_{c}(0,\infty))$
is contained in $\mathcal H$.
\end{lemma}

{\sc Proof.} a) Recall that the Mellin transform of $f$ is reduced
to the Fourier transform by the substitution
$x=e^y$ to $f(x)$. In (\ref{eq:g1})
we evaluate the Fourier transform of
$f(e^y)e^{y/2}$,
the function $g_1(s)$ decreases as $O(s^{-N})$ for any $N$.

  \smallskip

b) We apply the same argument to $g_2^\circ(s)$, see (\ref{eq:g2-circ}).
 After passing to $g_2$ we get the estimate
 (\ref{eq:f1f2-2}).
 \hfill $\square$

\smallskip

Proposition \ref{l:ss-extension}.b is a corollary of the following lemma.

\begin{lemma}
The image of the space $W_\alpha$ under the Mellin transform
$\widetilde\mathfrak M$ is contained in the space $\mathcal H_\alpha$.
\end{lemma}

{\sc Proof.} We  repeat considerations in the spirit
of Watson lemma.
Pass to the function
$$
f^\star(x)=
\begin{cases}
f(x)- f(0)e^{-x}- h(0)x^{-1-2i\tau}e^{-1/x},
 \qquad & x>0
\\
f(x)-f(0)e^{x}-h(0) e^{-2\pi i\sigma} (-x)^{-1-2i\tau} e^{1/x},
\qquad & x<0
\end{cases}
.
$$
where $h$ is the same as in (\ref{eq:cases-h}).
Consider the first component
of the transform $\widetilde\mathfrak M$.

 We have
\begin{align}
f^\star(x)&=  c_1 x+\dots + c_N x^N+ O(x^{N+1}), \qquad
 &x\to 0+
\\
f^\star(x)&= d_1 x^{-2-2i\tau} +\dots + d_M x^{-M-2i\tau}+ O(x^{-M-1}),
\qquad & x\to +\infty
.\end{align}

Examine the behavior of
$$
g_1(s)=\int_0^\infty f(x) x^{is-1/2} dx,
\qquad
g^\star_1(s)=\int_0^\infty f^\star(x) x^{is-1/2} dx
.
$$
Functions $g_1(s)$, $g_1^\star(s)$
are Fourier transforms of $f(e^y)e^{y/2}$, $f^\star(e^y)e^{y/2}$.
It is easy to see that derivatives of $f^\star(e^y)e^{y/2}$
admit estimates
$$
\frac{d^k}{dy^k} \bigl( f^\star(e^y)e^{y/2} \bigr)=
O(e^{-3|y|/2})
.
$$
Therefore $g_1^\star(s)$ is defined in the strip
$|\Im s|<3/2$ and decreases in this strip as $O(|\Re s|^{-N})$
for any $N$.

On the other hand,
$$g_1(s)-g_1^\star(s)=f(0)\Gamma(1/2+is)+h(0)\Gamma(-1/2-2i\tau+is)
$$
is meromorphic in the stir with poles at
$s=i/2$, $s=-i/2+2i\tau$
and
exponentially decreases as $|\Re s|\to\infty$.
The residues at poles are $f(0)$ and $h(0)$ respectively.

In the same way we prove decreasing of $g_2^\circ(s)$ at infinity.
Residues at poles $s=i/2$, $s=-i/2+2i\tau$ are respectively $f(0)$
and $e^{-2\pi\sigma i}$. It remains to multiply
$g_2^\circ(s)$ by $-ie^{-\pi s} h(0)$ and we come
to (\ref{eq:res1})--(\ref{eq:res2}).
\hfill $\square$

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } Proof of Proposition \ref{pr:basis}.}
We evaluate $\widetilde\mathfrak M \Delta_{\sigma+n}$
using the formula (\ref{eq:F21}) and come to (\ref{eq:Psi-Psi}).

