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\begin{document}
\title[Trace formulas]{Higher order trace formulas of the Buslaev-Faddeev
type for the half-line Schr\"{o}dinger operator with long-range potentials}
\date{Dec, 2002}
\author{S.M. Belov and A.V. Rybkin}
\address{Department of Mathematical Sciences\\
University of Alaska Fairbanks\\
PO Box 756660\\
Fairbanks, AK 99775}
\email{ftsmb@uaf.edu}
\address{Department of Mathematical Sciences\\
University of Alaska Fairbanks\\
PO Box 756660\\
Fairbanks, AK 99775}
\email{ffavr@uaf.edu}
\date{\today}
\subjclass{Primary 34L40; Secondary 47A55,47E05,81Q10}
\keywords{trace formulas, scattering phase, WKB asymptotics}
\maketitle
\begin{abstract}
We deal with trace formulas for half-line Schr\"{o}dinger operators with
long-range potentials. We generalize the Buslaev-Faddeev trace formulas to
the case of square integrable potentials. The exact relation between the
number of the trace formulas and the number of integrable derivatives of the
potential is also given. The main results are optimal.
\end{abstract}
\section{Introduction.}
The present paper continues to study trace formulas for half-line
Schr\"odinger operators with long-range potentials started previously by one
of the authors in \cite{RyTr}. The type of formulas we are concerned with
was discovered fifty years ago by Gel'fand-Levitan in their remarkable note
\cite{Gelfand} where it was shown that for the eigenvalues $\left\{ \lambda
_{n}\right\} $ of the regular Sturm-Liouville problem $\left( 0\leq x\leq
\pi \right) $
\begin{equation*}
\left\{
\begin{array}{c}
-u^{\prime \prime }+q\left( x\right) u=\lambda u \\
u\left( 0\right) =0=u\left( \pi \right) ,%
\end{array}%
\right.
\end{equation*}%
the following holds:%
\begin{equation}
\sum_{n\geq 1}\left\{ \lambda _{n}-n^{2}-\frac{1}{2\pi }\int_{0}^{\pi
}q\left( x\right) dx\right\} =-\frac{q(0)+q\left( \pi \right) }{4}+\frac{1}{%
2\pi }\int_{0}^{\pi }q(x)dx. \label{1.1}
\end{equation}%
Relation $\left( \ref{1.1}\right) $, commonly referred to as a trace
formula, links a properly regularized trace of the Sturm-Liouville operator
with some information on the potential $q$. Back in 1953 it was a completely
new type of formula and its explicit nature originated an intensive search
for similar relations eventually in all areas of mathematics dealing with
operators. The literature with the key word \textit{trace formula(s)} is
enormous and we will not even make an attempt to review it here. We restrict
ourselves to just mentioning some of the important stages of its development
pertinent to our work.
\bigskip The original Gel'fand-Levitan trace formula $\left( \ref{1.1}%
\right) $ belongs to the setting of operators with discrete spectrum (see
also Dikii \cite{Dikii} and Podol'skii-Sadovnichii \cite{Sadov} for a recent
account). In 1960 Buslaev-Faddeev \cite{BSTr} extended $\left( \ref{1.1}%
\right) $ to the case of a singular Sturm-Liouville problem
\begin{equation}
\left\{
\begin{array}{c}
-u^{\prime \prime }+q\left( x\right) u=\lambda u,x\geq 0, \\
u\left( 0\right) =0,%
\end{array}%
\right. \label{1.2}
\end{equation}%
with a short-range real potential $q$ (i.e. integrable on $\left( 0,\infty
\right) $ with the first moment). In this case, the Schr\"{o}dinger operator
$H$ associated with $\left( \ref{1.2}\right) $ has an absolutely continuous
spectrum filling $\left( 0,\infty \right) $ and only a finite number of
eigenvalues $\left\{ \lambda _{n}\right\} $ which are all negative. The
direct analog of $\left( \ref{1.1}\right) $ reads
\begin{equation}
-\sum_{n\geq 1}\lambda _{n}+\frac{2}{\pi }\int_{0}^{\infty }t\left( \theta
(t)-\frac{1}{2t}\int_{0}^{\infty }q(x)dx\right) dt=-\frac{1}{4}q(0).
\label{1.3}
\end{equation}%
The function $\theta $ in $\left( \ref{1.3}\right) $, called the limiting
phase, characterizes the absolutely continuous spectrum of $H$ and has a
scattering theoretical nature.
In 1971 Faddeev-Zakharov \cite{FZ} discovered an amazing connection between
trace formulas and the conservation laws for the Korteweg-de Vries equation
(see also \cite{Gest} for a more recent account). The natural Schr\"odinger
operator in this setting is $H=-d^{2}/dx^{2}+q\left( x\right) ,$ defined on
the whole line with a smooth, fast decaying potential $q$. The spectral
situation here is similar to that of the half-line case and the first
Faddeev-Zakharov trace formula reads
\begin{equation}
\sum_{n\geq 1}\left( -\lambda _{n}\right) ^{1/2}+\frac{1}{\pi }%
\int_{0}^{\infty }\ln \left\vert T\left( t\right) \right\vert dt=\frac{1}{4}%
\int_{-\infty }^{\infty }q(x)dx, \label{1.4}
\end{equation}%
where $T$ is the transmission coefficient.
In 1993 Gesztesy-Holden-Simon-Zhao \cite{GHSZ} put forward and consistently
studied (see, e.g. \cite{Xi}) local versions of trace formulas which are
valid eventually for any real potential. Their first trace relation is
\begin{equation}
E+\lim_{\varepsilon \rightarrow 0}\int_{E}^{\infty }e^{-\varepsilon
t}\left\{ 1-2\zeta \left( x,t\right) \right\} dt=q\left( x\right) ,
\label{1.5}
\end{equation}%
where$\ E$ is the low bound of the spectrum of $H=-d^{2}/dx^{2}+q\left(
x\right) $ and $\zeta \left( x,t\right) $ is the so-called xi-function (see
\cite{Xi} for the details). In the literature, formulas of type $\left( \ref%
{1.5}\right) $ are related to the trace approach to the inverse scattering
problem (see, e.g. \cite{RyInv} \ and the literature cited therein).
\bigskip The list of trace formulas given above is not nearly complete. We
mention only one recent development which can be regarded as a breakthrough.
In \cite{KilSim} Killip-Simon introduced a new type of trace formula for
Jacobi matrices that yields an exhaustive description of the spectrum of all
Hilbert-Schmidt perturbations of the free discrete Schr\"odinger operator,
which answered some open problems on orthogonal polynomials (see, also \cite%
{Zlatos}).
Note that the very term \textit{trace formulas} is due to the fact that they
are all related to computing regularized spectral traces of some operators.
Moreover, each of $\left( \ref{1.1}\right) $, $\left( \ref{1.3}\right) $-$%
\left( \ref{1.5}\right) $ actually represents the first formula in some
infinite chain of higher order trace relations representing regularized
moments of certain spectral characteristics. The full chain of trace
formulas corresponding to $\left( \ref{1.3}\right) $ will be given below
(formulas (\ref{4.6}),(\ref{4.7})).
