\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 115, pp. 1--21.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/115\hfil Oscillation of solutions]
{Oscillation and nonoscillation of solutions to even order self-adjoint
differential equations}

\author[Ond\v{r}ej Do\v{s}l\'y \& Simona Fi\v{s}narov\'a\hfil EJDE--2003/115\hfilneg]
{Ond\v{r}ej Do\v{s}l\'y \& Simona Fi\v{s}narov\'a} % in alphabetical order

\address{Department of Mathematics,
Masaryk University, Jan\'{a}\v{c}kovo n\'{a}m. 2a,
         CZ-662 95 Brno, Czech Republic}
\email[Ond\v{r}ej Do\v{s}l\'y]{dosly@math.muni.cz}
\email[Simona Fi\v{s}narov\'a]{simona@math.muni.cz}

\date{}
\thanks{Submitted September 30, 2003. Published November 25, 2003.}
\thanks{Research supported by grant 201/01/0079 from the Czech Grant Agency}
\subjclass[2000]{34C10}
\keywords{Self-adjoint differential equation, variational method,
\hfill\break\indent oscillation and nonoscillation criteria,
 conditional oscillation}

\begin{abstract}
 We establish oscillation and nonoscilation criteria for the linear
 differential equation
 $$
 (-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
 \frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y=q(t)y,\quad
 \alpha \not\in \{1, 3, \dots , 2n-1\},
 $$
 where
 $$
 \gamma_{n,\alpha}=\frac{1}{4^n}\prod_{k=1}^n(2k-1-\alpha)^2
 $$
 and $q$ is a real-valued continuous function.
 It is proved, using these criteria, that the equation
 $$
 (-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-\big(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}
 + \frac{\gamma}{t^{2n-\alpha}\lg^2 t}\big)y = 0
 $$
 is nonoscillatory if and only if
 $$
 \gamma \leq \tilde \gamma_{n,\alpha}:=
 \frac{1}{4^n}\prod_{k=1}^n(2k-1-\alpha)^2
              \sum_{k=1}^n\frac{1}{(2k-1-\alpha)^2}.
 $$
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{cor}[thm]{Corollary}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{conjecture}[thm]{Conjecture}
\newtheorem{remark}[thm]{Remark}


\section{Introduction}

In this paper we investigate the oscillatory behavior of the two term 
self-adjoint linear differential equation of the form
\begin{equation}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}=p(t)y, \quad
\alpha \not \in \{1, 3, \dots , 2n-1\},
 \label{p}
\end{equation}
where $p$ is a continuous function.

Oscillatory properties of equation \eqref{p} has been investigated in
several recent papers, see
\cite{D-MN-97,D-Olomouc,D-O-CZMJ,D-O-Zilina,Fiedler-1,Fiedler-2,F,M-Pf}
and the references given therein.
In these papers, equation \eqref{p} is seen
as a perturbation of the one term equation
$$
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}=0,
$$
or of the Euler-type equation
\begin{equation}                                      \label{Euler}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y=0,
\end{equation}
where
\begin{equation} \label{gamma}
\gamma_{n,\alpha}:=(-1)^n\prod_{k=0}^{n-1}(\lambda-k)(\lambda+\alpha-k-n)
                      |_{\lambda=\frac{2n-1-\alpha}{2}}
= \frac{1}{4^n}\prod_{k=1}^n(2k-1-\alpha)^2,
\end{equation}
i.e., \eqref{p} was considered in the form
\begin{equation}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y
=q(t)y,  \label{q}
\end{equation}
in the latter case.

If $n=1$ and $\alpha=0$, then \eqref{p} reduces to the second order
equation
\begin{equation}                                \label{2-p}
y''+p(t)y=0
\end{equation}
whose oscillation theory is deeply developed, see \cite{Sw}.
In the classical oscillation criteria, \eqref{2-p} is viewed as
a perturbation of the equation $y''=0$ and oscillatory nature
of \eqref{2-p} depends on ``how much $p$ is positive'', the last
vague expression being specified in the quantitative way in particular
(non)oscillation criteria. If \eqref{2-p} is viewed as a
perturbation of the Euler equation with the ``critical'' constant
$1/4$
\begin{equation*}                                  \label{2-Euler}
y''+\frac{1}{4t^2}y=0,
\end{equation*}
i.e., \eqref{2-p} is written in the form
\begin{equation}                                          \label{2-q}
y''+\frac{1}{4t^2}y+q(t)y=0,\quad q(t)=p(t)-\frac{1}{4t^2},
\end{equation}
one gets more refined criteria and (non)oscillation of \eqref{2-p} is
``measured'' by positivity of the difference $p(t)-\frac{1}{4t^2}$.
Generally, the second order equation with
iterated logarithms
\begin{equation}                                     \label{iterated}
y''+\frac{1}{4t^2}\big(1+\frac{1}{\lg^2 t}+\dots+
\frac{1}{\lg^2 t\lg_2^2 t\dots \lg^2_{n-1} t}+\frac{1}
{\lg^2 t\lg_2^2 t\cdots \lg_n^2 t}\big)y=0,
\end{equation}
where $\lg_2 t=\lg (\lg t)$, $\lg_n t=\lg (\lg_{n-1} t)$ and $\lg $
denotes the natural logarithm, is nonoscillatory and one can view
\eqref{2-p} as perturbation of \eqref{iterated}. The more logarithmic
terms are involved, the more refined oscillation criteria are obtained.

Here we follow this line in case of higher order equations.
Equation \eqref{p} (and hence also \eqref{q}) is viewed
as a perturbation of the equation
\begin{equation}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
\big(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}
+ \frac{\gamma}{t^{2n-\alpha}\lg^2t}\big)y=0   \label{ln}
\end{equation}
with
\begin{equation}                                     \label{tilde-gamma}
\gamma=\tilde{\gamma}_{n,\alpha} :=\gamma_{n,\alpha}
              \sum_{k=1}^n\frac{1}{(2k-1-\alpha)^2},
\end{equation}
$\gamma_{n,\alpha}$ being given by \eqref{gamma}.
We establish oscillation and nonoscillation criteria for \eqref{q}
and we show that \eqref{ln} is oscillatory for
$\gamma > \tilde{\gamma}_{n,\alpha}$
and nonoscillatory for $\gamma \leq \tilde{\gamma}_{n,\alpha}$.
This approach can be regarded as a generalization of the results of the papers
\cite{D-Olomouc,F}, dealing with the fourth order equations, i.e. with the
case $n=2$.

To study equation \eqref{q}, we use
methods based on the factorization of disconjugate operators,
variational technique and the relationship between self-adjoint equations and
linear Hamiltonian systems, similarly as in the papers mentioned above.
We also use some combinatorial identities to determine the exact value of the
oscillation constant $\tilde{\gamma}_{n,\alpha}$ of (\ref{ln}).

This paper is organized as follows. The next section contains
necessary definitions and some auxiliary results concerning
self-adjoint equations.
In section 3 we present the main results of the paper --
oscillation and nonoscillation
criteria for \eqref{q}. In Section 4 we discuss some open problems and
possibilities of the extension of our results.
The last section contains technical computations related to the
combinatorial identities used in (non)oscillation criteria of
Section 3.

\section {Preliminaries}

We start with a statement concerning factorization of the
formally self-adjoint differential operators
\begin{equation}                                         \label{S-L}
L(y):=\sum_{k=0}^n (-1)^k \left(r_k(t)y^{(k)}\right)^{(k)}=0,
\quad r_n(t)>0.
\end{equation}
Note that the differential operator generated by the left-hand side of
\eqref{q} is a special case of the operator $L$.

\begin{lem}[\cite{C}]                      \label{L:factor}
Suppose that equation \eqref{S-L} possesses a system of positive solutions
$y_1,\dots,y_{2n}$ such that Wronskians
$W(y_1,\dots,y_k)\ne 0$, $k=1,\dots, 2n$, for large $t$.
Then the operator $L$ admits  the factorization for large $t$
$$
L(y)=\frac{(-1)^n}{a_0(t)}\bigg(\frac{1}{a_1(t)}\bigg(\dots \frac{r_n(t)}
{a_n^2(t)}\Big(
\frac{1}{a_{n-1}(t)}\dots \frac{1}{a_1(t)}\big(\frac{y}{a_0(t)}\big)'
\dots\Big)'\dots\bigg)'\bigg)',
$$
where
%\begin{eqnarray*}
$$
a_0=y_1,\ a_1=\big(\frac{y_2}{y_1}\big)',\
a_i=\frac{W(y_1,\dots,y_{i+1})W(y_1,\dots,y_{i-1})}{W^2(y_1,\dots,y_i)},\
i=1,\dots,n-1,
$$
and $a_n=(a_0\cdots a_{n-1})^{-1}$.
\end{lem}

An important role is played in our investigation by the following
specification of the factorization formula to the differential
operator given by the left-hand side of \eqref{Euler}.

