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\title{Certain geometric properties of Generalized Bessel-Maitland function}
\author{Amit Soni  }
\address{Department of Mathematics, \\ Govt. Engineering College, Bikaner,\\
 334004, Rajasthan, India}
\email{aamitt1981@gmail.com}
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\author{Deepak Bansal}
\address{Department of Mathematics, \\ College of Engg. and Technology, Bikaner\\
334004, Rajasthan, India}
\email{deepakbansal\_79@yahoo.com}
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\subjclass{30C45}
\keywords{Univalent, Starlike, Convex and  Close-to-convex function, Subordination, Bessel functions, Bessel-Maitland functions}
\begin{abstract}
In the present study, we first introduce Generalized Bessel-Maitland function $\mathbb{J}^{\xi }_{\zeta,a}(z)$ and then derive sufficient conditions under which the Generalized Bessel-Maitland function $\mathbb{J}^{\xi}_{\zeta,a}(z)$ have geometric properties like univalency, starlikeness and convexity in the open unit disk $\mathscr{D}$. 

\end{abstract}
\maketitle

\section{ Introduction and preliminaries}
Let $\mathscr{H}$ denote the class of all functions analytic in the open unit disk ${\mathscr D} = \{z\in \mathbb{C} : |z| < 1\}$ and $\mathscr{A}$ be the class of all functions $f \in \mathscr{H}$ which are normalized by $f (0) = 0$ and $f'(0)=1$. Each $f(z)\in \mathscr{A}$ has a Maclaurin series expansion of the form:
\begin{equation}\label{begi}
f(z) = z + a_2z^2 + a_3z^3 +...\;  
\end{equation}
Let $g, h\in \mathscr{H}$, we say that $g$ is subordinated to $h$ in $\mathscr{D}$, and write
$g(z) \prec h (z)$, if there exists a function $\omega \in \mathscr{H}$ with $|\omega(z)| < |z|,\; z \in\mathscr{D}$, such that $g(z) = h(\omega(z))$ in
$\mathscr{D}$. In particular, if $h$ is univalent in $\mathscr{D}$, then we have: 
$$g(z)\prec h(z) \iff g(0)=h(0)\;{\rm and}\;g(\mathscr{D}) \subset h(\mathscr{D}).$$
For a given $0 \leq \beta <1$, a function $g \in \mathscr{A}$ is called starlike function of order $\beta$,  if $\Re \left({zg'(z)}/{g(z)} \right)>\beta, \,z \in \mathscr{D}$ class of such functions denoted by $\mathscr{S}^*(\beta)$. Similarly, for $0 \leq \beta <1$, a function $g \in \mathscr{A}$ is called convex function of order $\beta$ if $ \Re\left (1+ {zg''(z)}/{g'(z)}\right)>\beta, \,z \in \mathscr{D}$, class of such function denoted by $\mathscr{K}(\beta)$. It is customary that $\mathscr{S}^*(0)=\mathscr{S}^*$ and $\mathscr{K}(0)=\mathscr{K}.$ Moreover, a function $ g\in \mathscr{A}$ is said to be close-to-convex with respect to a fixed starlike function $h$, denoted by $\mathcal{C}_h$,  if $\Re \left({zg'(z)}/{h(z)} \right)>0, \,z \in \mathscr{D}.$ For more details one can refer \cite{duren}.\\
%=========================================================================
\indent  In the present perusal, we study some geometric properties of Generalized Bessel-Maitland function (see, e.g., \cite{mari},
Eq.(8.3)), ${J}^{\xi}_{\zeta}(z)$. This function is defined by the following series  representation:\\
\begin{equation}
J^{\xi}_{\zeta}(z)=\sum_{n=0}^\infty\dfrac{(-z)^n}{n!\Gamma({\xi n +\zeta+1})}\;(\Re(\xi)\geq 0,\; \Re(\zeta)\geq -1 \;{\rm and}\; z\in\mathscr{D}).
\end{equation}
It has many application in various research fields of Science and Engineering. For a comprehensive description of applications of Bessel functions and its generalization, the reader may be referred to \cite{watson}. Here in the present paper, we define a new (probably) generalization of Bessel-Maitland function called generalized   Bessel-Maitland function ${J}^{\xi}_{\zeta,c}(z)$, given by: 
\begin{equation}
J^{\xi}_{\zeta, a}(z)=\sum_{n=0}^\infty\dfrac{(-a)^nz^n}{n!\Gamma({\xi n +\zeta+1})}\;(a\in\mathbb{C}-\{0\}, \xi>0, \zeta>-1\; {\rm and}\;z\in\mathscr{D}).