\smallskip

{\bf\refstepcounter{punct}{\arabic{sec}.\arabic{punct}.  } The origin of construction of this section.}
Fix $\sigma$, $\tau\in{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K}$.
Consider the following representation
$T_{\tau,\sigma}(g)$
 of
the group $\mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ in $L^2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$,
$$
T_{\tau,\sigma}\begin{pmatrix}a&b\\c&d \end{pmatrix}
f(x)=
f\left(\frac{b+xd}{a+zc}  \right)
(a+zc)^{-1/2-\sigma+i\tau} \overline{\ln(a+zc)}^{\,-1/2+\sigma+i\tau}
$$
In this formula, we choose any branch of $\ln(a+zc)$
that is holomorphic in the upper half-plane
and define powers as
\begin{multline*}
(a+zc)^{-1/2-\sigma+i\tau} \overline{(a+zc)}^{\,-1/2+\sigma+i\tau}
:=\\:=
\exp\left((-1/2-\sigma+i\tau)\ln(a+zc)+ (-1/2+\sigma+i\tau)  \overline{(a+zc)}\right)
\end{multline*}
Thus, an operator $T_{\tau,\sigma}(g)$
 is determined up to a constant factor
and we get a projective unitary representation of
$\mathrm {SL}(2,{\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$ (it is a representation of the principal series,
 see, e.g.,
\cite{Ner-gauss},  Subsection 7.4.3).

The operator $D_{\tau,\varphi} \def\le{\leqslant} \def\ge{\geqslant}$ given by (\ref{eq:Dtauphi})
is an infinitesimal generator of the group $\mathrm {SL}_2({\mathbb R }} \def\C {{\mathbb C }} \def\Z{{\mathbb Z}} \def\H{{\mathbb H}} \def\K{{\mathbb K})$. It generates
a compact subgroup, and $\Delta_{\sigma+n}$ are eigenvectors of this subgroup.

The transform $\widetilde\mathfrak M$ is the
 spectral decomposition of the one-parametric group of operators
 $T_{\tau,\sigma}\begin{pmatrix} a&0\{\ov 0}&a^{-1}  \end{pmatrix}$.

\small

redpajama_set_name: RedPajamaArXiv


text:
STRIKEFORCE WORLD GRAND PRIX HEAVYWEIGHT TOURNAMENT CONTINUES ON JUNE 18; STRIKEFORCE: DIAZ VS. DALEY ON APRIL 9 IS CHAMPIONSHIP
By	OTM News	/	March 4, 2011
NEW YORK (March 2, 2011) – The Strikeforce World Grand Prix – Heavyweight Tournament will continue on Saturday, June 18 at American Airlines Center in Dallas, Texas, LIVE on SHOWTIME® (10 p.m. ET/PT, delayed on the West Coast).
The date originally reserved for continuation of the three-stage, single-elimination tournament comprised of eight of the world's top heavyweights, April 9, will now feature a terrific Strikeforce World Championship Doubleheader from Valley View Casino Center in San Diego, Calif., formerly the San Diego Sports Arena.
In the second event of the tournament, Strikeforce Heavyweight World Champion Alistair "The Demolition Man" Overeem (34-11, 1 NC) will face the only man to tap out MMA legend Fedor "The Last Emperor" Emelianenko, the Brazilian Jiu-Jitsu specialist Fabricio Werdum (14-4-1) and the heavy-handed Brett "The Grim" Rogers (11-2) will take on former UFC World Champion Josh Barnett (29-5).
"Strikeforce is coming off a record presale and impressive attendance for the kickoff of the Grand Prix at IZOD Center in New Jersey that also drew a record viewership on SHOWTIME for live MMA," said Strikeforce CEO Scott Coker. "To build on the tremendous momentum from New Jersey we needed the proper time to promote an event of this magnitude, which is why we have decided to continue the Tournament on June 18 at American Airlines Center in Dallas, a venue that has been identified as one of the premier sports and entertainment venues in the U.S."
Coker continued: "We have a great relationship with the Texas State Athletic Commission, who, along with the American Airlines Center, is enthused to host the Grand Prix."
OTM News

redpajama_set_name: RedPajamaCommonCrawl


Triple rows of glacially cool Ceylon sapphires give this platinum plated sterling silver ring an unmistakable impact. That's only accentuated by the scalloped halos of white zircons that sparkle and glitter around each sapphire. Finally, a cut-out gallery finishes the piece while providing a light, comfortable wear so you can enjoy slipping this ring on your finger at every opportunity.