\begin{lem}                                         \label{L:Euler-factor}
Let $\alpha\ne \{1,3,\dots,2n-1\}$.
Then we have for any sufficiently smooth function $y$
\begin{equation}
\begin{aligned}                                    \label{Euler-factor}
l(y)&:= (-1)^n \big(t^\alpha y^{(n)}\big)^{(n)}-
\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y \\
&=\frac{(-1)^n}{a_0(t)}\bigg(\frac{1}{a_1(t)}\bigg(\dots \frac{t^\alpha}
{a_n^2(t)}\Big(
\frac{1}{a_{n-1}(t)}\dots \frac{1}{a_1(t)}\big(\frac{y}{a_0(t)}\big)'
\dots\Big)'\dots\bigg)'\bigg)',
\end{aligned}
\end{equation}
where
\begin{equation*}                                %\label{alpha-formulas}
a_0(t)=t^{\alpha_0},\ a_k(t)=t^{\alpha_k-\alpha_{k-1}-1},\ k=1,\dots,n-1,\
a_n(t)=t^{(n-1)-\alpha_{n-1}},
\end{equation*}
with $\alpha_0=\frac{2n-1-\alpha}{2}$ and $\alpha_{1}<\dots<\alpha_{n-1}$
the first roots (ordered by their size) of the polynomial
\begin{equation}                                  \label{polynomial}
P(\lambda):=(-1)^n\prod_{i=0}^{n-1}(\lambda-i)(\lambda-n+\alpha-i)-
 \gamma_{n,\alpha}.
\end{equation}
\end{lem}

\begin{proof}
By a direct computation one can verify that the functions
$y=t^{\alpha_k}$, $k=0,\dots,n-1$, where $\alpha_k$ are the roots of
the polynomial $P$, are solutions of \eqref{Euler}.
Substituting into the formulas in Lemma \ref{L:factor} with
$y_1=t^{\alpha_0}$, $y_k=t^{\alpha_{k-1}}$, $k=2,\dots,n$,
we have
$$
a_0(t)=t^{\alpha_0}= t^{\frac{2n-1-\alpha}{2}},\quad
a_1(t)=\big(\frac{y_2}{y_1}\big)'=(\alpha_1-\alpha_0)t^{\alpha_1-\alpha_0-1}.
$$
In computing the remaining functions $a_k$, $k=2,\dots,n-1$, we use the
formula for computation of Wronskians of power functions
\begin{equation}                          \label{W-formula}
W(t^{\beta_1},\dots,t^{\beta_k})=
\prod_{1\leq i<j\leq k}(\beta_j-\beta_i)t^{\beta_1+\dots+\beta_k-
\frac{k(k-1)}{2}},
\end{equation}
see Lemma \ref{L:Wronskians} in the last section.
Using \eqref{W-formula}, for $k=2,\dots,n-1$,
\begin{align*}
&a_k(t)\\
&= \frac{W(t^{\alpha_0},\dots,t^{\alpha_k})W(t^{\alpha_0},\dots,
t^{\alpha_{k-2}})}{(W(t^{\alpha_0},\dots,t^{\alpha_{k-1}}))^2}
\\
&= \frac{t^{\alpha_0+\dots+\alpha_k-\frac{(k+1)k}{2}}t^{\alpha_0+\dots+
\alpha_{k-2}-\frac{(k-1)(k-2)}{2}}
\prod\limits_{0\leq i<j\leq k}^{}(\alpha_j-\alpha_i)
\prod\limits_{0\leq i<j\leq k-2}^{}(\alpha_j-\alpha_i)}
{\big(t^{\alpha_0+\dots+\alpha_{k-1}-\frac{(k-1)k}{2}}
\prod_{0\leq i<j\leq k-1}^{}(\alpha_j-\alpha_i)
\big)^2}
\\
&= \frac{\prod_{i=0}^{k-1}(\alpha_k-\alpha_i)}
{\prod_{i=0}^{k-2}
(\alpha_k-\alpha_i)}t^{\alpha_k-\alpha_{k-1}-1}
\\
&= \frac{(\alpha_k-\alpha_0)(\alpha_k-\alpha_1)\cdots (\alpha_k-\alpha_{k-1})}
{(\alpha_{k-1}-\alpha_0)(\alpha_{k-1}-\alpha_1)\cdots (\alpha_{k-1}-
\alpha_{k-2})} t^{\alpha_k-\alpha_{k-1}-1}
\end{align*}
and
$$
a_n=\frac{t^{(n-1)-\alpha_{n-1}}}
{(\alpha_{n-1}-\alpha_0)(\alpha_{n-1}-\alpha_1)\cdots
(\alpha_{n-1}-\alpha_{n-2})}.
$$
Since the product of all factors $(\alpha_j-\alpha_i)$
appearing in the functions $a_0,\dots,a_n$
equals $1$, we can neglect these terms and the factorization of the Euler
operator is really as stated.
\end{proof}

Now we recall basic oscillatory properties of self-adjoint
differential equations \eqref{S-L}.
These properties  can be investigated within the
scope of the oscillation theory of linear Hamiltonian systems (further LHS)
\begin{equation}                                         \label{LHS}
x'=A(t)x+B(t)u,\quad u'=C(t)x-A^T(t)u,
\end{equation}
where $A,B,C$ are $n\times n$ matrices with $B,C$ symmetric. Indeed,
if $y$ is a solution of \eqref{S-L} and we set
$$
x=\begin{pmatrix} y\\ y'\\ \vdots\\y^{(n-1)}\end{pmatrix},\quad
u=\begin{pmatrix} (-1)^{n-1}(r_ny^{(n)})^{(n-1)}+\dots+r_1y'\\
\vdots\\
-(r_ny^{(n)})'+r_{n-1}y^{(n-1)}\\
r_ny^{(n)}\end{pmatrix},
$$
then $(x,u)$ solves \eqref{LHS} with $A,B,C$ given by
\begin{gather*}                                            \label{ABC}
B(t)=\mathop{\rm diag}\{0,\dots,0,r^{-1}_n(t)\},
\quad
C(t)=\mathop{\rm diag}\{r_0(t),\dots,r_{n-1}(t)\},\\
A=A_{i,j}=\begin{cases} 1,& \mbox{if } j=i+1,\ i=1,\dots,n-1,\\
                0,&\mbox{elsewhere.}
\end{cases}
\end{gather*}
In this case we say that the solution $(x,u)$ of \eqref{LHS}
is generated by the solution $y$ of \eqref{S-L}. Moreover,
if $y_1,\dots,y_n$ are solutions of \eqref{S-L} and the columns of
the matrix solution $(X,U)$ of \eqref{LHS} are generated by the solutions
$y_1,\dots,y_n$, we say that the solution $(X,U)$  is generated
by the solutions $y_1,\dots,y_n$.

Recall that two different points $t_1,t_2$ are said to be {\it conjugate}
relative to system \eqref{LHS} if there exists a nontrivial solution
$(x,u)$ of this system such that $x(t_1)=0=x(t_2)$. Consequently,
by the above mentioned relationship between \eqref{S-L} and
\eqref{LHS}, these points are conjugate relative to \eqref{S-L} if
there exists a nontrivial solution $y$ of this equation such that
$y^{(i)}(t_1)=0=y^{(i)}(t_2)$, $i=0,1,\dots, n-1$.
System \eqref{LHS} (and hence also equation \eqref{S-L}) is said to
be {\it oscillatory} if for every $T\in \mathbb{R}$ there exists a pair of
points $t_1,t_2\in [T,\infty)$ which are conjugate
relative to \eqref{LHS} (relative to \eqref{S-L}), in the opposite
case \eqref{LHS} (or \eqref{S-L}) is said to be {\it nonoscillatory}.

Using the relation between \eqref{S-L}, \eqref{LHS} and the so-called
Roundabout Theorem for linear Hamiltonian systems (see e.g. \cite{Reid}),
one can easily prove the following variational lemma.

\begin{lem}[\cite{G}]                   \label{var-lemma}
Equation \eqref{S-L} is nonoscillatory if and only if there exists
$T\in \mathbb{R}$ such that
$$
{\mathcal F}(y;T,\infty):=\int_T^\infty
\Big[\sum_{k=0}^nr_k(t)(y^{(k)}(t))^2\Big]\,dt> 0
$$
for any nontrivial $y\in W^{n,2}(T,\infty)$ with compact support in
$(T,\infty)$.
\end{lem}

One of the main tools in our investigation is also the
following Wirtinger-type inequality.

\begin{lem}[\cite{G}]                 \label{Wirtinger}
Let $y\in W^{1,2}(T,\infty)$ have compact support in
$(T,\infty)$ and let $M$ be a positive differentiable function
such that $M'(t)\ne 0$ for $t\in [T,\infty)$. Then
$$
\int_T^{\infty}|M'(t)|y^2\,dt\leq 4 \int_T^{\infty}
\frac{M^2(t)}{|M'(t)|} y'{}^2\,dt.
$$
\end{lem}

Now we express the quadratic functional
associated with \eqref{Euler} in a way suitable for the application of the
Wirtinger inequality. This statement can be proved using
the repeated integration by parts, similarly as in \cite[Lemma 4]{D-Olomouc}.

\begin{lem}                              \label{Euler-Wirt}
Let $y\in W^{n,2}_0(T,\infty)$ have compact support in $(T,\infty)$.
Then
\begin{align*}
&\int_T^\infty \left[t^{\alpha}\big(y^{(n)}\big)^2-
\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y^2\right]dt\\
&=\int_T^\infty \frac{t^\alpha}{a_n}\bigg\{\bigg[\frac{1}{a_{n-1}}
\bigg(\frac{1}{a_{n-2}}\Big(\dots \frac{1}{a_1}\big(\frac{y}{a_0}\big)'
\Big)'\dots \bigg)'\bigg]'\bigg\}^2 dt,
\end{align*}
where $a_0,\dots, a_n$ are given in Lemma \ref{L:Euler-factor}.
\end{lem}


We finish this section with the concept of the principal system
of solutions of \eqref{S-L}.
%and one oscillation criterion based on this concept.
A conjoined basis $(X,U)$ of \eqref{LHS} (i.e. a matrix solution
of this system with $n\times n$ matrices $X,U$
satisfying $X^T(t)U(t)=U^T(t)X(t)$ and rank\,$(X^T,U^T)^T=n$)
is said to be the
{\it principal solution} of \eqref{LHS} if $X(t)$ is nonsingular for
large $t$ and for any other conjoined basis $(\bar X,\bar U)$ such that
the (constant) matrix $X^T\bar U- U^T\bar X$ is nonsingular
$\lim_{t\to \infty}\bar X^{-1}(t)X(t)=0$ holds. The last limit equals zero
if and only if
\begin{equation}                                         \label{principal}
\lim_{t\to \infty}\left(\int^t X^{-1}(s)B(s)X^{T-1}(s)\,ds\right)^{-1}=0,
\end{equation}
see \cite{Reid}. A principal solution of \eqref{LHS}
is determined uniquely up to a right multiple by a
constant nonsingular $n\times n$ matrix. If $(X,U)$ is the principal
solution,
any conjoined basis $(\bar X,\bar U)$ such that the matrix $X^T\bar U-
U^T\bar X$ is nonsingular is said to be a {\it nonprincipal solution}
of \eqref{LHS}.
Solutions $y_1,\dots,y_n$ of \eqref{S-L} are said to form the {\it principal
(nonprincipal) system of solutions} if the solution $(X,U)$ of the associated
linear Hamiltonian system generated by $y_1,\dots,y_n$ is a principal
(nonprincipal) solution. Note that if \eqref{S-L} possesses a
fundamental system of positive solutions $y_1,\dots,y_{2n}$
satisfying $y_i=o(y_{i+1})$ as $t\to \infty$, $i=1,\dots,2n-1$,
(the so-called {\it ordered system} of solutions), then the ``small''
solutions $y_1,\dots,y_n$  form the principal system of solutions of
\eqref{S-L}. In particular, if $L(y)=(-1)^n(t^\alpha y^{(n)})^{(n)}-
\gamma_{n,\alpha}t^{\alpha-2n}y$ and $\alpha_0,\dots,\alpha_{n-1}$
are the same as in Lemma \ref{L:Euler-factor}, then $y_k=t^{\alpha_k}$,
$k=1,\dots,n-1$, $y_n=t^{(2n-1-\alpha)/2}$ is the ordered
principal system of solutions of \eqref{Euler}.