\end{equation}
It can be easily seen that 
\begin{equation}
J^{\xi}_{\zeta,-1}(z)=W_{\xi,\zeta+1}(z)=\sum_{n=0}^\infty\dfrac{z^n}{n!\Gamma({\xi n +\zeta+1})}
\end{equation}
where $W_{\xi,\zeta+1}(z)$ is called Wright function and
\begin{equation}
J^{\xi}_{\zeta,1}(z)=J^{\xi}_{\zeta}(z)=\sum_{n=0}^\infty\dfrac{(-1)^nz^n}{n!\Gamma({\xi n +\zeta+1})}.
\end{equation}
%===========================================================================
%===========================================================================
Observe that the Generalized Bessel-Maitland function $J^{\xi}_{\zeta, a}(z)\notin \mathscr{A}$. We can consider the following two types of normalization of the Generalized Bessel-Maitland function:
\begin{equation}\label{main}
\mathbb{J}^{\xi}_{\zeta,a}(z)=z\Gamma{(\zeta+1)}J^{\xi}_{\zeta, a}(z)=z+\sum_{n=1}^\infty\dfrac{(-a)^n\Gamma{(\zeta+1)}z^{n+1}}{n!\Gamma({\xi n +\zeta+1})}
\end{equation}
and
\begin{equation*}
\mathcal{J}^{\xi}_{\zeta,a}(z)=\dfrac{\Gamma{(\xi+\zeta+1)}}{(-a)}\left(J^{\xi}_{\zeta, a}(z)-\frac{1}{\Gamma{(\zeta+1)}}\right)
\end{equation*}
\begin{equation}=\sum_{n=0}^{\infty}\dfrac{(-a)^n\Gamma{(\xi+\zeta+1)}z^{n+1}}{(n+1)!\Gamma{(\xi n+\xi+\zeta+1)}}.
\end{equation}
\[\left(\xi>0,\; \xi+\zeta>-1,\; a\in\mathbb{C}-\{0\},\;z\in\mathscr{D}\right)\]
Also note that
\begin{equation}
\mathbb{J}^{1}_{\zeta, 1}(z)=\mathbb{J}_{\zeta}(z)=\Gamma{(\zeta+1)z^{1-{\zeta/2}}}J_{\zeta}(2\sqrt{z})=\sum_{n=0}^{\infty}\frac{(-1)^{n}\Gamma{(\zeta+1)}z^{n+1}}{n!\Gamma{(n+\zeta+1})}.
\end{equation}
where $J_\zeta(z)$ is well known Bessel function of order $\zeta$ and $\mathbb{J}_{\zeta}(z)$ is the normalized Bessel function, studied recently for the various geometric properties (see \cite{prajapat}-\cite{sza2}).
Conversely, it can be easily seen that 
\[J_{\zeta}(z)=\frac{1}{\Gamma(\zeta+1)}\left(\frac{z}{2}\right)^{\zeta-2}\mathbb{J}^{1}_{\zeta,1}\left(\frac{z^{2}}{4}\right)=\sum_{n=0}^{\infty}\frac{(-1)^{n}(z/2)^{2n+\zeta}}{n!\Gamma{(n+\zeta+1)}}.\]
Additionally,  we observe that\\
\[
\mathbb{V}^{\xi}_{\zeta,a}(z)=\dfrac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}
=\frac{1}{z}\left[z+\sum_{n=1}^\infty\dfrac{(-a)^n\Gamma{(\zeta+1)}z^{n+1}}{n!\Gamma({\xi n +\zeta+1})}\right]\]
\[=1+\sum_{n=1}^\infty\dfrac{(-a)^n\Gamma{(\zeta+1)}z^{n}}{n!\Gamma({\xi n +\zeta+1})}\]
and
\[z(\mathbb{V}^{\xi}_{\zeta,a}(z))'=\sum_{n=1}^{\infty}\dfrac{(-a)^n\Gamma{(\zeta+1)}nz^n}{n!\Gamma{(\xi n+\zeta+1})}.\]

\noindent The following identity relations can be easily established:
\begin{equation}\label{id1}\xi z(\mathbb{J}^{\xi}_{\zeta,a}(z))'=(\zeta+1)\mathbb{J}^{\xi}_{\zeta,a}(z)+(\xi-\zeta-1)\mathbb{J}^{\xi}_{\zeta+1,a}(z)
\end{equation}
\begin{equation}\label{rec}z(\mathcal{J}^{\xi}_{\zeta,a}(z))'=\mathbb{J}^{\xi}_{\xi+\zeta,a}(z)
\end{equation}
{and}
\begin{equation}\label{derivativeofv} \left(\mathbb{V}^{\xi}_{\zeta,a}(z)\right)'=\frac{(-a)\Gamma{(\zeta+1)}}{\Gamma{(\xi+\zeta+1)}}\mathbb{V}^{\xi}_{\xi+\zeta,a}(z).
