**Theorem 3.1** *Suppose that* (1.2) *and* (1.3) *hold with* \beta >\gamma. *If*

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=-\mathrm{\infty}

(3.1)

*and*

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}[v(s)-H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=\mathrm{\infty}

(3.2)

*for every sufficiently large* *T*, *where* H(s):=(\beta /\gamma -1){[\gamma {p}_{2}(s)/\beta ]}^{\beta /(\beta -\gamma )}{p}_{1}^{\gamma /(\gamma -\beta )}(s), *then every solution of* (1.1) *is oscillatory*.

*Proof* Let *x* be a nonoscillatory solution of (1.1). Firstly, we suppose that *x* is an eventually positive solution of (1.1). Then there exists {T}_{1}>a such that x(t)>0 for t\ge {T}_{1}. Let s\ge {T}_{1} and take X={|x|}^{\gamma}(s), Y=\gamma {p}_{2}(s)/(\beta {p}_{1}(s)), u=\beta /\gamma and v=\beta /(\beta -\gamma ), then from Part (i) of Lemma 2.1 we conclude

\begin{array}{r}{p}_{2}(s){|x|}^{\gamma}(s)-{p}_{1}(s){|x|}^{\beta (}s)\\ \phantom{\rule{1em}{0ex}}=\frac{\beta {p}_{1}(s)}{\gamma}[{|x|}^{\gamma}(s)\frac{\gamma {p}_{2}(s)}{\beta {p}_{1}(s)}-\frac{1}{\beta /\gamma}{({|x|}^{\gamma}(s))}^{\beta /\gamma}]\\ \phantom{\rule{1em}{0ex}}=\frac{\beta {p}_{1}(s)}{\gamma}[XY-\frac{1}{u}{X}^{u}]\le \frac{\beta {p}_{1}(s)}{\gamma}\frac{1}{v}{Y}^{v}=H(s)\phantom{\rule{1em}{0ex}}\text{for}s\ge {T}_{1},\end{array}

(3.3)

where *H* is defined as in Theorem 3.1. From (1.4), (1.2), (1.3) and (3.3), we obtain

\begin{array}{rl}\mathrm{\Gamma}(q)x(t)=& \mathrm{\Gamma}(q)\sum _{k=1}^{m}\frac{{b}_{k}{(t-a)}^{q-k}}{\mathrm{\Gamma}(q-k+1)}+{\int}_{a}^{{T}_{1}}{(t-s)}^{q-1}[v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ +{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \le & \mathrm{\Phi}(t)+\mathrm{\Psi}(t,{T}_{1})+{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+{p}_{2}(s){x}^{\gamma}(s)-{p}_{1}(s){x}^{\beta (}s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \le & \mathrm{\Phi}(t)+\mathrm{\Psi}(t,{T}_{1})+{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\phantom{\rule{1em}{0ex}}\text{for}t\ge {T}_{1},\end{array}

(3.4)

where

\mathrm{\Phi}(t):=\mathrm{\Gamma}(q)\sum _{k=1}^{m}\frac{{b}_{k}{(t-a)}^{q-k}}{\mathrm{\Gamma}(q-k+1)}

(3.5)

and

\mathrm{\Psi}(t,{T}_{1}):={\int}_{a}^{{T}_{1}}{(t-s)}^{q-1}[v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.

(3.6)

Multiplying (3.4) by {t}^{1-q}, we have, for t\ge {T}_{1},

0<{t}^{1-q}\mathrm{\Gamma}(q)x(t)\le {t}^{1-q}\mathrm{\Phi}(t)+{t}^{1-q}\mathrm{\Psi}(t,{T}_{1})+{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.

(3.7)

Take {T}_{2}>{T}_{1}. Next, we consider the cases 0<q\le 1 and q>1, respectively.

Case (i). Let 0<q\le 1. Then we get m=1, \mathrm{\Phi}(t)={b}_{1}{(t-a)}^{q-1},

|{t}^{1-q}\mathrm{\Phi}(t)|=|{b}_{1}|{t}^{1-q}{(t-a)}^{q-1}\le |{b}_{1}|{\left(\frac{{T}_{2}}{{T}_{2}-a}\right)}^{1-q}:={c}_{1}({T}_{2})\phantom{\rule{1em}{0ex}}\text{for}t\ge {T}_{2}

(3.8)

and

\begin{array}{rl}|{t}^{1-q}\mathrm{\Psi}(t,{T}_{1})|& =\left|{t}^{1-q}{\int}_{a}^{{T}_{1}}{(t-s)}^{q-1}[v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\right|\\ \le {\int}_{a}^{{T}_{1}}{t}^{1-q}{(t-s)}^{q-1}|v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))|\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \le {\int}_{a}^{{T}_{1}}{\left(\frac{{T}_{2}}{{T}_{2}-s}\right)}^{1-q}|v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))|\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ :={c}_{2}({T}_{1},{T}_{2})\phantom{\rule{1em}{0ex}}\text{for}t\ge {T}_{2}.\end{array}

(3.9)

It follows from (3.7)-(3.9) that {t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>-[{c}_{1}({T}_{2})+{c}_{2}({T}_{1},{T}_{2})] for t\ge {T}_{2}. Therefore, we find {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\ge -[{c}_{1}({T}_{2})+{c}_{2}({T}_{1},{T}_{2})]>-\mathrm{\infty}, which contradicts (3.1).