\section{Main results -- oscillation and nonoscillation criteria}

We start this section with a statement where nonoscillation
of \eqref{q} is compared with nonoscillation of a certain
associated second order differential equation.

\begin{thm}                            \label{T:nonosc}
If the second order linear differential equation
\begin{equation}                        \label{second-eq}
(tz')'+\frac{1}{4\tilde \gamma_{n,\alpha}}t^{2n-1-\alpha}q(t)z=0
\end{equation}
is nonoscillatory, then equation \eqref{q} is also nonoscillatory.
\end{thm}

\begin{proof}
Let $T \in \mathbb R$ and $y \in W^{n,2}(T,\infty)$ be any function
having compact support in $(T, \infty).$
Using Lemma \ref{Euler-Wirt}, Wirtinger inequality (Lemma \ref{Wirtinger}),
which we apply $(n-1)$-times, and also Lemma \ref{L:nonosc_thm} from the last
section, we have
\begin{align*}
&\int_T^\infty \left[t^{\alpha}\big(y^{(n)}\big)^2-
\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}y^2\right]dt\\
&=
\int_T^\infty \frac{t^\alpha}{a_n}\bigg\{\bigg[\frac{1}{a_{n-1}}
\bigg(\frac{1}{a_{n-2}}\Big(\dots \frac{1}{a_1}\big(\frac{y}{a_0}\big)'
\Big)'\dots \bigg)'\bigg]'\bigg\}^2 dt\\
&\geq
\prod_{k=1}^{n-1}\Big(\frac{2n-1-\alpha}{2}-\alpha_k\Big)^2
\int_T^\infty t\big[\big(\frac{y}{a_0}\big)'\big]^2dt\\
&= 4 \tilde \gamma_{n,\alpha}
\int_T^\infty t\Big[\Big(\frac{y}{t^{(2n-1-\alpha)/2}}\Big)'\Big]^2dt.
\end{align*}
If we denote $z=y/t^{(2n-1-\alpha)/2}$ then, since
\eqref{second-eq} is nonoscillatory, it follows from Lemma \ref{var-lemma},
that
$$
\int_T^\infty \Big[t(z'(t))^2-\frac{1}{4\tilde \gamma_{n,\alpha}}
t^{2n-1-\alpha}q(t)z^2(t)\Big]dt>0.
$$
Summarizing
\begin{align*}
&\int_T^\infty \Big[t^{\alpha}\big(y^{(n)}\big)^2-
\left(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}+q(t)\right)y^2\Big]dt
\\
&\geq 4 \tilde \gamma_{n,\alpha}
\int_T^\infty t\Big\{\Big[\Big(\frac{y}{t^{(2n-1-\alpha)/2}}\Big)'\Big]^2
-\frac{1}{4\tilde \gamma_{n,\alpha}}q(t)y^2(t)\Big\}dt
\\
&=4 \tilde \gamma_{n,\alpha}
\int_T^\infty t\Big\{\Big[\Big(\frac{y}{t^{(2n-1-\alpha)/2}}\Big)'\Big]^2
-\frac{1}{4\tilde \gamma_{n,\alpha}}t^{2n-1-\alpha}q(t)
\big(\frac{y}{t^{(2n-1-\alpha)/2}}\big)^2\Big\}dt>0.
\end{align*}
The nonoscillation of \eqref{q} follows now from Lemma \ref{var-lemma}.
\end{proof}

In the proof of the next oscillatory counterpart of the previous
theorem, in addition to the constant $\tilde \gamma_{n,\alpha}$,
three other constants appeared, namely $K_{n,\alpha}$,
$\tilde K_{n,\alpha}$ and $L_{n,\alpha}$. To prove the statement of
this theorem, we needed the equalities $K_{n,\alpha}=\tilde K_{n,\alpha}$
and $L_{n,\alpha}=\tilde \gamma_{n,\alpha}$. The formulas which defined
these constants looked completely different on the first view
(compare below given formulas \eqref{E:K-def}, \eqref{E:L-def},
\eqref{E:tilde-K-def}) and the proof of the required equalities
leads to interesting combinatorial identites which are presented in
the last section.

\begin{thm}                            \label{T:osc}
Suppose that $q(t)\geq 0$ for large $t$ and
\begin{equation}                                   \label{integral-cond}
\int^\infty \Big(q(t)-\frac{\tilde\gamma_{n,\alpha}}{t^{2n-\alpha}\lg^2t}\Big)
t^{2n-1-\alpha}\lg t\,dt=\infty.
\end{equation}
Then equation \eqref{q} is oscillatory.
\end{thm}

\begin{proof}
Let $T\in \mathbb{R}$ be arbitrary, $T<t_0<t_1<t_2<t_3$ (these values will
be specified later). Further, let
$$
h(t)=t^{\frac{2n-1-\alpha}{2}}\sqrt{\lg t},
$$
$f\in C^n[t_0,t_1]$ be any function such that
$$
f^{(j)}(t_0)=0,\; f^{(j)}(t_1)=h^{(j)}(t_1),\; j=0,\dots,n-1,
$$
and $g$ be the solution of \eqref{Euler} satisfying the boundary conditions
\begin{equation}                                    \label{conditions}
g^{(j)}(t_2)=h^{(j)}(t_2),\; g^{(j)}(t_3)=0,\; j=0,\dots,n-1.
\end{equation}
We construct a function $0\not \equiv y\in W^{n,2}(T,\infty)$
with compact support in $(T,\infty),$ as follows
\begin{equation}                                  \label{E:y-def}
y(t)=\begin{cases}
 0,   & t\leq t_0, \\
 f(t),& t_0\leq t \leq t_1, \\
 h(t),& t_1\leq t \leq t_2, \\
 g(t),& t_2\leq t \leq t_3, \\
 0,   & t\geq t_3.
\end{cases}
\end{equation}
We show that for $t_2,\;t_3$ sufficiently large
$$
{\mathcal F}(y;T,\infty)
:=\int_T^\infty \Big[t^\alpha \big(y^{(n)}(t)\big)^2
-\left(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}
+q(t)\right)y^2(t)\Big] dt \leq0
$$
and hence \eqref{q} is oscillatory according to Lemma \ref{var-lemma}.
To this end, denote
$$
K:=\int_{t_0}^{t_1}\Big[t^\alpha \big(f^{(n)}(t)\big)^2
-\left(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}
+q(t)\right)f^2(t)\Big] dt.
$$
Concerning the interval $[t_1, t_2]$, let us compute $\left(h^{(n)}\right)^2.$
We use the usual convention that the value of a product
$\prod_{m}^n$ equals $1$ whenever the lower multiplication limit
$m$ is greater that the upper limit $n$.
Using Lemma \ref{L:lg-deriv} from the last section we have
\begin{align*}
&h^{(n)}(t)\\
&=\left(t^{\frac{2n-1-\alpha}{2}}\sqrt{\lg t}\right)^{(n)}\\
&= \frac{1}{2^n}\prod_{k=1}^{n}(2k-1-\alpha)
t^{\frac{-1-\alpha}{2}}\sqrt{\lg t}
\\
&+\sum_{j=1}^{n}\binom{n}{j}
\frac{1}{2^{n-j}}\prod_{k=j+1}^{n}(2k-1-\alpha)
\frac{(-1)^{j-1}}{t^j}
\Big[\frac{a_j}{\sqrt{\lg t}}+\frac{b_j}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big]t^{\frac{2j-1-\alpha}{2}}
\\
&=t^{\frac{-1-\alpha}{2}}\Big[\frac{1}{2^n}\prod_{k=1}^{n}(2k-1-\alpha)\sqrt{\lg t}
+\frac{A_n}{\sqrt{\lg t}}+\frac{B_n}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big],
\end{align*}
as $t\to \infty$, where
\begin{gather}
A_n=
\sum_{j=1}^{n}(-1)^{j-1}\binom{n}{j}
\frac{a_j}{2^{n-j}}\prod_{k=j+1}^{n}(2k-1-\alpha),\label{A_n}\\
B_n=
\sum_{j=1}^{n}(-1)^{j-1}\binom{n}{j}
\frac{b_j}{2^{n-j}}\prod_{k=j+1}^{n}(2k-1-\alpha).\label{B_n}
\end{gather}
Then we have
\begin{align*}
\big(h^{(n)}\big)^2
&=t^{-1-\alpha}\Big[\frac{\lg t}{4^n}\prod_{k=1}^{n}(2k-1-\alpha)^2
+\frac{A_n}{2^{n-1}}\prod_{k=1}^{n}(2k-1-\alpha)
\\
&\quad + \frac{B_n}{2^{n-1}\lg t}\prod_{k=1}^{n}(2k-1-\alpha)
+\frac{A^2_n}{\lg t}+2\frac{A_nB_n}{\lg^2t}+\frac{B^2_n}{\lg^3 t}
+ O\left(\lg^{-3}t\right)\Big]
\end{align*}
as $t\to \infty$. If we denote
\begin{gather}                                \label{E:K-def}
K_{n,\alpha}:=\frac{A_n}{2^{n-1}}\prod_{k=1}^{n}(2k-1-\alpha), \\
                              \label{E:L-def}
L_{n,\alpha}:=A^2_n+\frac{B_n}{2^{n-1}}\prod_{k=1}^{n}(2k-1-\alpha),
\end{gather}
we get
$$
\big(h^{(n)}\big)^2=t^{-1-\alpha}\Big[
\gamma_{n,\alpha}\lg t+K_{n,\alpha}+\frac{L_{n,\alpha}}{\lg t}
+O\left(\lg^{-2}t\right)\Big],
$$
as $t\to \infty$. Consequently,
$$
\int_{t_1}^{t_2} \Big[t^\alpha \left(h^{(n)}(t)\right)^2-\frac{\gamma_{n,\alpha}}
{t^{2n-\alpha}}h^2(t)\Big]dt=
K_{n,\alpha}\lg t_2 + L_{n,\alpha} \int_{t_1}^{t_2}\frac{dt}{t\lg t}+
L_1 + o(1),
$$
as $t\to \infty$, where $L_1$ is a real constant.