The present paper\ deals with trace formulas for long-range potentials (i.e.
nonintegrable on $\left( 0,\infty \right) $). In the context of the Coulomb
potential, such formulas were studied in 1972 by Yafaev \cite{Yaf} and by
Kvitsinsky in 1987 \cite{Kvit}. For three dimensional Schr\"{o}dinger
operators, a variant of trace formulas serving long-range potentials was put
forward in 1991 by Melin \cite{Mel} and recently by Bouclet \cite{Bou}. Our
goal is to derive a series of relations which are direct analogs to the
Buslaev-Faddeev trace formulas (with the first formula $\left( \ref{1.3}%
\right) $). In 1999 one of the authors \cite{RyTr} proved that
\begin{equation}
-\sum_{n\geq 1}\lambda _{n}^{2}+2\int_{0}^{\infty }\left( \eta (t)+\frac{1}{%
4\pi \sqrt{t}}\int_{0}^{\infty }q^{2}(x)dx\right) dt=\frac{1}{4}q^{2}(0),
\label{1.6}
\end{equation}%
where $\eta $ is the so-called modified spectral shift function that appears
in the Koplienko trace formula \cite{KoSh} ($\func{Im}\lambda \neq 0$):
\begin{eqnarray}
&&\text{tr}\left\{ (H-\lambda )^{-1}-(H_{0}-\lambda )^{-1}-(H_{0}-\lambda
)^{-1}V(H_{0}-\lambda )^{-1}\right\} \label{1.7} \\
&=&-\int_{-\infty }^{\infty }\frac{d^{2}}{dt^{2}}\left( t-\lambda \right)
^{-1}\eta \left( t\right) dt, \notag
\end{eqnarray}%
where $H_{0},H=H_{0}+V$ are selfadjoint operators such that $V\left\vert
H_{0}-i\right\vert ^{-1/2}$ is Hilbert-Schmidt. In the current paper we use
a simpler approach which does not rely on $\left( \ref{1.7}\right) $. We
obtain our trace formulas directly from Buslaev-Faddeev's ones using a
certain limiting procedure. This way, we also improve on \cite{RyTr} by
imposing \ in $\left( \ref{1.6}\right) ~\ $weaker conditions on $q.$
\textbf{Notation. }We will adhere standard notation: $\mathbb{N}%
=\{1,2,3,...\}$, $\mathbb{R}_{+}=(0,\infty )$, $\mathbb{C}_{+}=\{z\in
\mathbb{C}:\func{Im}z>0\}$, and $\overline{\mathbb{C}}_{+}=\mathbb{C}%
_{+}\cup \mathbb{R}$. All functions are assumed to be measurable and we will
use the following Lebesgue classes:\newline
\begin{equation*}
L_{p}:=\left\{ f:\left\Vert f\right\Vert _{p}^{p}\equiv \int_{\mathbb{R}_+
}\left\vert f(x)\right\vert ^{p}dx<\infty \right\} ,\ \ 1\leq p<\infty ,
\end{equation*}%
\begin{equation*}
L_{\infty }:=\left\{ f:\left\Vert f\right\Vert _{\infty }\equiv
ess\sup_{x\in \mathbb{R}_+ }{|f(x)|}<\infty \right\} ,
\end{equation*}
and Sobolev classes:
\begin{equation*}
W_{1}^{n}:=\left\{ f:\dsum\limits_{m=0}^{n}\int_{\mathbb{R}_+ }\left\vert
f^{\left( m\right) }(x)\right\vert dx<\infty \right\} .
\end{equation*}
Furthermore, for a linear operator $A$ we denote $\sigma (A)$ the spectrum
of $A$, $\left\{ \lambda _{n}(A)\right\} $ are its eigenvalues. By $%
\mathfrak{S}_{p},p\geq 1,$ as usual, we denote the ideal of compact
operators $A$, for which
\begin{equation*}
\left\Vert A\right\Vert _{p}^{p}:=\sum_{n}\lambda _{n}^{p}\left( \left\vert
A\right\vert \right) <\infty ,\left\vert A\right\vert :=\sqrt{A^{\ast }A}.
\end{equation*}%
The most important particular cases: $\mathfrak{S}_{1}$ is the space of
trace class operators, and $\mathfrak{S}_{2}$ is the Hilbert-Schmidt class.
\section{Some key asymptotics}
In this section we obtain a suitable representation of the solution to the
Schr\"{o}dinger equation satisfying the WKB type asymptotics at infinity.
Such representations are well-known but we need a relation between the
number of exact terms and the smoothness of the potential $q$. The
proposition below is a long-range generalization of a analogous statement
for $L_{1}$-potentials \cite{RySc}, \cite{Rymf}.
\begin{proposition}
If $q\in L_{2},q^{\prime }\in W_{1}^{N-1}(\mathbb{R}_{+})$ with some $N\in
\mathbb{N}$, then the problem
\begin{equation}
\left\{
\begin{array}{l}
{\displaystyle-u^{\prime \prime }+q(x)u=k}^{2}{u} \\
{\displaystyle\lim\limits_{x\rightarrow \infty }u(x,k)\exp \left( -ikx-\frac{%
1}{2ik}\int_{0}^{x}q(s)ds\right) =1}%
\end{array}%
\right. \label{2.1}
\end{equation}%
has a unique bounded solution ${u(x,k)}$ for all $k\in \mathbb{R},k\neq 0.$
Moreover, for $M(k):={u(0,k)}$ the representation
\begin{equation}
M(k)=\ 1+\sum_{j=2}^{N+1}\frac{m_{j}}{(-2ik)^{j}} \label{2.2}
\end{equation}%
\begin{equation*}
+\frac{1}{(-2ik)^{N+1}}\int_{0}^{\infty }\exp \left( 2ikx+\frac{1}{ik}%
\int_{0}^{x}{q}(s)ds\right) q^{(N)}(x)dx+O\left( \frac{1}{k^{N+2}}\right)
,k\rightarrow \infty ,
\end{equation*}%
holds with some coefficients $\{m_{j}\}$ dependant only on $q$.
\end{proposition}
\begin{proof}
It is enough to consider only the case $k\in {\mathbb{R}}_{+}$. Set
\begin{equation*}
\Theta (x,k):=\exp \left( ikx+\frac{1}{2ik}\int_{0}^{x}{q}(s)ds\right)
\end{equation*}
and observe that $\Theta (x,k)$ satisfies the equation
\begin{equation}
-\Theta ^{\prime \prime }+\left\{ q(x)-q_{1}(x,k)-k^{2}\right\} \Theta =0,
\label{2.3}
\end{equation}%
where
\begin{equation}
q_{1}(x,k)=\frac{q^{\prime }(x)}{-2ik}+\frac{q^{2}(x)}{4k^{2}}. \label{2.4}
\end{equation}%
Rewrite (\ref{2.1}) in the form
\begin{equation}
-u^{\prime \prime }+\left\{ q(x)-q_{1}(x,k)-k^{2}\right\} u=-q_{1}(x,k)u
\label{2.5}
\end{equation}%
and note that (\ref{2.3}) is the homogeneous equation for (\ref{2.5}).