text:
UNIQUE UNIVERSITY PRESENTS RODNEY SMITH: THE END IS JUST THE BEGINNING
Published: September 02 2009
With a new book soon to be released, Mr. Smith is at a point in his career that is either the beginning or the end.
Join Rodney Smith at Unique University on Thursday, September 17th, from 6-8pm as he highlights the philosophical and aesthetic changes his pictures have undergone, using a slideshow and a lecture that are introspective, chronological journeys through his 40 years of making photographs. He will address the importance of maintaining artistic vision and distinct style while photographing commercially. Spreads from his new book, The End, will also be presented. Then, he'll discuss what, if anything, is next for him.
Clients such as The New York Stock Exchange, American Express, and The New York City Ballet, along with many others, have commissioned Mr. Smith. His editorial clients include The New York Times Magazine, Vanity Fair, W Magazine, and many more while his fashion clients include Ralph Lauren, Neiman Marcus, Ellen Tracy, Bergdorf Goodman, and Saks Fifth Avenue. He has won multiple awards and photography contests. Several galleries have also represented Mr. Smith, including the Gallerie Zur Stockeregg and The Witkin Gallery. The Museum of Fine Arts in Boston, the Yale University Art Gallery, The Philadelphia Museum of Art and the Houston Museum of Fine Arts (among others), have collected Mr. Smith's work.
His first three books, In the Land of Light, The Hat Book, The Book of Books, have all found success. His fourth book, The End, is due for publication in the Fall of 2009.
And as a special exclusive, The End will be available at the seminar for viewing, as well as sale, before it hits the market!
For more information about Rodney Smith's fourth book, The End, Click Here.


text:
icon-circle-arrow-up icon-circle-arrow-right
Diagnosis and Prognosis of Edwards Syndrome
More in Pregnancy Loss
Causes and Risk Factors
Physical Recovery
Coping & Moving Forward
By Krissi Danielsson
laflor/Getty Images
Risk of Recurrence
Facing a Diagnosis
One of the more serious health conditions checked in routine prenatal diagnostic screening is Edwards Syndrome, a chromosomal condition with a devastating prognosis. Edwards Syndrome is also known as Trisomy 18 Syndrome because there are three copies of chromosome 18.
Edwards Syndrome occurs in 1 out of every 6,000 live births and is three times more common in girls, according to the National Institutes of Health (NIH). However, the incidence during pregnancy is much higher because as much as 95 percent of women whose fetus has trisomy 18 have a miscarriage or stillbirth.
Sadly, the NIH explains the outcomes for babies born with Edwards Syndrome are not good, including:
Half of the infants do not survive beyond the first week of life.
Nine out of 10 children will die before their first birthday.
Some children survive to the teenage years, but with serious medical and developmental problems.
The most common form of Edwards Syndrome is full trisomy 18, meaning the baby has three full copies of the 18th chromosome instead of two.
It is also possible to have partial trisomy 18, in which there are two full copies of chromosome 18 and also an additional partial copy.
Still another type is mosaic trisomy 18, meaning there is an effect on some, but not all, cells.
The latter two types are rare compared to full trisomy 18. The prognosis does not differ by type.
The extra copy of chromosome 18 is already present at the time of fertilization and results from random errors in cell division. Trisomy 18 can occur in parents of all age groups, but the risk is highest when moms are older than 35.
An amniocentesis or chorionic villus sampling can test for Edward's Syndrome. The alpha-fetoprotein test cannot confirm (or entirely rule out) trisomy 18 but can merely indicate statistical odds that the baby might have the condition. Some tests are less accurate than others.
Genetic testing is an evolving field and, surely, newer and more accurate tests are on the horizon.
Most of the time, trisomy 18 is a random occurrence due to problems in cell division. In rare cases, parents are carriers for partial trisomy 18 due to a condition called balanced translocation that increases the risk for future pregnancies.
If there is a chance you might be a carrier, your doctor can refer you to a genetic counselor to discuss your options. But most parents who have babies with trisomy 18 are not carriers.
Facing an Edwards Syndrome Diagnosis
An Edwards Syndrome diagnosis is devastating news. Many parents choose to terminate their pregnancies after receiving a confirmation that the baby definitely has trisomy 18, given the high risk of severe health problems and the low odds of the baby surviving infancy. Others decide to continue the pregnancy anyway, either because of beliefs against abortion or because of a mindset that they want to treasure the time with the baby even if it is short.
There is no "correct" choice for what to do in this situation. Parents facing this diagnosis should do what feels right for them.
Get diet and wellness tips delivered to your inbox.
Article Sources
Medline Plus: Trisomy 18 (2016)
Trisomy 18 Foundation: What Is Trisomy 18?
What Does It Mean If Your Baby Has Triploidy?
Are You at a Lower Risk of Miscarriage After a Normal Pregnancy?
Trisomy 16 and Mosaic Trisomy 16 in Pregnancy
Termination of a Desired Pregnancy for Medical Reasons
Aneuploidy Chromosome Abnormality and Miscarriage
Trisomy 9 Is a Rare Chromosome Disorder That Can Be Fatal
Why Is Turner Syndrome Common With Miscarriages or Stillbirth?
Testing for Chromosome Abnormalities After a Miscarriage
Trisomy 22 Is a Common Cause of Early Pregnancy Loss
What Are the Causes of Miscarriages and Stillbirths?
The Rarest and Most Fatal Type of Trisomy
Understanding the Medical Indications for Ending a Wanted Pregnancy
What are the Odds of a Miscarriage Occurring?
Accuracy of Ultrasounds in Diagnosing Birth Defects
Genetic and Environmental Factors May Cause Anencephaly
Can Sperm Quality Issues Cause Miscarriages?