\end{equation}
%=====================================================================
Lately, several researchers have studied innumerable special functions belonging to class $\mathscr{A}$ and found sufficient conditions such that the special functions belonging to class $\mathscr{A}$ have certain properties like univalency, starlikeness or convexity in $\mathscr{D}$. For the generalized hypergeometric functions one can refer \cite{miller1990, samy2001,ruscheweyh}, Bessel functions \cite{baricz2010, baricz2010-1, baricz2014, baricz2014-1} and Wright function \cite{prajap-2}.
In the present paper, we derive sufficient conditions for the same geometric properties for the functions $\mathbb{J}^{\xi}_{\zeta,a}(z)$ and $\mathcal{J}^{\xi}_{\zeta,a}(z)$ .
%%%%%%%%%%%%%%%%%%%%%%%%
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\section{Lemmas}

\setcounter{equation}{0}
\noindent
To prove main results, we requisite the following results:
%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{2.2lem}{\rm (see \cite{lemma2})}. Let $g\in\mathscr{A}$ satisfy the inequality $$|(g(z)/z)-1|<1\;(z\in\mathscr{D})$$ then $g$ is starlike in the disk $\mathscr{D}_{1/2}=\{z: |z|<1/2\}$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{2.3lem}{\rm (see \cite{lemma3})}. Let $g\in\mathscr{A}$ satisfy the inequality $$|g'(z)-1|<1\;(z\in\mathscr{D})$$ then $g$ is convex in the disk $\mathscr{D}_{1/2}$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{2.4lem}{\rm (see \cite{lemma4})}. Let $g\in\mathscr{A}$ satisfy $$|g'(z)-1|<2/\sqrt{5}\;(z\in\mathscr{D})$$ then $g$ is starlike in the disk $\mathscr{D}$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{oho}{\rm (see \cite{lemma5})}. Let $g\in\mathscr{A}$ satisfy the inequality
\[\left|\frac{zg'(z)}{g(z)}-1\right|<L,\;\;\;z\in\mathscr{D},
\]
where $L$ is solution of the equation $\cos L=L$, then $\Re(g'(z))>0$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{lemma}\label{2.6}{\rm (see \cite{lemma6})}. Let $\delta\in\mathbb{C}$ with $\Re(\delta)>0, d\in\mathbb{C}$ with $|d|\leq1,\;d\neq-1$. If $h\in\mathscr{A}$ satisfies
\[\left|d|z|^{2\delta}+(1-|z|^{2\delta})\frac{zh''(z)}{\delta h'(z)}\right|\leq 1,\;\;\;z\in\mathscr{D}\]
then the integral operator
\[\mathcal{C}_{\delta}(z)=\left\{\delta\int_{0}^{z}t^{\delta-1}h'(t)dt\right\}^{1/\delta}_{,}\;\;z\in\mathscr{D}\]
is analytic and univalent in $\mathscr{D}$.