Taking $\Theta (x,k)$ and $\Theta (x,k)\int_{0}^{x}\Theta ^{-2}(s,k)ds$ as
fundamental solutions to (\ref{2.3}), by variation of parameters, equation (%
\ref{2.1}) can be represented as
\begin{equation}
u(x,k)=\Theta (x,k)\left\{ 1+\int_{x}^{\infty }q_{1}(s,k)\Theta (s,k)\left(
\int_{x}^{s}\Theta ^{-2}(t,k)dt\right) u(s,k)ds\right\} . \label{2.6}
\end{equation}%
Setting $y=\Theta ^{-1}u$ and introducing the kernel
\begin{equation}
K(x,s,k)=q_{1}(s,k)\cdot \Theta ^{2}(s,k)\int_{x}^{s}\Theta ^{-2}(t,k)dt,
\label{2.7}
\end{equation}%
equation (\ref{2.6}) reads
\begin{equation}
y(x,k)=1+\int_{x}^{\infty }K(x,s,k)y(s,k)ds=1+(\mathbb{K}y)(x,k),
\label{2.8}
\end{equation}%
where $\mathbb{K}$ stands for the Volterra integral operator with the kernel
$K(x,s)$ (\ref{2.7}). We show that $\mathbb{K}$ maps $L_{\infty }$ to $%
L_{\infty }$ and then (\ref{2.8}) can be solved by iteration. Let us estimate
\begin{equation}
\int_{x}^{s}{\Theta^{-2}}(t,k)dt =\int_{x}^{s} e^{-2ikt}\cdot\exp\left(-%
\frac{1}{ik}\int_{0}^{t} q(z)dz\right)dt. \label{2.9}
\end{equation}
Integrating in (\ref{2.9}) twice by parts twice, on has
\begin{eqnarray}
&&\int_{x}^{s}{\Theta ^{-2}}(t,k)dt \label{2.10} \\
&=&\left. \left\{ \frac{\Theta ^{-2}(t,k)}{-2ik}+\frac{q(t)\Theta ^{-2}(t,k)%
}{-4ik^{3}}\right\} \right\vert _{t=x}^{t=s}+\frac{1}{4ik^{3}}%
\int_{x}^{s}\Theta ^{-2}(t,k)\left( q^{\prime }(t)-\frac{q^{2}(t)}{ik}%
\right) dt. \notag
\end{eqnarray}%
Since $q^{\prime }\in L_{1}$ implies $q\in L_{\infty }$, it follows from (%
\ref{2.10}) that
\begin{equation}
\left\vert \int_{x}^{s}{\Theta ^{-2}}(t,k)dt\right\vert \leq \frac{1}{k}+%
\frac{2\Vert q\Vert _{\infty }+\Vert q^{\prime }\Vert _{1}}{4k^{3}}+\frac{%
\Vert q\Vert _{2}}{4k^{4}}. \label{2.11}
\end{equation}%
Due to (\ref{2.11}) and our conditions on $q$
\begin{eqnarray}
& &\int_{x}^{\infty }\left\vert K(x,s,k)\right\vert ds \leq \int_{x}^{\infty
}|q_{1}(s,k)|\left\vert \int_{x}^{s}{\Theta ^{-2}}(t,k)dt\right\vert ds
\notag \\
& &\leq \frac{1}{k^{2}}\left( 1+\frac{2\Vert q\Vert _{\infty }+\Vert
q^{\prime }\Vert _{1}}{4k^{2}}+\frac{\Vert q\Vert _{2}}{4k^{3}}\right)
\left( \frac{1}{2}\int_{x}^{\infty }|q^{\prime }(s)|ds+\frac{1}{4k}%
\int_{x}^{\infty }|q(s)|^{2}ds\right) \label{2.12}
\end{eqnarray}
and we have:
\begin{eqnarray}
\Vert \mathbb{K}\Vert &=&\sup_{\Vert f\Vert _{\infty }=1}\Vert \mathbb{K}%
f\Vert _{\infty }\leq \left\Vert \int_{x}^{\infty }\left\vert
K(x,s,k)\right\vert ds\right\Vert _{\infty } \notag \\
&\leq& \frac{1}{k^{2}}\left( \frac{\Vert q^{\prime }\Vert _{1}}{2}+\frac{%
\Vert q\Vert _{2}}{4k}\right) \left( 1+\frac{2\Vert q\Vert _{\infty }+\Vert
q^{\prime }\Vert _{1}}{4k^{2}}+\frac{\Vert q\Vert _{2}}{4k^{3}}\right) .
\label{2.13}
\end{eqnarray}%
Estimate (\ref{2.13}) implies that for some positive constant $r_{q}$
dependent only on $q$
\begin{equation}
\Vert \mathbb{K}\Vert \leq 1/2\text{ for }k\geq r_{q}, \label{2.14}
\end{equation}%
Equation (\ref{2.8}) can then be solved by iteration:
\begin{equation}
y(x,k)=1+\sum_{n=1}^{\infty }\Omega _{j}(x,k),k\geq r_{q}, \label{2.15}
\end{equation}%
where $\Omega _{j}(x,k)=(\mathbb{K}^{j}1)(x,k)$, and
\begin{equation}
\Vert y\Vert_\infty \leq 2\text{ for }k\geq r_{q}. \label{2.16}
\end{equation}
This proves the existence.
Now we establish (\ref{2.2}). To this end, introduce the differential
operation
\begin{equation*}
D:=\frac{d}{dx}+\frac{q(x)}{ik}.
\end{equation*}
It is clear that if $q\in L_{2}$, $q^{\prime }\in W_{1}^{m-1}(\mathbb{R}%
_{+}) $ and $f\in W_{1}^{m}(\mathbb{R}_{+})$, for some $m\geq 1$, then
\begin{equation}
D^{j}f\in L_{1},\ \forall\ 1\leq j\leq m. \label{2.17}
\end{equation}
Denote
\begin{equation*}
G(x,s,k)=\Theta ^{2}(s,k)\int_{x}^{s}{\Theta ^{-2}(}t,k{)}dt.