Now we turn our attention to the interval $[t_2, t_3]$.
Using the assumption that
$q(t)\geq0$ for large $t$ we have
\begin{align*}
&\int_{t_2}^{t_3} \Big[t^\alpha \left(g^{(n)}(t)\right)^2
-\left(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}
+q(t)\right)g^2(t)\Big] dt \\
&\leq
\int_{t_2}^{t_3} \Big[t^\alpha \left(g^{(n)}(t)\right)^2
-\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}g^2(t)\Big] dt.
\end{align*}
Further, denote
$$
x=\begin{pmatrix} g\\ g'\\ \vdots\\g^{(n-1)}\end{pmatrix},\quad
u=\begin{pmatrix} (-1)^{n-1}(t^\alpha g^{(n)})^{(n-1)}\\
\vdots\\
-(t^\alpha g^{(n)})'\\
t^\alpha g^{(n)}\end{pmatrix},\quad
\tilde h=\begin{pmatrix} h\\ h'\\ \vdots\\h^{(n-1)}\end{pmatrix}.
$$
Since $g$ is a solution of \eqref{Euler}, we can use a relationship between
this equation and associated LHS \eqref{LHS}. In this case
\begin{gather*}
B(t)=\mathop{\rm diag}\{0,\dots,0,t^{-\alpha}\},
\quad
C(t)=\mathop{\rm diag}\{-\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}},0,\dots,0\},\\
A=A_{i,j}=\begin{cases}1, &\mbox{if } j=i+1,\ i=1,\dots,n-1,\\
                0,&\mbox{elsewhere.}
\end{cases}
\end{gather*}
This relationship, together with conditions \eqref{conditions} imply that
\begin{align*}
&\int_{t_2}^{t_3} \Big[t^\alpha \big(g^{(n)}\big)^2(t)-\frac{\gamma_{n,\alpha}}
{t^{2n-\alpha}}g^2(t)\Big]dt\\
&=\int_{t_2}^{t_3}[u^T(t) B(t)u(t)+x^T(t) C(t)x(t)] dt\\
&= \int_{t_2}^{t_3}[u^T(t)(x'(t)-Ax(t))+x^T(t) C(t)x(t)] dt\\
&= u^T(t) x(t) \big|_{t_2}^{t_3}
+  \int_{t_2}^{t_3}x^T(t)[-u'(t)-A^T u(t)+C(t)x(t)] dt\\
&= -u^T(t_2)x(t_2).
\end{align*}
Let $(X,U)$ be the principal solution of the LHS associated with
\eqref{Euler}.Then $(\bar{X},\bar{U})$ defined by
\begin{gather*}
\bar{X}(t)=X(t)\int_t^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds,\\
\bar{U}(t)=U(t)\int_t^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds - X^{T-1}(t)
\end{gather*}
is also a conjoined basis of this LHS, and according to \eqref{conditions}
we get
\begin{align*}
x(t)&= X(t)\int_t^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds
\Big( \int_{t_2}^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds \Big)^{-1}\\
&\quad\times X^{-1}(t_2) \tilde{h}(t_2),
\\
u(t)&= \Big( U(t)\int_t^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds -X^{T-1}(t)\Big)
\\
&\quad\times\Big( \int_{t_2}^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds \Big)^{-1}
X^{-1}(t_2) \tilde{h}(t_2)
\end{align*}
and hence
\begin{align*}
-u^T(t_2)x(t_2) &= \tilde{h}^T(t_2)X^{T-1}(t_2)
\left( \int_{t_2}^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds \right)^{-1}
X^{-1}(t_2) \tilde{h}(t_2)\\
&\quad- \tilde{h}^T(t_2)U(t_2)X^{-1}(t_2)\tilde{h}(t_2).
\end{align*}
Using the fact that the principal solution of LHS associated with
\eqref{Euler} is generated by
$y_1(t)=t^{\alpha_1},\; y_2(t)=t^{\alpha_2}, \dots,
y_{n-1}(t)=t^{\alpha_{n-1}},\;y_n(t)=t^{\alpha_0}=t^{\frac{2n-1-\alpha}{2}}$,
where $\alpha_k,$ $k=0,\dots,n-1,$ are the roots of \eqref{polynomial},
we have
\[
X(t)=\begin{pmatrix}
  t^{\alpha_1} & \cdots & t^{\alpha_{n-1}} & t^{\frac{2n-1-\alpha}{2}} \\
  \alpha_1t^{\alpha_1-1} & \cdots & \alpha_{n-1}t^{\alpha_{n-1}-1}
  & \frac{2n-1-\alpha}{2}t^{\frac{2n-3-\alpha}{2}}\\
  \vdots &  &  & \vdots \\
  \scriptstyle{\prod\limits_{k=0}^{n-2}(\alpha_1-k)t^{\alpha_1-n+1}}& \cdots
  & \scriptstyle{\prod\limits_{k=0}^{n-2}(\alpha_{n-1}-k)t^{\alpha_{n-1}-n+1}}&
  \scriptstyle{\frac{1}{2^{n-1}}\prod\limits_{k=1}^n(2k-1-\alpha)
  t^{\frac{1-\alpha}{2}}}
\end{pmatrix}
\]
$$
U(t)=\left(
\begin{array}{ccc}\ell^{[1,1]}_{n,\alpha} t^{\alpha_1+\alpha-2n+1}&
\dots & \ell^{[1,n]}_{n,\alpha}t^{\frac{\alpha-2n+1}{2}}\\
\vdots &  &\vdots \\
\ell^{[n,1]}_{n,\alpha} t^{\alpha_1+\alpha-n}&\dots &
\ell^{[n,n]}_{n,\alpha} t^{\frac{\alpha-1}{2}}
\end{array}
\right),
$$
where
\begin{align*}
\ell^{[1,1]}_{n,\alpha}&=(-1)^{n-1}
\prod_{j=0}^{n-1}(\alpha_1-j)\prod_{j=0}^{n-2}(\alpha_1+\alpha-n-j),
\\
\ell^{[1,n]}_{n,\alpha}&= \frac{2n-1-\alpha}{2}
\prod_{j=0}^{n-1}\left(\frac{2n-1-2j-\alpha}{2}\right)^2,
\\
\ell^{[n,1]}_{n,\alpha}&= \prod_{j=0}^{n-1}(\alpha_1-j),
\\
\ell^{[n,n]}_{n,\alpha}&= \prod_{j=0}^{n-1}\left(\frac{2n-1-2j-\alpha}{2}\right),
\end{align*}
and
$$
\tilde h(t)=
\begin{pmatrix}
  t^{\frac{2n-1-\alpha}{2}}\sqrt{\lg t} \\
  t^{\frac{2n-3-\alpha}{2}}\left(\frac{2n-1-\alpha}{2}\sqrt{\lg t}
  +O(\lg^{-\frac{1}{2}}(t))\right) \\
  \vdots \\
  t^{\frac{1-\alpha}{2}}
  \left(\frac{1}{2^{n-1}}\prod_{k=1}^n(2k-1-\alpha)\sqrt{\lg t}
  +O(\lg^{-\frac{1}{2}}t)\right)
\end{pmatrix},\quad \text{as } t\to \infty
$$
Next we compute the asymptotic formula for
$(U^T\tilde h)^TX^{-1}\tilde h\big|_{t=t_2}$. Using Lemma
\ref{L:Wronskians} we have
\begin{gather*}
(X^{-1}\tilde h)_i\sim t^{\alpha_i-\frac{2n-1-\alpha}{2}}
\frac{1}{\sqrt{\lg t}},\quad i=1, \dots,n-1,
\\
(X^{-1}\tilde h)_n=\sqrt{\lg t}\left(1+O(\lg^{-1}t)\right),\quad
\text{as}\ t\to \infty.
\end{gather*}
Here $f_1\sim f_2$ for a pair of functions $f_1,f_2$ means that
$\lim_{t\to \infty}\frac{f_1(t)}{f_2(t)}=L$ exists and $0<L<\infty$.
By a direct computation, for $i=1,\dots,n-1$,
\begin{equation*}
(U^T\tilde h)_i=l_{n,\alpha}^{[i]}t^{\alpha_i-\frac{2n-1-\alpha}{2}}\sqrt{\lg t},
\end{equation*}
the constants $l^{[i]}_{n,\alpha}$ can be computed explicitly,
but their values are not important. As for $(U^T\tilde h)_n$
denote by $u_n$ the last column of $U$.
Then, again by a direct computation,
$$
(U^T\tilde h)_n=u_n^T\tilde h=\tilde K_{n,\alpha}\sqrt{\lg t}
(1+O(\lg^{-1} t)),\quad \text{as}\ t\to \infty,
$$
where
\begin{equation}                                   \label{E:tilde-K-def}
\tilde K_{n,\alpha}=\frac{2}{4^n}\prod_{k=1}^n(2k-1-\alpha)^2
\sum_{k=1}^n\frac{1}{(2k-1-\alpha)}.
\end{equation}
Consequently,
$$
\tilde{h}^T(t_2) U(t_2)X^{-1}(t_2)\tilde h(t_2)=\tilde K_{n,\alpha}\lg{t_2}+
L_2+o(1),\quad \text{as\ }t_2\to \infty,
$$
where $L_2$ is a real constant.