\end{lemma}
%%%%%%%%%%%%%%%%%%%%%%%%
\noindent
For $\delta=1$ and $d=0$, Lemma \ref{2.6} is equivalent to Becker's criterion for univalency \cite{backer}, which shows that, if $f\in\mathscr{A}$ satisfy the inequality $(1-|z|^2)|zf''(z)/f'(z)|\leq 1$ for each $z\in\mathscr{D}$, then $f$ is one-to-one (univalent) in $\mathscr{D}$.\\
%%%%%%%%%%%%%%%%%%%%%%
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\section{\bf {Main Results}}
\setcounter{equation}{0}
\noindent
%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}\label{first}
\noindent {\rm (i)}Let $\xi\geq 1$ and $\zeta\geq \dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$, then $\mathbb{J}^{\xi}_{\zeta,a}$ is starlike in $\mathscr{D}$.\\
{\rm (ii) } Let $\xi\geq 1$ and $\xi+\zeta\geq\dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$, then $\mathbb{V}^{\xi}_{\zeta,a}$ is convex in $\mathscr{D}$.\\
{\rm (iii)} Let $\xi\geq 1$ and $\xi+\zeta\geq\dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$, then $\mathcal{J}^{\xi}_{\zeta,a}$ is convex in $\mathscr{D}$.
\end{theorem}
%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%
\begin{proof}
Let $q(z)$ be a function defined by the equality $q(z)=z(\mathbb{J}^{\xi}_{\zeta,a}(z))'/\mathbb{ J}^{\xi}_{\zeta,a}(z)$\;\;$z\in\mathscr{D}$.
Since $\mathbb{J}^{\xi}_{\zeta,a}(z)/z\neq 0,\;\;z\in\mathscr{D}$, the function $q$ is analytic in $\mathscr{D}$ and $q(0)=1$. To prove the result, we need to show that $\Re(q(z))>0$ which follows if we show $|q(z)-1|<1$.\\
\indent For $\xi\ge 1$, it is easy to see that $\Gamma{(\zeta+n+1)}\leq\Gamma{(\xi n+\zeta+1)}$,  $n\in{\mathbb{N}}$, holds and is equivalent to 
\begin{equation}\label{th1}
\dfrac{1}{(\zeta+1)(\zeta+2)...(\zeta+n)}\geq\dfrac{\Gamma{(\zeta+1})}{\Gamma{(\xi n +\zeta+1})},\;\;n\in\mathbb{N}.
\end{equation}
If $z\in\mathscr{D}$, then using \eqref{main} and \eqref{th1}, we obtain 
\[
\left|(\mathbb{J}^{\xi}_{\zeta,a}(z))'-\dfrac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}\right|=\left|\sum_{n=1}^{\infty}\dfrac{(-a)^n nz^n\Gamma{(\zeta+1)}}{n!\Gamma{(\xi n+\zeta+1)}}\right|
\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
\[\leq\sum_{n=1}^{\infty}\frac{|a|^nn}{n!(\zeta+1)(\zeta+2)...(\zeta+n)}\]
\[<\frac{|a|}{(\zeta+1)}\sum_{n=0}^{\infty}\left(\dfrac{|a|}{\zeta+2}\right)^n\;\;\;\;\;\;\;\;\;\;\;\]
\begin{equation}\label{2.2}=\frac{|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{equation}
and
\[\left|\dfrac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}\right|\geq1-\left|\sum_{n=1}^{\infty}\dfrac{(-a)^nz^n\Gamma{(\zeta+1)}}{n!\Gamma{(\xi n+\zeta+1)}}\right|\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
\[\;\;\;\;\;\geq 1-\sum_{n=1}^{\infty}\dfrac{|a|^n}{n!(\zeta+1)(\zeta+2)...(\zeta+n)}\]
\[\;\;\;>1-\frac{|a|}{(\zeta+1)}\sum_{n=0}^{\infty}\left(\frac{|a|}{\zeta+2}\right)^n\;\;\;\;\;\;\;\;\]
\[\;\;\;\;=1-\dfrac{|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
\begin{equation}\label{2.3}\;\;\;\;\;\;\;=\dfrac{(\zeta+1)(\zeta+2-|a|)-|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)}.\;\;\;\;\;\end{equation}
From \eqref{2.2} and \eqref{2.3}, we have for $z\in\mathscr{D}$
\begin{eqnarray}\label{lo2.4}
|q(z)-1|=\left|\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{\mathbb{J}^{\xi}_{\zeta,a}(z)}-1\right|&=&
\left|\dfrac{(\mathbb{J}^{\xi}_{\zeta,a}(z))'-\frac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}}{\frac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}}\right|\notag \\
&<&{\small\dfrac{|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)-|a|(\zeta+2)}\;.}
\end{eqnarray}
This implies that if 
$(\zeta+1)(\zeta+2-|a|)-|a|(\zeta+2)\geq |a|(\zeta+2),$
then $\Re(q(z))>0$, hence $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is satrlike in $\mathscr{D}$, but the inequality $(\zeta+1)(\zeta+2-|a|)-2|a|(\zeta+2)\geq0$ is a consequence of the hypothesis $\zeta\geq \dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$. This shows that $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is starlike in $\mathscr{D}$.