\end{equation*}
Integrating by part, one can easily verify that
\begin{equation}
\int_{x}^{\infty }G(x,s,k)f(s)ds=-\frac{1}{2ik}\int_{x}^{\infty }f(s)ds-%
\frac{1}{2ik}\int_{x}^{\infty }G(x,s,k)Df(s)ds \label{2.18}
\end{equation}%
Equation (\ref{2.18}) is an integration by parts type formula. Since $%
q_{1}\in W_{1}^{N-1}$, by (\ref{2.17}), all $Dq_{1},D^{2}q_{1},\cdots
,D^{N-1}q_{1}$ are from $L_{1},$ (\ref{2.18}) can be applied $N-1$ times and
the remaining integral will still be absolutely convergent. Setting in (\ref%
{2.15}) $x=0$ we have
\begin{equation}
M(k)=1+\sum_{j=1}^{\infty }\Omega _{j}(k), \label{2.19}
\end{equation}%
where $\Omega _{j}(k)=\Omega _{j}(0,k)=(\mathbb{K}^{j}1)(0,k)$. Let us now
evaluate each term in (\ref{2.19}). Consider $\Omega _{1}$ first. Apply (\ref%
{2.18}) $N-1$ times to
\begin{equation}
\Omega _{1}(x,k) =\int_{x}^{\infty }G(x,s,k)q_{1}(s,k)ds \label{2.20}
\end{equation}
\begin{equation*}
=-\sum_{j=1}^{N-1}\frac{1}{(-2ik)^{j}}\int_{x}^{\infty }D^{j}q_{1}(s,k)ds+%
\frac{1}{(-2ik)^{N-1}}\int_{x}^{\infty }G(x,s,k)D^{N-1}q_{1}(s,k)ds \notag
\end{equation*}
\begin{equation*}
=\sum_{j=2}^{N}\frac{\omega _{j}^{(1)}(x)}{(-2ik)^{j}}+\frac{1}{%
(-2ik)^{N-1}}\int_{x}^{\infty }G(x,s,k)D^{N-1}q_{1}(s,k)ds+O\left( \frac{1}{%
k^{N+2}}\right) , \notag
\end{equation*}
where $\omega _{j}^{(1)}(x)$ are some continuous functions whose expressions
are inessential. By (\ref{2.10}) one can has that
\begin{equation*}
G(x,s,k)=\frac{1-\Theta ^{2}(s,k)\Theta ^{-2}(x,k)}{-2ik}+O\left( \frac{1}{%
k^{2}}\right)
\end{equation*}%
and (\ref{2.20}) reads
\begin{eqnarray}
\Omega _{1}(x,k) &=&\sum_{j=2}^{N+1}\frac{\omega _{j}^{(1)}(x)}{(-2ik)^{j}}+%
\frac{\Theta ^{-2}(x,k)}{(-2ik)^{N+1}}\int_{x}^{\infty }q^{(N)}(s)ds
\label{2.21} \\
&&+\frac{\Theta ^{-2}(x,k)}{(-2ik)^{N+1}}\int_{x}^{\infty }\Theta
^{2}(s,k)q^{(N)}(s)ds+O\left( \frac{1}{k^{N+2}}\right) \notag \\
&=&\sum_{j=2}^{N+1}\frac{\omega _{j}^{(1)}(x)}{(-2ik)^{j}}-\frac{%
q^{(N-1)}(x)\Theta ^{-2}(x,k)}{(-2ik)^{N+1}} \notag \\
&&+\frac{\Theta ^{-2}(x,k)}{(-2ik)^{N+1}}\int_{x}^{\infty }\Theta
^{2}(s,k)q^{(N)}(s)ds+O\left( \frac{1}{k^{N+2}}\right) . \notag
\end{eqnarray}%
Setting in (\ref{2.21}) $x=0$ we finally have
\begin{equation}
\Omega _{1}(k)=\sum_{j=2}^{N+1}\frac{\omega _{j}^{(1)}}{(-2ik)^{j}}+\frac{1}{%
(-2ik)^{N+1}}\int_{0}^{\infty }\Theta ^{2}(s,k)q^{(N)}(s)ds+O\left( \frac{1}{%
k^{N+2}}\right) . \label{2.22}
\end{equation}%
Similarly,
\begin{equation}
\Omega _{2}(k)=\sum_{j=4}^{N+1}\frac{\omega _{j}^{(2)}}{(-2ik)^{j}}+O\left(
\frac{1}{k^{N+2}}\right) ,\ k\rightarrow \infty . \label{2.23}
\end{equation}%
For the general $1\leq n\leq \left[ \frac{N+1}{2}\right] ,$ where $\left[ x%
\right] $ is the greatest integer $n\leq x,$ we get
\begin{equation}
\Omega _{n}(k)=\sum_{j=2n}^{N+1}\frac{\omega _{j}^{(n)}}{(-2ik)^{j}}+O\left(
\frac{1}{k^{N+2}}\right) ,\ k\rightarrow \infty , \label{2.24}
\end{equation}%
and
\begin{equation*}
\Omega _{\left[ \frac{N+1}{2}\right] +1}(k)=O\left( \frac{1}{k^{N+2}}\right)
,\ k\rightarrow \infty
\end{equation*}%
Since in $\Omega _{j}$, $j>1,$ all terms containing $\int_{x}^{\infty
}\Theta ^{2}(s,k)q^{(N)}(s)ds$ are included in the error term $o\left( \frac{%
1}{k^{N+1}}\right) $, we can get the $N+2$ term if and only if $%
\int_{0}^{\infty }\Theta ^{2}(x,k)q^{(N)}(x)dx=O\left( \frac{1}{k}\right)
,k\rightarrow \infty $. From (\ref{2.13}) and (\ref{2.14}) is easy to see
that
\begin{equation*}
\left\vert \sum_{n\geq \left[ \frac{N+1}{2}\right] +1}\Omega
_{n}(k)\right\vert \leq \sum_{n\geq \left[ \frac{N+1}{2}\right]
+1}\left\Vert \mathbb{K}\right\Vert ^{n}\leq 2\left\Vert \mathbb{K}%
\right\Vert ^{\left[ \frac{N+1}{2}\right] +1}\leq \frac{C_{q}}{k^{N+2}},
\end{equation*}%
where $C_{q}$ is a constant dependant only on $q$. Thus
\begin{equation}
\sum_{n\geq \left[ \frac{N+1}{2}\right] +1}\Omega _{n}(k)=O\left( \frac{1}{%
k^{N+2}}\right) ,\ k\rightarrow \infty , \label{2.25}
\end{equation}%
and combining (\ref{2.22})-(\ref{2.25}) we arrive at
\begin{equation*}
M(k)=1+\sum_{n\geq 1}\Omega _{n}(k)
\end{equation*}%
\begin{equation*}
=1+\sum_{j=2}^{N+1}\frac{m_{j}}{(-2ik)^{j}}+\frac{1}{(-2ik)^{N+1}}%
\int_{0}^{\infty }\Theta ^{2}(s,k)q^{(N)}(s)ds+O\left( \frac{1}{k^{N+2}}%
\right) ,\ k\rightarrow \infty ,
\end{equation*}%
that completes the proof.
\end{proof}
\begin{remark}
Since, by condition, $q^{(N)}\in L_{1}$ we conclude that
\begin{equation*}
\int_{0}^{\infty }\exp \left( 2ikx+\frac{1}{ik}\int_{0}^{x}{q}(s)ds\right)
q^{(N)}(x)dx=o(1),k\rightarrow \infty .
\end{equation*}%
(the proof is parallel to that of the Riemann-Lebesgue lemma) and hence the
integral term in (\ref{2.2}) is $o\left( k^{-N-1}\right) $. The fact that we
were able to extract an explicit 'intermediate' term between the exact terms
$\sum_{j=2}^{N+1}(-2ik)^{-j}m_{j}$ and $O\left( k^{-N-2}\right) $ will be
important in Section 4. Proposition 1 actually relates the number of exact
terms in (\ref{2.2}) with the smoothness of $q$ but does not provide a
recipe to evaluate coefficients $\{m_{j}\}$. Explicit formulas for $%
\{m_{j}\}$ can be derived e.g. from \cite{BSTr}.
\end{remark}
\section{The modified Jost function}
Introduce the half-line Schr\"odinger operator $H=-d^{2}/dx^{2}+q(x)$ with a
Dirichlet boundary condition $u\left( 0\right) =0.~$It is well-known that if
$q\in L_{1}$ then for any $k>0$ the equation $-u^{\prime \prime
}+q(x)u=k^{2}u$ has the so-called Jost solution $u_{0}\left( x,k\right) $,
i.e. a solution subject to the condition that $\lim\limits_{x\rightarrow
\infty }e^{-ikx}u_{0}\left( x,k\right) =1.$ The function $M_{0}\left(
k\right) :=u_{0}\left( 0,k\right) $ is commonly referred to as the Jost
function. It is continuous on $\mathbb{R}\backslash \left\{ 0\right\} $ and
analytic in $\mathbb{C}_{+}$ except for isolated poles at $i\sqrt{-\lambda
_{n}}$, where $\left\{ \lambda _{n}\right\} $ are (necessarily negative)
eigenvalues of $H$ accumulating only at $0$. The function $M(k),$ defined in
Proposition 1, can then be called \textit{the modified Jost function}. Note
that if $q\in L_{1}$ then
\begin{equation}
M(k)=M_{0}\left( k\right) \exp \left\{ \frac{1}{2ik}\dint\limits_{0}^{\infty
}q\left( x\right) dx\right\} . \label{3.1}
\end{equation}%
Given potential $q$, let $\widetilde{q}$ be its smooth cut-off approximation
defined by
\begin{equation}
\widetilde{q}\left( x,a\right) =q\left( x\right) \chi _{a}\left( x\right) ,
\label{3.2}
\end{equation}%
where $\chi _{a}$ is smooth such that $\chi _{a}\left( x\right) =1,0\leq
x\leq a,$ and $\chi _{a}\left( x\right) =0,x>a+1.$
It is clear that%
\begin{equation*}
\widetilde{q}\left( x,a\right) \rightarrow q\left( x\right) ,a\rightarrow
\infty ,
\end{equation*}%
which we agree to write just as $\widetilde{q}\rightarrow q.$ In the sequel,
we will drop $a$ in $\ \widetilde{q}\left( x,a\right) $ and put $\
\widetilde{}$ on top of every object related to $\widetilde{q}.$
The following lemma will be a key ingredient in our consideration.