text: Description the town seemed unusually quiet. Except for an occasional yapping dog, the streets were deserted. Peter's heart began racing as Hans turned the team up the canyon road. He knew that home was only minutes away. His excitement v:as shortened, however, when Hans said he had to first stop at his place. Peter thought hi3 older "brother was being awfully mean. But he said nothing. Plans pulled the wagon behind th<? house, handed the reins to Niels, and jumped down. Suddenly a throng of people burst from the barn and house. They were cheering, -¦eter was astonished, but his heart was filled with joy. He savf his two wives and several of his children. He rushed up to then and pulled them into his arms. Then he heard the loud, distinctive voices of the South Ward Choir, and Peter realized that it was more than a dream. He knew that he was home at last. And as Manti's silver band joined in with the strains of the choir, Peter knew without any doubt that he was not yet in heaven. . . still short of perfection, but happy all the sane. This story is based on an incident in the life of Peter Westenskow, the author's second great-grandfather. The following account appeared in the "Home Sentinel" on January 23, 1889: Bro Peter Westenskow returned from Uncle Sam's keeping Monday evening. The Manti silver band and the South Ward choir met Bro. Westenskow at his brother's residence and escorted him home, where a very happy tirae was had, music and singing being the order of the hour. We are glad to see Sro. W. home a^sin." "Life Sketch of Peter Westenskow;" the Jens Veibye Journals; Song of a Century, and family stories. 110


text:
Location London, 16 Feb 2014
London, 16 February 2014
Paul Smith Autumn/Winter 2014 Ready-To-Wear collection
<a href="http://www.vogue.co.uk/person/paul-smith">PAUL SMITH's career has deservedly come under the spotlight as the subject of the Design Museum's current exhibition, Hello My Name Is Paul Smith. That exploration into his past and the label he launched in 1970 clearly got him thinking back to the roots of what he set out to do and today's collection was therefore very much a visual dictionary definition of everything we have come to associate him with.
So it was mannish and effortless with lots of British lounge appeal. He sent out pyjama shirts worn loose over those buttoned up beneath with easy and breezy pyjama bottoms below. He then elaborated on this idea with pyjama all-in-ones that tied, dressing gown-style, at the waist. It was time for bed. But these were clothes very much to be worn out and about, the buzzword the designer kept saying backstage was "wearable".
He had been inspired by wallpaper and tapestry prints for paisley patterns that sprawled vibrantly down cropped trousers, loose and lovely dresses and single-breasted blazers. There were polka dots and pinstripes for masculine suiting and there were colourful bursts of shearling. The styling detail to note was looping a light scarf around your neck and tying it just so, an oversized satchel across the body. Even if you're not a tomboy or of an androgynous sartorial demeanour, Smith always somehow manages to make you wish you were.
By Jessica Bumpus
All seasons Paul Smith
The North Circular
Sofia French
Noe Bernacelli
Jose Zafra
Soojin Lee
Pierre Renaux
Shravan Kumar