Summarizing all the above computations
\begin{align*}
\mathcal{F}(y;T,\infty)
&\leq K+K_{n,\alpha}\lg t_2 + L_{n,\alpha}
 \int_{t_1}^{t_2}\frac{dt}{t\lg t}+ L_1 + o(1)-\int_{t_1}^{t_2}q(t)h^2(t)dt
\\
&\quad +\tilde{h}^T(t_2)X^{T-1}(t_2)
\left( \int_{t_2}^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds \right)^{-1}
X^{-1}(t_2) \tilde{h}(t_2)
\\
&\quad -\tilde K_{n,\alpha}\lg{t_2}-L_2-o(1),\quad \text{as}\ t_2\to \infty.
\end{align*}
It follows from Lemma \ref{L:Kn=tildeKn} that $K_{n,\alpha}=
\tilde K_{n,\alpha}$
and $L_{n,\alpha}=\tilde \gamma_{n,\alpha}$ according to Lemma \ref{L:gamma=On}.
Using \eqref{integral-cond} let $t_2>t_1$ be such that
\begin{align*}
L_{n,\alpha} \int_{t_1}^{t_2}\frac{dt}{t\lg t}-\int_{t_1}^{t_2}q(t)h^2(t)dt
&=-\int_{t_1}^{t_2} \left(q(t)-\frac{\tilde\gamma_{n,\alpha}}{t^{2n-\alpha}\lg^2t}\right)
t^{2n-1-\alpha}\lg tdt\\
&\leq-(K+L_1-L_2+2).
\end{align*}
Since $(X, U)$ is  the principal solution, it is possible to choose
$t_3>t_2$ such that
$$
\tilde{h}^T(t_2)X^{T-1}(t_2)
\left( \int_{t_2}^{t_3} X^{-1}(s)B(s) X^{T-1}(s) ds \right)^{-1}
X^{-1}(t_2) \tilde{h}(t_2)\leq 1.
$$
Finally, if $t_2$ is so large that the sum of all the terms
$o(1)$ is less then 1,
then for these $t_2,\;t_3$ we have
$${\mathcal F}(y;T,\infty)\leq K -(K+L_1-L_2+2)+L_1+1+1-L_2=0,$$
which means that \eqref{q} is oscillatory.
\end{proof}

\begin{cor}                                 \label{cor-ekviv}
The equation \eqref{ln} is nonoscillatory if and only if
$\gamma\leq \tilde \gamma_{n,\alpha}.$
\end{cor}

\begin{proof}
If $\gamma\leq \tilde \gamma_{n,\alpha},$ then the second order equation
$$
(tz')'+\frac{1}{4 \tilde \gamma_{n,\alpha}}t^{2n-1-\alpha}
\frac{\gamma}{t^{2n-\alpha}\lg^2t}z=0
$$
is nonoscillatory, which follows from the fact, that the equation
$$
(tz')'+\frac{\mu}{t\lg^2t}z=0
$$
is nonoscillatory for $\mu\leq\frac{1}{4}.$
Hence, \eqref{ln} is nonoscillatory according to Theorem \ref{T:nonosc}.
Conversely, if $\gamma > \tilde \gamma_{n,\alpha},$ then for
$q(t)=\frac{\gamma}{t^{2n-\alpha}\lg^2t}$ condition \eqref{integral-cond} holds
and we have oscillation of \eqref{ln} using Theorem \ref{T:osc}.
\end{proof}

\section{Remarks and open problems}

(i) The oscillation criterion given in Theorem \ref{T:osc} is proved
under the assumption
\begin{equation}                      \label{q-geq-0}
q(t)\geq 0\quad  \mbox{for large }t.
\end{equation}
This restriction has been successfully removed in oscillation criteria
presented in
some recent papers \cite{D-MN-94,D-JMAA,D-MN-97,F,M-Pf}.
In particular, it was proved for equation \eqref{p} with $n=2$ that the
function $g/h$ (the function $h,g$ apper in \eqref{E:y-def})
is monotonically decreasing
and this fact enabled to remove the assumption \eqref{q-geq-0}
via the second mean value theorem of integral
calculus, see \cite{D-JMAA}. The computations
proving monotonicity of $g/h$ are rather complex even for
$n=2$ and we have not been able to prove this monotonicity in the
general case yet. However, we believe that the function $g/h$ is monotonic
also in the general case treated in our paper and we conjecture
that Theorem \ref{T:osc} remains valid without
assumption \eqref{q-geq-0}.
\smallskip

(ii) In \cite{D-EJDE} we have discussed the problem of the value of the best
constants in oscillation and nonoscillation criteria for
equations of the form \eqref{p} and \eqref{q}. In particular, it is known
(see \cite{D-EJDE,D-O-CZMJ}) that equation \eqref{q} is oscillatory if
\begin{equation}                                      \label{lg-osc}
M:=\lim_{t\to \infty} \lg t \int_t^\infty q(s)
s^{2n-1-\alpha}\,ds >\omega_{n,\alpha}
\end{equation}
and it is nonoscillatory if the above limit is less than
$\omega_{n,\alpha}/4$, where
\begin{equation}                                  \label{omega-def}
\omega_{n,\alpha}= \frac{(-1)^n\prod_{i=0}^{n-1}(\lambda-i)
(\lambda-n+\alpha-i)-\gamma_{n,\alpha}}{\left(\lambda-
\frac{2n-1-\alpha}{2}\right)^2}\Big|_{\lambda=\frac{2n-1-\alpha}{2}}.
\end{equation}
An open problem remained what is the oscillatory nature of \eqref{q}
if the limit in \eqref{lg-osc} is between $\omega_{n,\alpha}/4$
and $\omega_{n,\alpha}$. Here we answer this question by showing that the ``right''
oscillation constant is $\omega_{n,\alpha}/4$, i.e. \eqref{q}
is oscillatory if the limit in \eqref{lg-osc} is greater than
this constant.

Observe that $\omega_{n,\alpha}/4=\tilde \gamma_{n,\alpha}$,
this identity is proved in Lemma \ref{L:gamma-omega} of the last section.
If $q(t)=\frac{\lambda}{t^{2n-\alpha}\lg^2 t}$, the next statement is in
the full agreement with Corollary \ref{cor-ekviv}.

\begin{thm}                           \label{T:improvement}
Suppose that \eqref{q-geq-0} holds.
Equation \eqref{q} is oscillatory if the limit $M$ in \eqref{lg-osc}
is greater than $\tilde \gamma_{n,\alpha}$ and it is nonoscillatory
if it is less than this constant.
\end{thm}

\begin{proof}
If $M>\omega_{n,\alpha}$ in \eqref{lg-osc}, equation \eqref{q}
is oscillatory by \cite[Theorem 4.1]{D-O-CZMJ}.
Hence we suppose that $\tilde \gamma_{n,\alpha}<M\leq
\omega_{n,\alpha}=4\tilde \gamma_{n,\alpha}$. In this case we use
Theorem \ref{T:osc}.
Since $M>\tilde \gamma_{n,\alpha}$, there exist $\varepsilon>0$
and $T\in \mathbb{R}$ such that
$$
\int_t^\infty q(s)s^{2n-1-\alpha}\,ds >\frac{\tilde \gamma_{n,\alpha}+
\varepsilon}{\lg t}\quad \text{for }t\geq T,
$$
and hence, multiplying the last inequality by $\frac{1}{t}$ and integrating
it from $T$ to $b$ we get
$$
\int_T^b \frac{1}{t}\int_t^\infty q(s)s^{2n-1-\alpha}\,ds
>(\tilde \gamma_{n,\alpha}+\varepsilon)\lg\big(
\frac{\lg b}{\lg T}\big)
$$
for $b>T$. Integration by parts yields
\begin{align*}
&\int_T^b \Big(q(s)-\frac{\tilde \gamma_{n,\alpha}}{t^{2n-\alpha}
\lg^2t}\Big)t^{2n-1-\alpha}\lg t\,dt
\\
&= \int_T^b q(t)t^{2n-1-\alpha}\lg t\,dt
- \tilde \gamma_{n,\alpha}\lg \big(\frac{\lg b}{\lg T}\big)
\\
&=-\lg t \int_t^\infty q(s)s^{2n-1-\alpha}\,ds
\Big|_T^b + \int_T^b \frac{1}{t}\Big(\int_t^\infty
q(s)s^{2n-1-\alpha}\,ds\Big)dt
-\tilde \gamma_{n,\alpha}\lg \big(\frac{\lg b}{\lg T}\big)
\\
&>-\lg t\int_t^\infty q(s)s^{2n-1-\alpha}\,ds
\Big|_{T}^b+ (\tilde \gamma_{n,\alpha}+\varepsilon-\tilde \gamma_{n,\alpha})
 \lg \big(\frac{\lg b}{\lg T}\big)\to \infty
\end{align*}
as $b\to \infty$ since the first term in the last line of the previous
computation is bounded as $t\to \infty$.
Hence, by Theorem \ref{T:osc} equation \eqref{q} is oscillatory.
\end{proof}

Note that assumption \eqref{q-geq-0} in the oscillatory part of
Theorem \ref{T:improvement}
can be removed
if the conjecture formulated in the previous remark turns out to be true.
\smallskip

(iii) Let $L$ be a formally self-adjoint differential operator
given by \eqref{S-L} and consider the equation
\begin{equation}                            \label{w-equation}
L(y)=\lambda w(t)y,
\end{equation}
where $w$ is a positive continuous function. This equation is
said to be {\it conditionally oscillatory} if there exists
a constant $\lambda _0$, the so called {\it oscillation constant},
such that \eqref{w-equation} is oscillatory for $\lambda >\lambda _0$
and nonoscillatory for $\lambda <\lambda _0$.
If we put now
\begin{equation}                                  \label{L-special}
L(y):=(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
\left(\frac{\gamma_{n,\alpha}}{t^{2n-\alpha}}+
\frac{\tilde \gamma_{n,\alpha}}{t^{2n-\alpha}\lg^2 t}\right)y
\end{equation}
a natural question is for what fuction $w$ equation \eqref{w-equation}
with $L$ given by \eqref{L-special} is  conditionally oscillatory
equation. Theorem \ref{T:nonosc} and oscillatory behavior of the
second order equation \eqref{second-eq} lead to the conjecture
(whose proof is a subject of the present investigation)
that this term is
$$
w(t)=\frac{1}{t^{2n-\alpha} \lg^2 t \lg^2(\lg t)}
$$
and that the oscillation constant is
$\lambda _0=\tilde \gamma_{n,\alpha }$.