Case (ii). Let q>1. Then we have m\ge 2,

\begin{array}{rl}|{t}^{1-q}\mathrm{\Phi}(t)|& =|{t}^{1-q}\mathrm{\Gamma}(q)\sum _{k=1}^{m}\frac{{b}_{k}{(t-a)}^{q-k}}{\mathrm{\Gamma}(q-k+1)}|\\ \le \mathrm{\Gamma}(q)\sum _{k=1}^{m}\frac{|{b}_{k}|{t}^{1-q}{(t-a)}^{q-k}}{\mathrm{\Gamma}(q-k+1)}\le \mathrm{\Gamma}(q)\sum _{k=1}^{m}\frac{|{b}_{k}|{({T}_{2}-a)}^{1-k}}{\mathrm{\Gamma}(q-k+1)}\\ :={c}_{3}({T}_{2})\phantom{\rule{1em}{0ex}}\text{for}t\ge {T}_{2}\end{array}

(3.10)

and

\begin{array}{rl}|{t}^{1-q}\mathrm{\Psi}(t,{T}_{1})|& =\left|{t}^{1-q}{\int}_{a}^{{T}_{1}}{(t-s)}^{q-1}[v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\right|\\ \le {\int}_{a}^{{T}_{1}}{t}^{1-q}{(t-s)}^{q-1}|v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))|\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \le {\int}_{a}^{{T}_{1}}|v(s)+{f}_{2}(s,x(s))-{f}_{1}(s,x(s))|\phantom{\rule{0.2em}{0ex}}\mathrm{d}s:={c}_{4}({T}_{1})\phantom{\rule{1em}{0ex}}\text{for}t\ge {T}_{2}.\end{array}

(3.11)

From (3.7), (3.10) and (3.11), we conclude {t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s>-[{c}_{3}({T}_{2})+{c}_{4}({T}_{1})] for t\ge {T}_{2}. Hence, we obtain {lim\hspace{0.17em}inf}_{t\to \mathrm{\infty}}{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\ge -[{c}_{3}({T}_{2})+{c}_{4}({T}_{1})]>-\mathrm{\infty}, which contradicts (3.1).

Finally, we assume that *x* is an eventually negative solution of (1.1). Then a similar argument leads to a contradiction with (3.2). The proof is complete. □

**Remark 3.1** In [20], the plus sign ‘+’ in (2.9) in Theorem 2.2, (2.13) in Theorem 2.3, (2.17) in Theorem 2.4, (3.6) in Theorem 3.2, (3.8) in Theorem 3.3 and (3.10) in Theorem 3.4 should be the minus sign ‘−’.

**Remark 3.2** Theorems 2.2 and 2.3 in [20] are the special cases of our Theorem 3.1 with \beta >1=\gamma and \beta =1>\gamma >0, respectively. Our Theorem 3.1 improves and extends the results of Theorems 2.2-2.4 in [20] since these theorems did not include the cases \beta >\gamma >1 and 1>\beta >\gamma >0 for (1.1).

The following example shows that the condition (3.1) cannot be dropped.

**Example 3.1** Consider the Riemann-Liouville fractional differential equation

\begin{array}{l}\begin{array}{rl}({D}_{0}^{q}x)(t)+{x}^{5}(t)ln(\mathrm{e}+t)=& \frac{2{t}^{2-q}}{\mathrm{\Gamma}(3-q)}+({t}^{10}-{t}^{2/3})ln(\mathrm{e}+t)\\ +{x}^{1/3}(t)ln(\mathrm{e}+t),\phantom{\rule{1em}{0ex}}t>0,\end{array}\\ {lim}_{t\to {0}^{+}}({I}_{0}^{1-q}x)(t)=0,\end{array}\},

(3.12)

where 0<q<1.

In (3.12), a=0, m=1, {f}_{1}(t,x)={x}^{5}ln(\mathrm{e}+t), v(t)=\frac{2{t}^{2-q}}{\mathrm{\Gamma}(3-q)}+({t}^{10}-{t}^{2/3})ln(\mathrm{e}+t), {f}_{2}(t,x)={x}^{1/3}ln(\mathrm{e}+t) and {b}_{1}=0. Taking {p}_{1}(t)={p}_{2}(t)=ln(\mathrm{e}+t), \beta =5 and \gamma =1/3, we find that the conditions (1.2) and (1.3) are satisfied. But the condition (3.1) is not satisfied since for every sufficiently large T\ge 1 and all t\ge T, we have v(t)>0 and

\begin{array}{rl}\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s& \ge \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}H(s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ =\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}{15}^{-15/14}14ln(\mathrm{e}+s)\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ \ge \underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}{15}^{-15/14}14\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\\ =\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}\frac{{15}^{-15/14}14{t}^{1-q}{(t-T)}^{q}}{q}\\ =\mathrm{\infty},\end{array}

where *H* is defined as in Theorem 3.1. Taking x(t)={t}^{2}, by Definition 2.1 we get

\left({I}_{0}^{1-q}x\right)(t)=\frac{1}{\mathrm{\Gamma}(1-q)}{\int}_{0}^{t}{(t-s)}^{-q}{s}^{2}\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.