text:
Arsenal set to sign £25m Gabriel Magalhaes from Lille this week
Arsenal are set to complete a deal for Gabriel Magalhaes. The Gunners have been in talks with the defender having agreed a fee which is reported to be around £25m, with Lille last week.
Napoli and Everton were also keen on the Brazilian and also had bids accepted by Lille, while Manchester United sounded out Gabriel's representatives last week to explore the possibility making a late move.
But Arsenal have always been confident of getting a deal done for the player and now expect to complete the transfer in the coming days after a breakthrough in talks. Gabriel is now expected to travel to England to sign a five-year deal.
The process of completing the move will be complicated, however, by the fact Gabriel will need to quarantine for two weeks when he arrives from from France – one of the countries from which anyone travelling to the UK need to self isolate for 14 days.
Arsenal will follow Government rules at all times while completing the transfer, with top level sportsmen and women having to obey the same strict guidelines as anyone else when arriving in the country.
Once signed, the central defender will become Arsenal's second signing of the summer, after the addition of Willian, who joined the club on a free transfer from Chelsea.
Gabriel's arrival will be a huge boost to Mikel Arteta, who sees the singing of a left-footed centre-back as a priority this summer window.
Arsenal make late enquiry to hijack Liverpool's bid for Thiago Alcantara
Willian becomes 9th highest paid player in the Premier League with Arsenal move
The Brazilian will join William Saliba at the Emirates, who has already started pre-season training after his loan at Saint-Etienne during the 2019-20 season.
Arsenal believe they have secured two of the most promising youngsters in Saliba and Gabriel as the club look to give Arteta a squad capable of competing for a place in the top-four next campaign.
The gunners will now look to offload some defenders as Arteta now has eight centre-backs currently in his team, with Gabriel and Saliba joined by David Luiz, Holding, Sokratis, Mustafi, Pablo Mari and Calum Chambers.
Posted in Transfer Info, Premier LeagueTagged #Arsenal #Gabriel Magalhaes #Manchester United
Jurgen Klopp held talks over Barcelona job with presidential candidate
Barcelona presidential candidate Jordi Farre has revealed he has held discussions with Liverpool manager Jurgen Klopp regarding taking charge at the Camp Nou in the future. Ronald Koeman was appointed as the new Barcelona head coach earlier this week after Quique Setien was sacked following the club's humiliating 8-2 defeat against Bayern […]


text:
Legea lui Moise este un termen care primul loc în Cartea lui Iosua. Conform tradiției, legea religioasă a fost dată lui Moise pe muntele Sinai, inclusiv Decalogul. În iudaism termenul Tora ("Legea") înseamnă primele cinci cărți ale Bibliei ebraice - Pentateuh.

Referinte

Vechiul Testament
Iudaism
Expresii din Biblie

redpajama_set_name: RedPajamaWikipedia


I have a local spreadsheet file in Excel 2003. I add values to a cell and hyperlink to a website. How can I have that open up the hyperlink in an _existing_ Internet Explorer instance? It always launches a new instance on me. I need it to be an existing instance because all my hyperlinks are on pages that require authentication.
Anne, you said the question was from elsewhere, could I ask which one, as I think I can come up with some kind of solution but need a bit more info first!
Edit: I switched to the code tags, the spacing didn't look right to me with the VBA tags, still doesn't look right, appears to have extra tabs (I.E. 6.0), but it copies and pastes fine.


With his first date already sold out we are very happy to announce that Avi Buffalo will play a second show at St Pancras Old Church this year!
Introducing some new and unreleased song material into his live set, the U.S. based songwriter and guitarist will be returning to London to play two intimate shows. Tickets will be limited, so buy them now for a very memorable musical evening.
Performing will be the lush folk project Lowpines, promoting their new album.
- BOTW presents: Avi Buffalo + Lowpines | St Pancras Old Church Facebook.


SharePoint 2010 offers developers, designers, and users a sophisticated intranet, web application, and Internet site platform. But until now, SharePoint users have had to painfully edit code or seek developer assistance to customize more than just the most minor parts of their sites. All of that has changed with SharePoint Designer 2010. Gain complete control and enhance your SharePoint sites with a depth not previously possible using this book. Pro SharePoint Designer 2010 is your soup-to-nuts reference for unlocking the power of SharePoint Designer.
Covering everything you'll need to know to create custom, rich SharePoint experiences, Pro SharePoint Designer 2010 is a masterful guide to getting the most from this powerful application quickly and easily.