\begin{lemma}
Let $q$ be real and $q\in L_{2}$, $q^{\prime }\in L_{1}$ and $\widetilde{q}$
be defined by (\ref{3.2}) then:
\begin{description}
\item[(i)] for every $k\in \overline{\mathbb{C}}_{+},\ |k|>r_{q},$ with some
$r_{q}>0$ dependent only on $q$,%
\begin{equation*}
\widetilde{M}(k)\rightarrow M(k),\text{ \ \ }\widetilde{q}\rightarrow q;
\end{equation*}
\item[(ii)] for any $p\geq 3/2$%
\begin{equation*}
\sum_{n}|\widetilde{\lambda }_{n}|^{p}\rightarrow \sum_{n}|\lambda _{n}|^{p},%
\text{ \ \ }\widetilde{q}\rightarrow q.
\end{equation*}
\end{description}
\end{lemma}
\begin{proof}
(i) Assume for the time being that $k>0.$ In the notation of Proposition 1,
by representation (\ref{2.15}) we have: for any natural number $N$
\begin{equation}
|M(k)-\widetilde{M}(k)|=|u(0,k)-\widetilde{u}(0,k)|=|y(0,k)-\widetilde{y}%
(0,k)| \label{3.3}
\end{equation}%
\begin{eqnarray*}
&\leq &\sum_{n=1}^{N}\left\vert (\mathbb{K}^{n}1)(0,k)-(\widetilde{\mathbb{K}%
}^{n}1)(0,k)\right\vert +\left\vert (\mathbb{K}^{N+1}y)(0,k)-(\widetilde{%
\mathbb{K}}^{N+1}\widetilde{y})(0,k)\right\vert \\
&\leq &\sum_{j=1}^{N}\left\Vert \mathbb{K}^{n}-\widetilde{\mathbb{K}}%
^{n}\right\Vert +\left\Vert \mathbb{K}^{N+1}-\widetilde{\mathbb{K}}%
^{N+1}\right\Vert \left\Vert y\right\Vert +\left\Vert \mathbb{K}^{N+1}(y-%
\widetilde{y})\right\Vert .
\end{eqnarray*}%
Observing that for any $n\in \mathbb{N}$
\begin{equation*}
\mathbb{K}^{n}-\widetilde{\mathbb{K}}^{n}=\sum_{m=0}^{n-1}\widetilde{\mathbb{%
K}}^{m}\left( \mathbb{K}-\widetilde{\mathbb{K}}\right) \mathbb{K}^{n-m-1},
\end{equation*}%
and by (\ref{2.16}), estimate (\ref{3.3}) for $k\geq r_{q}$ can be continued
\begin{eqnarray}
|M(k)-\widetilde{M}(k)| &\leq &\sum_{n=1}^{N}\sum_{m=0}^{n-1}\left\Vert
\widetilde{\mathbb{K}}\right\Vert ^{m}\left\Vert \mathbb{K}-\widetilde{%
\mathbb{K}}\right\Vert \left\Vert \mathbb{K}\right\Vert ^{n-m-1} \notag \\
&&+\left\Vert y\right\Vert \sum_{n=0}^{N}\left\Vert \widetilde{\mathbb{K}}%
\right\Vert ^{n}\left\Vert \mathbb{K}-\widetilde{\mathbb{K}}\right\Vert
\left\Vert \mathbb{K}\right\Vert ^{N-n}+\left\Vert \mathbb{K}\right\Vert
^{N+1}\left( \left\Vert y\right\Vert +\left\Vert \widetilde{y}\right\Vert
\right) . \label{3.4}
\end{eqnarray}%
But for $k>\max \left\{ r_{q},r_{\widetilde{q}}\right\} =r_{q},$ by (\ref%
{2.14}) and (\ref{2.16}), $\left\Vert \widetilde{\mathbb{K}}\right\Vert
,\left\Vert \mathbb{K}\right\Vert \leq 1/2$ and $\left\Vert y\right\Vert
,\left\Vert \widetilde{y}\right\Vert \leq 2$ and (\ref{3.4}) finally yields
\begin{eqnarray}
|M(k)-\widetilde{M}(k)| &\leq &\sum_{n=1}^{N-1}n2^{-n}\cdot \left\Vert
\mathbb{K}-\widetilde{\mathbb{K}}\right\Vert +N2^{-N+1}\cdot \left\Vert
\mathbb{K}-\widetilde{\mathbb{K}}\right\Vert +2^{-N+1} \notag \\
&\leq &2\left( 1+N2^{-N+1}\right) \left\Vert \mathbb{K}-\widetilde{\mathbb{K}%
}\right\Vert +2^{-N+1}. \label{3.5}
\end{eqnarray}%
Letting in (\ref{3.5}) $N\rightarrow \infty $, we have
\begin{equation}
|M(k)-\widetilde{M}(k)|\leq 2\left\Vert \mathbb{K}-\widetilde{\mathbb{K}}%
\right\Vert . \label{3.6}
\end{equation}%
Show now that $\left\Vert \mathbb{K}-\widetilde{\mathbb{K}}\right\Vert
\rightarrow 0$ as $a\rightarrow \infty .$ It follows from (\ref{2.4}) and (%
\ref{2.7}) that
\begin{equation*}
\widetilde{K}\left( x,s,k\right) =K\left( x,s,k\right) \text{ for }x\leq
s\leq a
\end{equation*}%
and hence by (\ref{2.12}) one obtains
\begin{eqnarray}
\left\Vert \mathbb{K}-\widetilde{\mathbb{K}}\right\Vert &\leq &\sup_{x\geq
a}\int_{x}^{\infty }\left\vert K(x,s,k)-\widetilde{K}(x,s,k)\right\vert ds
\label{3.7} \\
&\leq &\sup_{x\geq a}\int_{x}^{\infty }\left\{ \left\vert
K(x,s,k)\right\vert +\left\vert \widetilde{K}(x,s,k)\right\vert \right\}
ds\rightarrow 0,a\rightarrow \infty . \notag
\end{eqnarray}%
Combining (\ref{3.6}), and (\ref{3.7}) we prove (i) with real $k$ and then
using Phragm\'{e}n-Lindel\H{o}f type arguments we arrive at (i).