(iv) The previous remark, again together with Theorem \ref{T:nonosc}, lead
to the following conjecture.

\begin{conjecture} Let
\begin{equation*}                               %\label{K-limit}
K:=\lim_{t\to \infty} \lg(\lg t)\int_t^\infty
\Big(q(s)-\frac{\tilde \gamma_{n,\alpha }}{s^{2n-\alpha }\lg^2 s}\Big)
s^{2n-\alpha -1}\lg s\,ds.
\end{equation*}
There exists a constant $\hat \gamma$ {\rm (}presumably $\hat \gamma=
\tilde \gamma_{n,\alpha }${\rm )} such that \eqref{q} is oscillatory
provided $K>\hat \gamma$.
\end{conjecture}

Note that the nonoscillatory complement of the previous
conjecture is true by Theorem
\ref{T:nonosc}. Indeed, the second order equation \eqref{second-eq}, when
written in the form
\begin{equation}                                \label{E:second-4}
(tu')' +\frac{1}{4t\lg^2 t}u+\Big(\frac{t^{2n-1-\alpha}q(t)}
{4\tilde \gamma_{n,\alpha}}-\frac{1}{4t\lg^2 t}\Big)u=0
\end{equation}
is nonoscillatory, provided
\begin{equation*}
\frac{t^{2n-1-\alpha}q(t)} {4\tilde \gamma_{n,\alpha}}-\frac{1}{4t\lg^2 t}
\geq 0
\end{equation*}
for large $t$ and
\begin{equation}                              \label{lg-lg}
\lim_{t\to \infty} \lg(\lg t)\int_t^\infty
\Big(\frac{s^{2n-1-\alpha}q(s)}{4\tilde\gamma_{n,\alpha}}-\frac{1}{4s\lg^2 s}
\Big)\lg s\,ds<\frac{1}{4}
\end{equation}
and the last conditions just the condtion $K<\tilde \gamma_{n,\alpha }$.
Condition \eqref{lg-lg} follows from the Hille nonoscillation criterion
which states that the second order differential equation
\begin{equation*}
(r(t)x')'+c(t)x=0
\end{equation*}
with $c(t)\geq 0$, $\int^\infty c(t)\,dt<\infty$ and $\int^\infty
r^{-1}(t)\,dt=\infty$ is nonoscillatory provided
\begin{equation*}
\lim_{t\to \infty}\Big(\int^t r^{-1}(s)\,ds\Big)
\Big(\int_t^\infty c(s)\,ds\Big)<\frac{1}{4}.
\end{equation*}
The transformation $u=\sqrt{\lg t}\,v$ transforms \eqref{E:second-4}
into the equation
\begin{equation*}
\left(t\lg t v'\right)'+
\left(\frac{t^{2n-1-\alpha}q(s)}{4\tilde\gamma_{n,\alpha}}-
\frac{1}{4t\lg^2 t} \right)\lg t\,v=0
\end{equation*}
and Hille's criterion applied to this equation gives \eqref{lg-lg}.
\smallskip

(v) Throughout the paper we consider the case
$\alpha \not\in \{1,3,\dots,2n-1\}$ only. The reason is
that for $\alpha \in \{1,3,\dots,2n-1\}$ the Euler equation
\begin{equation}                             \label{E:Euler}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
\frac{\lambda }{t^{2n-\alpha }}y=0
\end{equation}
is no longer conditionally oscillatory and
one has to consider the equation
\begin{equation}                           \label{alpha-crit}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
\frac{\lambda }{t^{2n-\alpha }\lg^2 t}y=0.
\end{equation}
According to \cite{D-EJDE} and \cite{D-O-Zilina}, equation
\eqref{alpha-crit} is oscillatory for the values $\lambda >\nu_{n,m}:=
[m!(n-m-1)!]^2/4$ and $m:=(2n-1-\alpha )/2$,
and nonoscillatory in the opposite case. In the proof of Theorem
\ref{T:osc} we have defined the function $g$ as the solution of
\eqref{Euler} (satisfying certain boundary conditions) and we have
used the fact that we know solutions (even if with generally
unknown exponents) of \eqref{Euler}. Concerning equation \eqref{alpha-crit},
we do not know solutions explicitly even for $n=2$ and
$\lambda=\nu_{2,m}$, so we cannot apply
directly the method used in the proof of Theorem \ref{T:osc}.
Nevertheless, we conjecture that the equation
\begin{equation}                          \label{alpha-perturb}
(-1)^n\big(t^\alpha y^{(n)}\big)^{(n)}-
\frac{\nu_{n,m}}{t^{2n-\alpha }\lg^{2}t}y=q(t)y,\quad
\alpha \in \{1,3,\dots,2n-1\},
\end{equation}
is oscillatory provided $q(t)\geq 0$ for large $t$ and
\begin{equation}                                      \label{nu}
\lim_{t\to \infty}\lg(\lg t)\int_t^\infty q(s)s^{2n-1-\alpha}\lg s\,ds>
\nu_{n,m}.
\end{equation}
Note that by \cite[Theorem 3.1]{D-O-Zilina} equation \eqref{alpha-perturb}
is nonoscillatory provided the second order equation
\begin{equation}                                   \label{2-q-lg}
(tu')'+\frac{t^{2m}}{4\nu_{n,m}}\Big(q(t)+\frac{\nu_{n,m}}
{t^{2n-\alpha}\lg^2 t}\Big)u=0
\end{equation}
is nonoscillatory. The application of the nonoscillation criterion
\eqref{lg-lg} with $\tilde\gamma_{n,\alpha}$ replaced by $\nu_{n,m}$
to this equation gives nonoscillation of \eqref{2-q-lg}
if the limit in \eqref{nu} is less than $\nu_{n,m}$.


\section{Technical results}

In this section we present some technical lemmata needed in the
proofs of our main results.

\begin{lem}                                   \label{L:nonosc_thm}
Let $\alpha_k$, $k=1,\dots, n-1$,
be the first $n-1$ roots (ordered by size) of the polynomial
\eqref{polynomial} and $\alpha_0=\frac{2n-1-\alpha}{2}$.
Then
$$
4 \tilde \gamma_{n,\alpha}
=\prod_{k=1}^{n-1}\Big(\frac{2n-1-\alpha}{2}-\alpha_k\Big)^2,
$$
where $\tilde \gamma_{n,\alpha}$ is given by \eqref{tilde-gamma}.
\end{lem}

\begin{proof}
Denote $\beta_k:=\frac{2n-1-\alpha}{2}-\alpha_k$, $k=1,\dots, n-1.$
Then, since the roots of \eqref{polynomial} are
$\alpha_k$, $2n-1-\alpha-\alpha_k$, $k=1,\dots,n-1$,
$\alpha_0=(2n-1-\alpha)/2$ ($\alpha_0$ is the double root),
we can write them in the form
$\alpha_k=\frac{2n-1-\alpha}{2}-\beta_k,\;
2n-1-\alpha-\alpha_k=\frac{2n-1-\alpha}{2}+\beta_k.$
The substitution $\mu=\frac{2n-1-\alpha}{2}-\lambda$
converts the polynomial
$$
P(\lambda):=(-1)^n\prod_{i=0}^{n-1}(\lambda-i)(\lambda-n+\alpha-i)-
\gamma_{n,\alpha}.
$$
into the polynomial
\begin{align*}
\tilde P(\mu)
&=(-1)^n\prod_{i=1}^{n}\Big(\frac{2i-1-\alpha}{2}-\mu\Big)
\Big(\frac{-2i+1+\alpha}{2}-\mu\Big)-\gamma_{n,\alpha}\\
&=(-1)^n\prod_{i=1}^{n}\Big[\mu^2-\Big(\frac{2i-1-\alpha}{2}\Big)^2\Big]
-\gamma_{n,\alpha}.
\end{align*}
The coefficient by $\mu^2$ in $(-1)^n \tilde P(\mu)$ is
\begin{align*}
& \frac{1}{4^{n-1}}\left[(2n-3-\alpha)^2(2n-5-\alpha)^2\cdots(1-\alpha)^2\right.\\
&+(2n-1-\alpha)^2(2n-5-\alpha)^2\dots(1-\alpha)^2+\cdots\\
&+\left.(2n-1-\alpha)^2(2n-3-\alpha)^2\cdots(3-\alpha)^2\right]\\
&= \frac{1}{4^{n-1}}\prod_{k=1}^n(2k-1-\alpha)^2
              \sum_{k=1}^n\frac{1}{(2k-1-\alpha)^2}
= 4 \tilde \gamma_{n,\alpha}.
\end{align*}
On the other hand, according to the above substitution, since the roots
of $\tilde P(\mu)$ are
$\pm\beta_k,\; k=1,\dots,n-1,\; \beta_0=0$ (double root),
it is possible to express $(-1)^n \tilde P(\mu)$ in the form
$$
(-1)^n \tilde P(\mu)=\mu^2(\mu^2-\beta_1^2)(\mu^2-\beta_2^2)
\cdots (\mu^2-\beta_{n-1}^2)
$$
and the coefficient by $\mu^2$ in $\tilde P(\mu)$ is
$\prod_{k=1}^{n-1}\beta_k^2.$
Comparing the both expressions by $\beta^2$ we have the result of this lemma.
\end{proof}