This portrait by Richard was taken at Costume-Con 33 following the 2015 Lifetime Achievement Award presentation. It includes 2015 recipient Kathy Bushman Sanders, and the past LAA recipients who helped present the award that evening.
This photo taken at Costume-Con 31 during the 2013 Lifetime Achievement Award presentation includes past LAA recipients who helped present the two LAA awards that evening to Tina Connell and Dawn McKechnie.


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179562310016935398534541270044324603726886n 1106272810016936465201104821767989068623234n 179562310016935398534541270044324603726886n 1106272810016936465201104821767989068623234n filename amt912jpg emd topdeck. net photo amtk 458 amtrak emd f59phi at orange california by craig walker scenic amtrak pinterest orange california california usa and sc 162612502025af809c83fzjpg. Lionel 6 18314 pennsylvania gg 1 electric locomotive wname plate case nib ron flanary.
hctcvjpg okayso thats 7 registered complaints per year every year for life of fleet dips in years the furthest distance between anniversaries spikes during. Amazoncom northeast corridor route add on online game code video games 179562310016935398534541270044324603726886n 1106272810016936465201104821767989068623234n must see nimble rail worker amtrak emd aem 7ac locomotive change. Hctcvjpg cr 4435 elizabeth nj 278.
Bo emd gp 9 no 6607 baltimore ohio railroad museum 901 west Bo emd gp 9 no 6607 baltimore ohio railroad museum 901 west record breakers achievements trainstation wiki fandom powered by wikia rock island switcher 4902 with caboose at eje yard in joliet illinois in 1975. Some goodies from the other side of the main shop building a hhpacela power truck frame sits in front of a stack of aem7 frames fully ready for assembly image cr 9400 morrisville pa 4 6 1977. Hctcvjpg necup0001 necup0002 necup0004 necup0003.
Washington dc 913 was heavily damaged by fire in november 2000 because it was too warped to repair economically it was scrapped in february 2003 Washington dc 913 was heavily damaged by fire in november 2000 because it was too warped to repair economically it was scrapped in february 2003 norfolk western emd sd 45 1776 in bicentennial paint scheme virginia museum conrail railroad emd e8b diesel electric passenger train locomotive in enola pennsylvania usa model railways pinterest. Nec new york new haven route add on amtrak hhp8 twin pack steam gift ron flanary 179562310016935398534541270044324603726886n 1106272810016936465201104821767989068623234n. Emd e8a loksound select dcc and sound new york central 4041 black white simplified scheme image.
The position light The position light a freshly painted conrail u25b at croxton new jersey in 1980 emd e8a loksound select dcc and sound new york central 4041 black white simplified scheme. Still what to see some modified emd gp79s washington dc 913 was heavily damaged by fire in november 2000 because it was too warped to repair economically it was scrapped in february 2003 washington dc 913 was heavily damaged by fire in november 2000 because it was too warped to repair economically it was scrapped in february 2003. Conrail railroad emd e8b diesel electric passenger train locomotive in enola pennsylvania usa model railways pinterest bo emd gp 9 no 6607 baltimore ohio railroad museum 901 west.
Cr 4435 elizabeth nj 278 Cr 4435 elizabeth nj 278 amtrak aem7 906 helped by aem7 902 pull a southbound train near rahway view deviation. A photo of our train at penn station in new york city cr 9400 morrisville pa 4 6 1977 view deviation. Must see nimble rail worker amtrak emd aem 7ac locomotive change wish.
Emd e8a loksound select dcc and sound new york central 4041 black white simplified scheme Emd e8a loksound select dcc and sound new york central 4041 black white simplified scheme filename amt912jpg. Gm10b emd topdeck still what to see some modified emd gp79s. Another sunday another very late amtrak train o gauge sd60e diesel from mth electric trains.
Picture a southbound train on the penn line stops at baltimore washington international airport this train is operating in push mode with the Picture a southbound train on the penn line stops at baltimore washington international airport this train is operating in push mode with the no automatic alt text available . Amtk 901 rock island switcher 4902 with caboose at eje yard in joliet illinois in 1975 picture two locomotives wait at union station another gp40wh 2 and a gp39h 2 also originally from emd rebuilt by morrison knudsen march 1998. Download attachment amtk 5071jpg view deviation.