(ii). Let $H_{0}=-d^{2}/dx^{2}$ with $u\left( 0\right) =0.$ For the Green's
function $G\left( x,s,\lambda \right) $ of $H_{0}$ one has
\begin{equation*}
G\left( x,y,\lambda \right) =\frac{1}{i\sqrt{\lambda }}\left( e^{i\sqrt{%
\lambda }(x+y)}-e^{i\sqrt{\lambda }|x-y|}\right) .
\end{equation*}%
and hence for $\func{Im}\lambda \neq 0$
\begin{equation*}
\left\Vert (q-\widetilde{q})(H_{0}-\lambda I)^{-1}\right\Vert _{2}^{2}
\end{equation*}%
\begin{equation*}
=\frac{1}{|\lambda |}\int_{0}^{\infty }\int_{0}^{\infty }|q(x)-\widetilde{q}%
(x)|^{2}\left\vert e^{i\sqrt{\lambda }(x+y)}-e^{i\sqrt{\lambda }%
|x-y|}\right\vert ^{2}dxdy
\end{equation*}%
\begin{equation}
\leq \frac{5}{|\lambda |\func{Im}\sqrt{\lambda }}\Vert q-\widetilde{q}\Vert
_{2}^{2}\rightarrow 0,\widetilde{q}\rightarrow q. \label{3.8}
\end{equation}%
Due to%
\begin{equation*}
(H_{0}-\lambda )^{-1}=(\widetilde{H}-\lambda )^{-1}(I+\widetilde{q}%
(H_{0}-\lambda )^{-1})
\end{equation*}%
we have
\begin{equation}
\left\Vert (H-\lambda )^{-1}-(\widetilde{H}-\lambda )^{-1}\right\Vert
_{2}=\left\Vert (H-\lambda )^{-1}(\widetilde{H}-H)(\widetilde{H}-\lambda
)^{-1}\right\Vert _{2} \label{3.9}
\end{equation}%
\begin{equation*}
\leq \left\Vert (H-\lambda )^{-1}\right\Vert \cdot \left\Vert (q-\widetilde{q%
})(\widetilde{H}-\lambda )^{-1}\right\Vert _{2}
\end{equation*}
\begin{equation*}
=\left\Vert (H-\lambda%
)^{-1}\right\Vert \cdot \left\Vert (q-\widetilde{q})(H_{0}-\lambda )^{-1}(I+%
\widetilde{q}(H_{0}-\lambda )^{-1})^{-1}\right\Vert _{2} \notag
\end{equation*}%
\begin{equation*}
\leq \left\Vert (H-\lambda )^{-1}\right\Vert \left\Vert (I+\widetilde{q}%
(H_{0}-\lambda )^{-1})^{-1}\right\Vert \cdot \left\Vert (q-\widetilde{q}%
)(H_{0}-\lambda )^{-1}\right\Vert _{2}.
\end{equation*}%
Since for $\func{Im}\lambda \neq 0$ the first two norms on the right side of
(\ref{3.9}) are finite and, by (\ref{3.8}), the third one goes to zero and
we conclude that
\begin{equation}
\left\Vert (H-\lambda )^{-1}-(\widetilde{H}-\lambda )^{-1}\right\Vert \leq
\left\Vert (H-\lambda )^{-1}-(\widetilde{H}-\lambda )^{-1}\right\Vert
_{2}\rightarrow 0,\widetilde{q}\rightarrow q. \label{3.10}
\end{equation}%
One can now employ some arguments from \cite{Gest}. Let $\Delta =\left[ a,b%
\right] \subset (-\infty ,0):a<\inf \left\{ \sigma \left( H\right) ,\sigma
\left( \widetilde{H}\right) \right\} ,-10.
\label{3.11}
\end{equation}%
Consider $\left( p\geq 3/2\right) $
\begin{equation}
\left\vert \sum_{n}\left( |\widetilde{\lambda }_{n}|^{p}-|\lambda
_{n}|^{p}\right) \right\vert \leq \left\vert \sum_{\widetilde{\lambda }%
_{n},\lambda _{n}b}|\widetilde{%
\lambda }_{n}|^{p}+\sum_{\lambda _{n}>b}|\lambda _{n}|^{p}. \label{3.12}
\end{equation}%
Denote by $q_{-}=\frac{1}{2}\left( q-\left\vert q\right\vert \right) $ the
negative part of $q$. Since $q_{-}(x)\leq \widetilde{q}(x)$ we have $%
H_{-}=-d^{2}/dx^{2}+q_{-}(x)\subseteq \widetilde{H}.$ Hence, by the standard
fact of perturbation theory, $\lambda _{n}\left( H_{-}\right) \leq
\widetilde{\lambda }_{n}$ and (all eigenvalues are negative)
\begin{equation}
\sum_{n}|\widetilde{\lambda }_{n}|^{p}\leq \sum_{n}\left\vert \lambda
_{n}\left( H_{-}\right) \right\vert ^{p}. \label{3.13}
\end{equation}%
From the Lieb-Thirring bounds (see, e.g. \cite{Lieb})
\begin{equation}
\sum_{n}|\lambda _{n}\left( H_{-}\right) |^{3/2},\text{ }\sum_{n}|\lambda
_{n}\left( H\right) |^{3/2}\leq \frac{3}{16}\Vert q_{-}\Vert _{2}^{2}\leq
\frac{3}{16}\Vert q\Vert _{2}^{2}, \label{3.14}
\end{equation}%
and hence for $p\geq 3/2$ by (\ref{3.13}) and taking into account that $%
\left\vert b\right\vert <1$, we get%
\begin{equation}
\sum_{\widetilde{\lambda }_{n}>b}|\widetilde{\lambda }_{n}|^{p}\leq \sum_{%
\widetilde{\lambda }_{n}>b}|\widetilde{\lambda }_{n}|^{3/2}\leq \sum_{n}|%
\widetilde{\lambda }_{n}|^{3/2}\leq \sum_{n}|\lambda _{n}\left( H_{-}\right)
|^{3/2}\leq \frac{3}{16}\Vert q\Vert _{2}^{2}<\infty . \label{3.15}
\end{equation}%
Due to (\ref{3.11}), (\ref{3.14}) and (\ref{3.15}), by the standard $%
\varepsilon /3$ argument (\ref{3.12}) yields (ii) and the lemma is proved.
\end{proof}
\begin{remark}
Since $M\left( k\right) $ is analytic in $\mathbb{C}_{+}$, (i) of Lemma 1
actually holds in $\overline{\mathbb{C}}_{+}\backslash \left\{ 0\right\} .$
\end{remark}
\begin{remark}
Lemma 1 improves on one result by Koplienko \cite{KoSh}. Namely, if $\left(
H,H_{0}\right) $: $H_{0}=-d^{2}/dx^{2}$ with $u\left( 0\right) =0$ and $%
H=H_{0}+q$ with $q\in L_{2},q^{\prime }\in L_{1},$ then the modified
spectral shift function $\eta \left( t\right) $ for $\left( H,H_{0}\right) $
appearing in (\ref{1.7}) is differentiable and
\begin{equation}
\eta ^{\prime }\left( t\right) =\pi ^{-1}\func{Im}\ln M\left( t\right) ,t>0.
\label{3.16}
\end{equation}%
In \cite{KoSh} relation (\ref{3.16}) was proven under stronger conditions
involving $q^{\prime \prime }.$
\end{remark}
\section{Trace formulas}
This section contains the main results of the paper, the Buslaev-Faddeev
trace formulas for certain long-range potentials. Relation (\ref{3.1}),
Proposition 1, Lemma 1, and some limiting arguments will let us not only
adjust the original Buslaev-Faddeev trace formulas \cite{BSTr} to the
long-range setting but also provide sharp conditions on potentials.