Integrating by parts twice, we obtain

\left({I}_{0}^{1-q}x\right)(t)=\frac{1}{\mathrm{\Gamma}(1-q)}\frac{2{t}^{3-q}}{(1-q)(2-q)(3-q)}.

(3.13)

Therefore, by Definition 2.2 we conclude

\left({D}_{0}^{q}x\right)(t)=\frac{\mathrm{d}}{\mathrm{d}t}\left({I}_{0}^{1-q}x\right)(t)=\frac{1}{\mathrm{\Gamma}(1-q)}\frac{2{t}^{2-q}}{(1-q)(2-q)}=\frac{2{t}^{2-q}}{\mathrm{\Gamma}(3-q)},

which implies that x(t)={t}^{2} satisfies the first equality in (3.12). From (3.13) we get {lim}_{t\to {0}^{+}}({I}_{0}^{1-q}x)(t)=0, which yields that x(t)={t}^{2} satisfies the second equality in (3.12). Hence, x(t)={t}^{2} is a nonoscillatory solution of (3.12).

Next, we consider the case when (1.5) holds, which was not considered in [20].

**Theorem 3.2** *Let* q\ge 1 *and suppose that* (1.2) *and* (1.5) *hold with* \beta <\gamma. *If*

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}sup}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=\mathrm{\infty}

(3.14)

*and*

\underset{t\to \mathrm{\infty}}{lim\hspace{0.17em}inf}{t}^{1-q}{\int}_{T}^{t}{(t-s)}^{q-1}[v(s)-H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s=-\mathrm{\infty}

(3.15)

*for every sufficiently large* *T*, *where* *H* *is defined as in Theorem * 3.1, *then every bounded solution of* (1.1) *is oscillatory*.

*Proof* Let *x* be a bounded nonoscillatory solution of (1.1). Then there exist constants {M}_{1} and {M}_{2} such that

{M}_{1}\le x(t)\le {M}_{2}\phantom{\rule{1em}{0ex}}\text{for}t\ge a.

(3.16)

Firstly, we suppose that *x* is a bounded eventually positive solution of (1.1). Then there exists {T}_{1}>a such that x(t)>0 for t\ge {T}_{1}. Similar to the proof of (3.3), by Part (ii) of Lemma 2.1 we find

{p}_{2}(s){|x|}^{\gamma}(s)-{p}_{1}(s){|x|}^{\beta (}s)\ge H(s)\phantom{\rule{1em}{0ex}}\text{for}s\ge {T}_{1},

(3.17)

where *H* is defined as in Theorem 3.1. Define Φ and Ψ as in (3.5) and (3.6), respectively. Similar to the proof of (3.7), from (1.4), (1.2), (1.5) and (3.17), we get, for t\ge {T}_{1},

{t}^{1-q}\mathrm{\Gamma}(q)x(t)\ge {t}^{1-q}\mathrm{\Phi}(t)+{t}^{1-q}\mathrm{\Psi}(t,{T}_{1})+{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s.

(3.18)

Take {T}_{2}>{T}_{1}. Next, we consider the cases q=1 and q>1, respectively.

Case (i). Let q=1. Then (3.8) and (3.9) are still true. From (3.16), (3.8), (3.9) and (3.18), we conclude {M}_{2}\mathrm{\Gamma}(q)\ge -{c}_{1}({T}_{2})-{c}_{2}({T}_{1},{T}_{2})+{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s for t\ge {T}_{2}. Thus, we see {lim\hspace{0.17em}sup}_{t\to \mathrm{\infty}}{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\le {c}_{1}({T}_{2})+{c}_{2}({T}_{1},{T}_{2})+{M}_{2}\mathrm{\Gamma}(q)<\mathrm{\infty}, which contradicts (3.14).

Case (ii). Let q>1. Then (3.10) and (3.11) are still valid. From (3.16), (3.10), (3.11) and (3.18), we conclude {M}_{2}\mathrm{\Gamma}(q){t}^{1-q}\ge -{c}_{3}({T}_{2})-{c}_{4}({T}_{1})+{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s for t\ge {T}_{2}. Since {lim}_{t\to \mathrm{\infty}}{t}^{1-q}=0, we obtain {lim\hspace{0.17em}sup}_{t\to \mathrm{\infty}}{t}^{1-q}{\int}_{{T}_{1}}^{t}{(t-s)}^{q-1}[v(s)+H(s)]\phantom{\rule{0.2em}{0ex}}\mathrm{d}s\le {c}_{3}({T}_{2})+{c}_{4}({T}_{1})<\mathrm{\infty}, which contradicts (3.14).

Finally, we suppose that *x* is a bounded eventually negative solution of (1.1). Then a similar argument leads to a contradiction with (3.15). The proof is complete. □