\begin{lem}                                     \label{L:lg-deriv}
For arbitrary $j\in \mathbb N$
\begin{equation*}                              \label{lg-deriv}
\left(\sqrt{\lg t}\right)^{(j)}=\frac{(-1)^{j-1}}{t^j}
\Big(\frac{a_j}{\sqrt{\lg t}}+\frac{b_j}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big),
\end{equation*}
where $a_j,\;b_j$ are given by recursion
\begin{equation}                                \label{koef-recursion}
a_1=\frac{1}{2},\;a_{k+1}=ka_k;\quad b_1=0,\;b_{k+1}=kb_k+\frac{a_k}{2}.
\end{equation}
\end{lem}

\begin{proof}
If $j=1$, then
$\left(\sqrt{\lg t}\right)'= \frac{1}{2t\sqrt{\lg t}}$,
and hence $a_1=\frac{1}{2},\;b_1=0.$
By induction
\begin{align*}
\big(\sqrt{\lg t}\big)^{(k+1)}
&=\Big[\frac{(-1)^{k-1}}{t^k}
\Big(\frac{a_k}{\sqrt{\lg t}}+\frac{b_k}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big)\Big]' \\
&= \frac{(-1)^kk}{t^{k+1}}
\Big(\frac{a_k}{\sqrt{\lg t}}+\frac{b_k}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big)\\
&\quad +\frac{(-1)^k}{t^{k+1}}
\Big(\frac{a_k}{2\sqrt{\lg^3 t}}+\frac{3b_k}{2\sqrt{\lg^5t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big)\\
&=\frac{(-1)^k}{t^{k+1}}
\Big(\frac{ka_k}{\sqrt{\lg t}}+\frac{kb_k+a_k/2}{\sqrt{\lg^3t}}
+o\big(\lg^{-\frac{3}{2}}t\big)\Big).
\end{align*}
\end{proof}

\begin{remark}  \rm                   \label{koef-a_n,b_n}
Using \eqref{koef-recursion} we have $a_{n+1}=na_n$, which implies
$$
a_n=a_1\prod_{j=1}^{n-1}j=\frac{1}{2}(n-1)!
$$
and $b_{n+1}=nb_n+\frac{1}{2}a_n=nb_n+\frac{1}{4}(n-1)!.$
Solving the last difference equation using variation of parameters
method, we obtain
$$
b_n=\frac{(n-1)!}{4}\sum_{j=1}^{n-1}\frac{1}{j},\;n\geq2.
$$
\end{remark}

The next lemma presents basic rules for computation of Wronskians.

\begin{lem}                                   \label{L:Wronskians}
Let $W(f_1,\dots,f_n)$ denote the Wronskian of the functions in
brackets. Then the following statements hold.
\begin{itemize}
\item[(i)] We have (with a function $r$)
$W(rf_1,\dots,rf_n)=r^nW(f_1,\dots,f_n)$.
In particular, if $f_i(t)\ne 0$ for some $i\in \{1,\dots,n\}$, then
$$
W(f_1,\dots,f_n)=(-1)^{i-1}f_i^nW\Big(\big(\frac{f_1}{f_i}\big)',\dots,
\big(\frac{f_{i-1}}{f_i}\big)',\big(\frac{f_{i+1}}{f_{i}}\big)',\dots,
\big(\frac{f_n}{f_i}\big)'\Big).
$$
\item[(ii)] Let $f_1=t^{\beta_1},\dots, f_n=t^{\beta_n}$, then
$$
W(f_1,\dots,f_n)=t^{\sum_{i=1}^n\beta_i-\frac{n(n-1)}{2}}
\prod_{1\leq j<i\leq n}(\beta_i-\beta_j)
$$
\item[(iii)] Let $X$ be the Wronskian matrix of the functions
$f_1,\dots,f_n$, and let $\tilde h=(h,h',\dots,h^{(n-1)})^T$. Then
$$
(X^{-1}\tilde h)_i=\frac{W(f_1,\dots,f_{i-1},h,f_{i+1},\dots,f_n)}
{W(f_1,\dots,f_n)}.
$$
\item[(iv)] If $f_i$, $i=1,\dots,n$, are the same as in (ii)
with $\beta_i\ne \beta_j$, $i\ne j$, $X$ is their Wronskian matrix and
$h(t)=t^{\beta_n}\sqrt{\lg t}$, then
\begin{gather*}
(X^{-1}\tilde h)_i\sim t^{\beta_i-\beta_n}\frac{1}{\sqrt{\lg t}},\quad
i=1,\dots,n-1,
\\
(X^{-1}\tilde h)_n=\sqrt{\lg t}(1+O(\lg^{-1} t)),
\end{gather*}
here $f_1\sim f_2$ for a pair of functions $f_1,f_2$ means
that $\lim_{t\to \infty}\frac{f_1(t)}{f_2(t)}=L$ exists and
$0<L<\infty$.
\end{itemize}
\end{lem}

\begin{proof} The statements (i) and (iii) are proved in
\cite[Chap. III]{C}. The statement (ii) can be found
e.g. in \cite{D-O-GJM} and the claim (iv) can be proved by
a direct computation using the rules (i)--(iii).
\end{proof}

\begin{lem}                                    \label{L:gamma-omega}
Let $\omega_{n,\alpha}$ and $\tilde \gamma_{n,\alpha}$ be given
by \eqref{omega-def} and \eqref{tilde-gamma}, respectively.
Then $\omega_{n,\alpha}/4=\tilde \gamma_{n,\alpha}$
\end{lem}

\begin{proof} Using the substitution $\mu=\lambda -\frac{2n-1-\alpha}{2}$
and then $i=n-1-j$ in the product formula for $\omega_{n,\alpha}$, we have
\begin{align*}
\omega_{n,\alpha}&= \frac{(-1)^n\prod_{i=0}^{n-1}(\lambda-i)
(\lambda-n+\alpha-i)-\gamma_{n,\alpha}}{\left(\lambda-
\frac{2n-1-\alpha}{2}\right)^2}\Big|_{\lambda=\frac{2n-1-\alpha}{2}}
\\
&=\frac{(-1)^n\prod_{j=0}^{n-1}\left(\mu+\frac{2n-1-\alpha-2j}{2}\right)
\left(\mu-\frac{2n-1-\alpha-2j}{2}\right)-\gamma_{n,\alpha}}{\mu^2}
\Big|_{\mu=0}
\\
&=\frac{(-1)^n\prod_{j=0}^{n-1}\left(\mu^2-\frac{(2n-1-\alpha-2j)^2}{4}\right)
-\gamma_{n,\alpha}}{\mu^2}\Big|_{\mu=0}
\\
&=4\prod_{i=0}^{n-1}\Big(\frac{2n-1-\alpha-2i}{2}\Big)^2
\Big\{\frac{1}{(2n-1-\alpha)^2}+\dots+\frac{1}{(1-\alpha)^2}\Big\}
=4\tilde \gamma_{n,\alpha}.
\end{align*}
\end{proof}

The proofs of the next two combinatorial results can be found in \cite{Koutsky}

\begin{lem}                                 \label{Kaucky-sum}
Let $x,\;z\in \mathbb R$  be arbitrary, $n\in \mathbb N$. Then
\begin{gather}                             \label{sum-x}
\sum_{j=0}^n(-1)^j\binom{n}{j}\binom{j+x}{j}^{-1}=\frac{x}{n+x},
\\                        \label{sum-x2}
\sum_{j=1}^n(-1)^{j-1}\binom{n}{j}\binom{j+x}{j}^{-1}
\Big(\sum_{i=1}^j\frac{1}{i+x}\Big)=\frac{n}{(x+n)^2},
\\                              \label{sum-xz}
\begin{aligned}
&\sum_{j=1}^n\binom{n}{j}\binom{z}{j}\binom{j+x}{j}^{-1}
\Big(\sum_{i=1}^j\frac{1}{i+x}\Big)\\
&=\binom{x+z+n}{n}\binom{x+n}{n}^{-1}\Big[\sum_{i=1}^n\frac{1}{i+x}-
\sum_{i=1}^n\frac{1}{i+x+z}\Big].
\end{aligned}
\end{gather}
\end{lem}

\begin{lem}                                  \label{Kaucky-seq}
Let $n\in \mathbb N$ and $\{F_n\},\;\{f_n\}$ be sequences such that
$F_n=\sum_{j=1}^n \binom{n}{j}f_j$.
Then
$$f_n=(-1)^n\sum_{j=1}^n (-1)^j\binom{n}{j}F_j.$$
\end{lem}

\begin{lem}                               \label{L:Kn=tildeKn}
Let $n\in \mathbb N$. Then
\begin{equation*}
\frac{A_n}{2^{n-1}}\prod_{k=1}^{n}(2k-1-\alpha)
=\frac{2}{4^n}\prod_{k=1}^n(2k-1-\alpha)^2
\sum_{k=1}^n\frac{1}{(2k-1-\alpha)},
\end{equation*}
where $A_n$ is given by \eqref{A_n}.
\end{lem}