Set in (\ref{3.1}) $M_{0}=A_{0}e^{i\theta _{0}},$ where $A_{0}=|M_{0}|$ is
the scattering amplitude and $\theta _{0}$ is the scattering phase.
Similarly, $M=Ae^{i\theta }$ where $A=|M|$ can called \textit{the modified
scattering amplitude} and $\theta _{0}$ \textit{the modified scattering phase%
}. Relation (\ref{3.1}) then implies that for $q\in L_{1}$ and $k>0$
\begin{equation}
\theta (k)=\theta _{0}(k)-\frac{1}{2k}\int_{0}^{\infty }q(t)dt,~A\left(
k\right) =A_{0}\left( k\right) \label{4.1}
\end{equation}%
Following \cite{BSTr}, introduce the sequence $\left\{ Q_{n}\right\} _{n\geq
1}$:%
\begin{equation}
Q_{n}=\lim_{x\rightarrow \infty }\left( V_{n-1}\left( x\right)
+\sum_{j=1}^{n-2}\frac{j}{n}V_{n-j-1}\left( x\right) Q_{j}\right) ,
\label{4.2}
\end{equation}%
where
\begin{equation}
V_{n}\left( x\right) =q^{(n-1)}(0)+\sum_{m=1}^{n-1}\left(
\begin{array}{c}
l-1 \\
m%
\end{array}%
\right) \int_{0}^{x}V_{m}(s)q^{(n-m-1)}(s)ds,V_{0}\left( x\right) =0.
\label{4.3}
\end{equation}
The following statement is our main result.
\begin{theorem}
Suppose that for $q\in L_{2}\left( 0,\infty \right) $ and $q^{\prime }\in
W_{1}^{N-1}\left( 0,\infty \right) ,N\in \mathbb{N}$. Then for \textit{the
modified scattering phase }$\theta \left( k\right) $ and \textit{modified
scattering amplitude} $A\left( k\right) $ the following trace formulas are
valid:
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\lambda _{l}\right) ^{n }+\frac{2n }{%
\pi }\int_{0}^{\infty }k^{2n -1}\left\{ \theta (k)-\sum_{l=1}^{n -1}\frac{%
(-1)^{l+1}}{(2k)^{2l+1}}Q_{2l+1}\right\} dk \notag \\
&=&(-1)^{n }\frac{n }{2^{2n }}Q_{2n },\ \ \ 2\leq n \leq \frac{N}{2};
\label{4.4}
\end{eqnarray}%
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }(-\lambda _{l})^{n +1/2}-\frac{2n +1}{\pi }%
\int_{0}^{\infty }k^{2n }\left\{ \ln A(k)-\sum_{l=1}^{n }\frac{(-1)^{l+1}}{%
(2k)^{2l}}Q_{2l}\right\} dk \notag \\
&=&(-1)^{n }\frac{2n +1}{2^{2n +1}}Q_{2n +1},\ \ \ 1\leq n \leq \frac{N-1}{2}%
; \label{4.5}
\end{eqnarray}%
where coefficients $\left\{ Q_{n}\right\} $ are computed by (\ref{4.2}) and
all integrals are absolutely convergent. In order to extract one additional
formula in (\ref{4.4})-(\ref{4.5}) with the absolutely convergent
integral it is necessary and sufficient that
\begin{equation*}
\frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-\frac{1}{k}\int_{0}^{x}q(s)ds%
\right) q^{(N)}(x)dx\in L_{1}\left( r,\infty \right) ,
\end{equation*}%
for some $r>0$.
\end{theorem}
\begin{proof}
Introduce $\widetilde{q}(x)$ by (\ref{3.2}). Since $\widetilde{q}\in
L_{1}\left( \left( 1+x\right) dx,\mathbb{R}_{+}\right) $, then
Faddeev-Buslaev trace formulas \cite{BSTr} for $\widetilde{\theta }_{0}(k)$
and $\widetilde{A}_{0}(k)$ apply:
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\widetilde{\lambda }_{l}\right) ^{n }+%
\frac{2n }{\pi }\int_{0}^{\infty }k^{2n -1}\left[ \widetilde{\theta }%
_{0}(k)-\sum_{l=0}^{n -1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}\widetilde{Q}_{2l+1}%
\right] dk \label{4.6} \\
&=&(-1)^{n }\frac{n }{2^{2n }}\widetilde{Q}_{2n },\ \ \ 1\leq n \leq \frac{N%
}{2}; \notag
\end{eqnarray}
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\widetilde{\lambda }_{l}\right) ^{n
+1/2}-\frac{2n +1}{\pi }\int_{0}^{\infty }k^{2n }\left[ \ln \widetilde{A}%
_{0}(k)-\sum_{l=1}^{n }\frac{(-1)^{l+1}}{(2k)^{2l}}\widetilde{Q}_{2l}\right]
dk \label{4.7} \\
&=&(-1)^{n }\frac{2n +1}{2^{2n +1}}\widetilde{Q}_{2n +1},\ \ \ 1\leq n \leq
\frac{N-1}{2}. \notag
\end{eqnarray}
In view of (\ref{4.1}), relations (\ref{4.6}) and (\ref{4.7}) transform into%
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\widetilde{\lambda }_{l}\right) ^{n }+%
\frac{2n }{\pi }\int_{0}^{\infty }k^{2n -1}\left[ \widetilde{\theta }%
(k)-\sum_{l=1}^{n -1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}\widetilde{Q}_{2l+1}%
\right] dk \label{4.8} \\
&=&(-1)^{n }\frac{n }{2^{2n }}\widetilde{Q}_{2n },\ \ \ 2\leq n \leq \frac{N%
}{2}; \notag
\end{eqnarray}
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\widetilde{\lambda }_{l}\right) ^{n
+1/2}-\frac{2n +1}{\pi }\int_{0}^{\infty }k^{2n }\left[ \ln \widetilde{A}%
(k)-\sum_{l=1}^{n }\frac{(-1)^{l+1}}{(2k)^{2l}}\widetilde{Q}_{2l}\right] dk
\label{4.9} \\
&=&(-1)^{n }\frac{2n +1}{2^{2n +1}}\widetilde{Q}_{2n +1},\ \ \ 1\leq n \leq
\frac{N-1}{2}. \notag
\end{eqnarray}
Since (\ref{4.8}) and (\ref{4.9}) are similar we treat only (\ref{4.8}). Let
now in (\ref{4.8}) $\widetilde{q}\rightarrow q$. Due to (\ref{4.2}) and (\ref%
{4.3}) one easily has that $\widetilde{Q}_{j}\rightarrow Q_{j}$,$j\leq N.$
It follows from Lemma 1 that $\sum_{l}\left( -\widetilde{\lambda }%
_{l}\right) ^{n }$ $\rightarrow \sum_{l}\left( -\lambda _{l}\right) ^{n } $
and (\ref{4.8}) becomes
\begin{eqnarray}
&&(-1)^{n }\sum_{l=1}^{\infty }\left( -\lambda _{l}\right) ^{n }+\frac{2n }{%
\pi }\lim_{\widetilde{q}\rightarrow q}\int_{0}^{\infty }k^{2n -1}\left[
\widetilde{\theta }(k)-\sum_{l=1}^{n -1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}%
\widetilde{Q}_{2l+1}\right] dk \label{4.10} \\
&=&(-1)^{n }\frac{n }{2^{2n }}Q_{2n },\ \ \ 2\leq n \leq \frac{N}{2}. \notag
\end{eqnarray}
It is only left to pass to the limit under the integral sign in (\ref{4.10}%
). To this end, represent this integral as the sum of the integrals over $%