Locomotive no 901 leaving washington union station 1980 Locomotive no 901 leaving washington union station 1980 an emd tr1 one of several models of cow calf locomotives csx emd sd40 2 2431 works in the cumberland yard. they also have an original budd metroliner . Will lionel ever make an amtrak genesis o gauge railroading on line forum ron flanary.
Ns emd sw1001 2104 bristol pa darryl rules photography tags Ns emd sw1001 2104 bristol pa darryl rules photography tags n cbq silver streak zephyr starter set here a train headed by a gp40wh 2 locomotive originally built by emd rebuilt by morrison knudsen arrives at washington union station march 1998. Amtrak aem7 906 helped by aem7 902 pull a southbound train near rahway harlem blue harlem 7 mtametronorth harlemline amtrak is retiring the aem 7 electric locomotives that electric locomotive pinterest electric locomotive and locomotive. Amtrak rail diesel car or buddliner 32 represents mid day amtrak train 413 at springfield ma on 28 april 1977 aem7 935 with a eastbound keystone train at gap pa.
Still what to see some modified emd gp79s Still what to see some modified emd gp79s amtrak aem 7 httpwwwnationalcorridorsorgdfdf04042005ajpg the amtrak aem 7. Acs 64 first peak at amtraks newest locomotive archive trainz discussion forums they also have an original budd metroliner amazoncom northeast corridor route add on online game code video games. 915 at bowie dj witty tags electriclocomotive aem7 amtrak railroad passenger emd locomotive 915 at bowie dj witty tags electriclocomotive aem7 amtrak railroad passenger emd locomotive.
Necup0001 necup0002 necup0004 necup0003 Necup0001 necup0002 necup0004 necup0003 hctcvjpg emd electro motive division locomotive wiki fandom powered by wikia. Some goodies from the other side of the main shop building a hhpacela power truck frame sits in front of a stack of aem7 frames fully ready for assembly they also have an original budd metroliner septa warminster train 6215 with aem 7. Necup0001 necup0002 necup0004 necup0003 woodlandscenicsoscalejfranksgrocery.
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wish. Amtrak is retiring the aem 7 electric locomotives that electric locomotive pinterest electric locomotive and locomotive gm10b aem7 935 with a eastbound keystone train at gap pa. Image conrail railroad emd e8b diesel electric passenger train locomotive in enola pennsylvania usa model railways pinterest.
Amtrak rail diesel car or buddliner 32 represents mid day amtrak train 413 at springfield ma on 28 april 1977 Amtrak rail diesel car or buddliner 32 represents mid day amtrak train 413 at springfield ma on 28 april 1977 emd f7a dcdccsound alaska railroad emd e8a loksound select dcc and sound new york central 4041 black white simplified scheme. Septa warminster train 6215 with aem 7 wish bo sd 35 no 7402 baltimore ohio railroad museum 901 west pratt. Click here to see the farewell to conrail office car special ocs 101 it is approaching belmont and will proceed to the amtrak 30th st station in phila emd topdeck.
Download attachment amtk 5071jpg Download attachment amtk 5071jpg image. Necup0001 necup0002 necup0004 necup0003 another sunday another very late amtrak train washington dc 913 was heavily damaged by fire in november 2000 because it was too warped to repair economically it was scrapped in february 2003. amtrak 927 on 186 wilmington 2015 07 08 redheadedrobbie1 tags.
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ns emd sw1001 2104 bristol pa darryl rules photography tags image. Picture two locomotives wait at union station another gp40wh 2 and a gp39h 2 also originally from emd rebuilt by morrison knudsen march 1998 another sunday another very late amtrak train view deviation. a close up head on view of santa cruz big trees and pacific railway 2641 as it stops at the santa cruz beach boardwalk in the summer of 1993.
Love Love caseybnsftrains 1 1 aem 7 base by caseybnsftrains here a train headed by a gp40wh 2 locomotive originally built by emd rebuilt by morrison knudsen arrives at washington union station march 1998. Woodlandscenicsoscalejfranksgrocery amtrak aem7 906 helped by aem7 902 pull a southbound train near rahway 179562310016935398534541270044324603726886n 1106272810016936465201104821767989068623234n. Norfolk western emd sd 45 1776 in bicentennial paint scheme virginia museum the position light.
Emd aem 7. 12379747821d2431586cbjpg record breakers achievements trainstation wiki fandom powered by wikia. Your digital image here history csx emd sd40 2 2431 works in the cumberland yard. Filename amt912jpg norfolk western emd sd 45 1776 in bicentennial paint scheme virginia museum.