\begin{proof}
Using \eqref{A_n} and Remark \ref{koef-a_n,b_n} we get
\begin{align*}
A_n&=\sum_{j=1}^{n}\Big[(-1)^{j-1}\binom{n}{j}
\frac{(j-1)!}{2^{n-j+1}}\prod_{k=j+1}^{n}(2k-1-\alpha)\Big]
\\
&=\frac{1}{2^n}\prod_{j=1}^n(2j-1-\alpha)
\sum_{j=1}^{n}(-1)^{j-1}\binom{n}{j}
\frac{2^{j-1}(j-1)!}{\prod_{k=1}^{j}(2k-1-\alpha)}\\
&=\frac{1}{2^n}\prod_{j=1}^n(2j-1-\alpha)
\sum_{j=0}^{n-1}(-1)^{j}\binom{n}{j+1}
\frac{2^jj!}{\prod_{k=1}^{j+1}(2k-1-\alpha)}.
\end{align*}
We have to show that
\begin{equation}                         \label{Cn}
A_n=\frac{1}{2^n}\prod_{k=1}^n(2k-1-\alpha)
\sum_{k=1}^n\frac{1}{2k-1-\alpha},
\end{equation}
i.e.
\begin{equation*}
\sum_{j=0}^{n-1}(-1)^{j}\binom{n}{j+1}
\frac{2^jj!}{\prod_{k=1}^{j+1}(2k-1-\alpha)}
=\sum_{k=1}^{n}\frac{1}{2k-1-\alpha}.
\end{equation*}
Denote $C_n$ the expression on the left-hand side and $\tilde C_n$
the expression on the right-hand side of the last formula.
If $n=2$, then $C_2=\frac{2(2-\alpha)}{(1-\alpha)(3-\alpha)}=\tilde C_2.$
By induction, it remains to show that $\Delta C_n=\Delta \tilde C_n,$
where
\begin{align*}
\Delta C_n&=C_{n+1}-C_n\\
&=\sum_{j=0}^{n}(-1)^{j}\left[\binom{n+1}{j+1}-\binom{n}{j+1}\right]
\frac{2^jj!}{\prod_{k=1}^{j+1}(2k-1-\alpha)}\\
&=\sum_{j=0}^{n}(-1)^{j}\binom{n}{j}
\frac{2^jj!}{\prod_{k=1}^{j+1}(2k-1-\alpha)},\\
\Delta \tilde C_n &=\tilde C_{n+1}- \tilde C_n=\frac{1}{2n+1-\alpha}.
\end{align*}
Substituting $x=(1-\alpha)/2$ into \eqref{sum-x} we obtain
$$
\sum_{j=0}^n(-1)^j\binom{n}{j}\binom{j+\frac{1-\alpha}{2}}{j}^{-1}
=\frac{1-\alpha}{2n+1-\alpha}
$$
and since
\begin{equation}                                 \label{binom}
\binom{j+\frac{1-\alpha}{2}}{j}=\frac{\prod_{k=2}^{j+1}(2k-1-\alpha)}{2^jj!},
\end{equation}
we have $\Delta C_n=\Delta \tilde C_n.$
\end{proof}

\begin{lem}                                   \label{L:gamma=On}
Let $A_n$, $B_n$ be given by \eqref{A_n}, \eqref{B_n} respectively.
Then for arbitrary $n\in \mathbb N$
$$
\tilde \gamma_{n,\alpha}=A_n^2+\frac{B_n}{2^{n-1}}
\prod_{k=1}^n(2k-1-\alpha).
$$
\end{lem}

\begin{proof}
Similarly as in the proof of Lemma \ref{L:Kn=tildeKn}, using
Remark \ref{koef-a_n,b_n} we can express $B_n$ in the following form
\begin{align*}
B_n&=\sum_{j=1}^{n}(-1)^{j-1}\binom{n}{j}\frac{b_j}{2^{n-j}}
\frac{\prod_{k=1}^n(2k-1-\alpha)}{\prod_{k=1}^j(2k-1-\alpha)}\\
&=\frac{\prod_{k=1}^n(2k-1-\alpha)}{2^n}
\sum_{j=2}^n(-1)^{j-1}\binom{n}{j}\frac{2^jb_j}{\prod_{k=1}^j(2k-1-\alpha)}\\
&=\frac{\prod_{k=1}^n(2k-1-\alpha)}{2^n}
\sum_{j=2}^n(-1)^{j-1}\binom{n}{j}\frac{2^j(j-1)!}{4\prod_{k=1}^j(2k-1-\alpha)}
\Big(\sum_{i=1}^{j-1}\frac{1}{i}\Big)\\
&=\frac{\prod_{k=1}^n(2k-1-\alpha)}{2^n}
\sum_{j=1}^{n-1}(-1)^{j}\binom{n}{j+1}\frac{2^{j-1}j!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big).
\end{align*}
Using this formula and \eqref{Cn}, we obtain
\begin{align*}
A_n^2+\frac{B_n}{2^{n-1}}\prod_{k=1}^n(2k-1-\alpha)
&= \frac{\prod_{j=1}^n(2j-1-\alpha)^2}{4^n}
\Big[\Big(\sum_{k=1}^{n}\frac{1}{2k-1-\alpha}\Big)^2\\
&\quad +\sum_{j=1}^{n-1}(-1)^{j}\binom{n}{j+1}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big)\Big].
\end{align*}
>From the definition of $\tilde \gamma_{n,\alpha}$, it suffices to show that
$D_n=\tilde D_n,$
where we have denoted
\begin{align*}
D_n&=\sum_{j=1}^{n-1}(-1)^{j}\binom{n}{j+1}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big),\\
\tilde D_n&=\sum_{k=1}^n\frac{1}{(2k-1-\alpha)^2}
- \Big(\sum_{k=1}^{n}\frac{1}{2k-1-\alpha}\Big)^2.
\end{align*}
One can see that
$D_2=-\frac{2}{(3-\alpha)(1-\alpha)}=\tilde D_2$ and we verify equality
$\Delta D_n=\Delta \tilde D_n.$
We have
\begin{align*}
\Delta D_n
&=\sum_{j=1}^{n}(-1)^{j}\binom{n+1}{j+1}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big)\\
&\quad -\sum_{j=1}^{n-1}(-1)^{j}\binom{n}{j+1}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big)\\
&=\sum_{j=1}^{n-1}(-1)^{j}\binom{n}{j}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big)\\
&\quad+(-1)^{n}\frac{2^nn!}
{\prod_{k=1}^{n+1}(2k-1-\alpha)}\Big(\sum_{i=1}^n\frac{1}{i}\Big)\\
&=\sum_{j=1}^{n}(-1)^{j}\binom{n}{j}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}\Big(\sum_{i=1}^j\frac{1}{i}\Big)
\end{align*}
and
\begin{align*}
\Delta \tilde D_n
&=\sum_{k=1}^{n+1}\frac{1}{(2k-1-\alpha)^2}
-\sum_{k=1}^{n}\frac{1}{(2k-1-\alpha)^2}\\
&\quad-\Big[\Big(\sum_{k=1}^{n+1}\frac{1}{2k-1-\alpha}\Big)^2
-\Big(\sum_{k=1}^{n}\frac{1}{2k-1-\alpha}\Big)^2\Big]\\
&=\frac{1}{(2n+1-\alpha)^2}
-\frac{1}{2n+1-\alpha}\Big[2\sum_{k=1}^n\frac{1}{2k-1-\alpha}
+\frac{1}{2n+1-\alpha}\Big]\\
&=-\frac{2}{2n+1-\alpha}\sum_{k=1}^n\frac{1}{2k-1-\alpha}.
\end{align*}
Using \eqref{sum-x2} for $x=(1-\alpha)/2$ and \eqref{binom} we get
the following identity
\begin{equation}                           \label{identity_pf}
\sum_{j=1}^{n}(-1)^{j}\binom{n}{j}\frac{2^jj!}
{\prod_{k=1}^{j+1}(2k-1-\alpha)}
\Big(\sum_{i=1}^j\frac{1}{i+\frac{1-\alpha}{2}}\Big)
=-\frac{n}{(1-\alpha)\left(\frac{1-\alpha}{2}+n\right)^2}.
\end{equation}
Next we substitute $x=\frac{1-\alpha}{2}$, $z=\frac{\alpha-1}{2}$
into \eqref{sum-xz}. Using \eqref{binom} and the fact that
$$\binom{\frac{\alpha-1}{2}}{j}\binom{\frac{1-\alpha}{2}+j}{j}^{-1}
=(-1)^j\frac{1-\alpha}{2j+1-\alpha},$$
we obtain
\begin{align*}
&\sum_{j=1}^{n}(-1)^{j}\binom{n}{j}\frac{1}{2j+1-\alpha}
\Big(\sum_{i=1}^j\frac{1}{1+\frac{1-\alpha}{2}}\Big)\\
&=\frac{2^nn!}{\prod_{k=1}^{n+1}2k-1-\alpha}
\Big[\sum_{i=1}^n\frac{1}{i+\frac{1-\alpha}{2}}-
\sum_{i=1}^n\frac{1}{i}\Big].
\end{align*}
Next we use Lemma \ref{Kaucky-seq} in case
$f_j:=(-1)^j\frac{1}{2j+1-\alpha}
\big(\sum_{i=1}^j\frac{1}{1+\frac{1-\alpha}{2}}\big)$
and $F_n$ indicates the term on the right-hand side of the above equality.
Then
$$
\frac{1}{2n+1-\alpha}
\Big(\sum_{i=1}^n\frac{1}{1+\frac{1-\alpha}{2}}\Big)
=\sum_{j=1}^n(-1)^j\binom{n}{j}\frac{2^jj!}
{\prod_{k=1}^{j+1}2k-1-\alpha}
\Big[\sum_{i=1}^j\frac{1}{i+\frac{1-\alpha}{2}}-
\sum_{i=1}^j\frac{1}{i}\Big]
$$
and hence, using \eqref{identity_pf}
$$
\frac{1}{2n+1-\alpha}
\Big(\sum_{i=1}^n\frac{1}{i+\frac{1-\alpha}{2}}\Big)
=-\frac{n}{(1-\alpha)\left(\frac{1-\alpha}{2}+n\right)^2}-\Delta D_n,
$$
which implies
\begin{align*}
\Delta D_n&=-\frac{2}{2n+1-\alpha}\sum_{i=1}^n\frac{1}{2i+1-\alpha}
-\frac{n}{(1-\alpha)\left(\frac{1-\alpha}{2}+n\right)^2}\\
&=-\frac{2}{2n+1-\alpha}\sum_{i=1}^n\frac{1}{2i-1-\alpha}
=\Delta \tilde D_n.
\end{align*}
\end{proof}

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\end{document}