(0,r)$ and $(r,\infty )$ with any $r\geq $ $r_{q},$ where $r_{q}$ is as in
Lemma 1. It can be derived from \cite{KoSh-1} that $\dint\limits_{0}^{k}%
\widetilde{\theta }(t)dt\rightarrow \dint\limits_{0}^{k}\theta (t)dt$ in $%
L_{1}(0,r)$ and hence $\ k\widetilde{\theta }(k)\rightarrow k\theta (k)$ in $%
L_{1}(0,r)$-sense. Thus%
\begin{equation}
\lim_{\widetilde{q}\rightarrow q}\int_{0}^{r}k^{2n-1}\left\{ \widetilde{%
\theta }(k)-\sum_{l=1}^{n-1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}\widetilde{Q}%
_{2l+1}\right\} dk \label{4.11}
\end{equation}
\begin{equation*}
=\int_{0}^{r}k^{2n-1}\left\{ \theta (k)-\sum_{l=1}^{n-1}%
\frac{(-1)^{l+1}}{(2k)^{2l+1}}Q_{2l+1}\right\} dk,
\end{equation*}
since $k=0$ is not a singularity of the integrands in (\ref{4.11}). Consider
now the integral over $(r,\infty )$. Using the elementary fact
\begin{equation*}
f(x)=\dsum\limits_{m\geq 1}a_{m}x^{m}\Rightarrow \ln \left( 1+f(x)\right)
=\dsum\limits_{m\geq 1}b_{m}x^{m},b_{m}=a_{m-1}-\dsum\limits_{j\geq 1}\frac{j%
}{m}a_{m-j-1}bj,
\end{equation*}%
it follows form Proposition 1 that for $k\geq r_{q}$ the function
\begin{eqnarray*}
\widetilde{g}\left( k\right) &:&=k^{2n-1}\left\{ \widetilde{\theta }%
(k)-\sum_{l=1}^{n-1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}\widetilde{Q}%
_{2l+1}\right\} \\
&&-\frac{(-1)^{n}}{2^{2n}}\cdot \frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-%
\frac{1}{k}\int_{0}^{x}\widetilde{q}(s)ds\right) \widetilde{q}^{(N)}(x)dx
\end{eqnarray*}%
has a majorant from $L_{1}\left( r,\infty \right) $ and by the dominated
convergence theorem
\begin{equation*}
g\left( k\right) =\lim_{\widetilde{q}\rightarrow q}\widetilde{g}\left(
k\right) =k^{2n-1}\left\{ \theta (k)-\sum_{l=1}^{n-1}\frac{(-1)^{l+1}}{%
(2k)^{2l+1}}Q_{2l+1}\right\}
\end{equation*}%
\begin{equation*}
-\frac{(-1)^{n}}{2^{2n}}\cdot \frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-%
\frac{1}{k}\int_{0}^{x}q(s)ds\right) q^{(N)}(x)dx\in L_{1}\left( r,\infty
\right) .
\end{equation*}%
The latter means that \newline
\begin{equation}
\lim_{\widetilde{q}\rightarrow q}\int_{r}^{\infty }k^{2n-1}\left\{
\widetilde{\theta }(k)-\sum_{l=1}^{n-1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}%
\widetilde{Q}_{2l+1}\right\} dk \label{4.12}
\end{equation}
\begin{equation*}
=\int_{r}^{\infty }k^{2n-1}\left\{ \theta
(k)-\sum_{l=1}^{n-1}\frac{(-1)^{l+1}}{(2k)^{2l+1}}Q_{2l+1}\right\} dk,
\end{equation*}%
and the integral on the right side of (\ref{4.12}) converges absolutely if
and only if
\begin{equation*}
\frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-\frac{1}{k}\int_{0}^{x}q(s)ds%
\right) q^{(N)}(x)dx\in L_{1}\left( r,\infty \right)
\end{equation*}%
with some positive $r$. Plug (\ref{4.11}) and (\ref{4.12}) into (\ref{4.10})
and the theorem is proved.
\end{proof}
\begin{corollary}
If $q\in L_{2}\left( 0,\infty \right) $, $q^{\prime },q^{\prime \prime }\in
L_{1}\left( 0,\infty \right) ,$ and
\begin{equation}
\frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-\frac{1}{k}\int_{0}^{x}q(s)ds%
\right) q^{^{\prime \prime }}(x)dx\in L_{1}\left( r,\infty \right) ,r>0,
\label{4.13}
\end{equation}
then
\begin{equation}
\sum_{l\geq 1}(-\lambda _{l})^{3/2}+\frac{3}{\pi }\int_{0}^{\infty }\left\{
k^{2}\ln A(k)-\frac{1}{4}q(0)\right\} dk \label{4.14}
\end{equation}
\begin{equation*}
=\frac{3}{8}q^{\prime }(0)+\frac{3}{8%
}\int_{0}^{\infty }q^{2}(x)dx.
\end{equation*}%
If also $q^{\prime \prime \prime }\in L_{1}\left( 0,\infty \right) $ and%
\begin{equation}
\frac{1}{k}\int_{0}^{\infty }\sin \left( 2kx-\frac{1}{k}\int_{0}^{x}q(s)ds%
\right) q^{^{\prime \prime \prime }}(x)dx\in L_{1}\left( r,\infty \right)
,r>0, \label{4.15}
\end{equation}
then
\begin{equation}
\sum_{l=1}^{\infty }\lambda _{l}^{2}+\frac{4}{\pi }\int_{0}^{\infty }\left\{
k^{3}\theta (k)-\frac{1}{8}\left( q^{\prime }(0)+\int_{0}^{\infty
}q^{2}(x)dx\right) \right\} dk \label{4.16}
\end{equation}
\begin{equation*}
=\frac{1}{8}q^{\prime \prime }(0)-\frac{1}{4}%
q^{2}(0).
\end{equation*}%
Conditions (\ref{4.13}) and (\ref{4.15}) are necessary for absolute
convergence of the integrals in (\ref{4.14}) and (\ref{4.16}) respectively.
\end{corollary}
\begin{remark}
It is easy to verify that the function $\eta $ defined by%
\begin{equation*}
\eta \left( t\right) =\frac{1}{\pi }\dint\limits_{0}^{t}\left( \theta (k)+%
\frac{1}{2k}\int_{0}^{\infty }q(x)\cos 2kxdx\right) dk
\end{equation*}%
is actually the modified spectral shift function (see (\ref{1.6}) and Remark
3) and formula (\ref{4.16}) transforms into (\ref{1.6}). Relation (\ref%
{1.6}) was previously proven in \cite{RyTr} by different methods under
slightly stronger conditions on $q.$ Note that Corollary 1 gives optimal
condition on $q$ for (\ref{1.6}) to hold with the absolutely convergent
integral. In the short-range setting necessary and sufficient conditions for
absolute summability of trace relations are recently given in \cite{RyNes}.
\end{remark}
\begin{remark}
When it comes to the long-range potentials there is a certain flexibility in
choosing the limiting phase $\theta .$ With a proper choice of $\theta $
some of the formulas (\ref{4.4}) may actually simplify.
\end{remark}
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\end{document}