\end{proof}  
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\noindent (ii) In view of the hypothesis the inequality $\Gamma{(n+\xi+\zeta+1)}\leq\Gamma{(\xi n+\xi+\zeta+1)}$ holds. If $z\in\mathscr{D}$, then a calculation similar to \eqref{2.2} and \eqref{2.3} gives\\
\[{\left|z\left(\mathbb{V}^{\xi}_{\xi+\zeta,a}(z)\right)'\right|}<\dfrac{|a|(\xi+\zeta+2)}{(\xi+\zeta+1)(\xi+\zeta+2-|a|)}\] and \[\left|\mathbb{V}^{\xi}_{\xi+\zeta,a}(z)\right|>\dfrac{(\xi+\zeta+1)(\xi+\zeta+2-|a|)-|a|(\xi+\zeta+2)}{(\xi+\zeta+1)(\xi+\zeta+2-|a|)}.\]
From these inequalities and \eqref{derivativeofv}, we obtain
\[\left|\dfrac{z\left(\mathbb{V}^{\xi}_{\zeta,a}(z)\right)''}{\left(\mathbb{V}^{\xi}_{\zeta,a}(z)\right)'}\right|=\left|\dfrac{z\left(\mathbb{V}^{\xi}_{\xi+\zeta,a}(z)\right)'}{\left(\mathbb{V}^{\xi}_{\xi+\zeta,a}(z)\right)}\right|\]\[<\dfrac{|a|(\xi+\zeta+2)}{(\xi+\zeta+1)(\xi+\zeta+2-|a|)-|a|(\xi+\zeta+2)}\;(z\in\mathscr{D}).\]
This means that, if $(\xi+\zeta+1)^2+(\xi+\zeta+1)(1-3|a|)-2|a|\geq 0$, then by definition $\mathbb{V}^{\xi}_{\zeta,a}$ is convex in $\mathscr{D}$. But this inequality is true under the condition $\xi+\zeta>\dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$. Hence, $\mathbb{V}^{\xi}_{\zeta,a}$ is convex in $\mathscr{D}$.\\
{\noindent(iii)} The function $\mathcal{J}^{\xi}_{\zeta,a}(z)$ is convex iff  $z(\mathcal{J}^{\xi}_{\zeta,a}(z))'$ is starlike, but from \eqref{rec} \\$z(\mathcal{J}^{\xi}_{\zeta,a}(z))'=\mathbb{J}^{\xi}_{\xi+\zeta,a}(z)$. This in view of part (i) of the theorem completes the proof.
\newline
%=====================================================================
\begin{remark}
If we put $a=1$ and $\xi=1$, then we obtain part(a) and part(b) of Corollary 2.8 of \cite{prajap-2}. Similarly if we put $a=-1$ and $\xi=1$ and using Lemma \ref{oho},  we obtain part(c) of Corollary 2.8 of \cite{prajap-2}.
\end{remark}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{theorem}\label{2.9}
{\rm {(i)}} Let $\xi\geq 1$ and $\zeta\geq \frac{-3+2|a|+\sqrt{1+4|a|^2}}{2}$, then $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is starlike in the disk $\mathscr{D}_{1/2}$.\\
{\noindent \rm (ii)} Let $\xi\geq 1$ and  $\zeta\geq\dfrac{3(|a|-1)+\sqrt{9|a|^2+2|a|+1}}{2}$, then $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is convex in the disk $\mathscr{D}_{1/2}$.\\
{\noindent \rm (iii)} Let $\xi\geq 1$ and $\zeta>\zeta^*$, where $\zeta^*$ is positive root of the equation $\zeta^2+\zeta(3-(1+\sqrt{5})|a|)+(2-(1+2\sqrt{5})|a|)=0,$ then $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is starlike in the disk $\mathscr{D}$.
\end{theorem}
\begin{proof}
On performing calculations, we have
\[\left|\frac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}-1\right|=\left|\frac{1}{z}\left\{z+\sum_{n=1}^{\infty}\frac{(-a)^n\Gamma{(\zeta+1)}z^{n+1}}{n!\Gamma{(\xi n+\zeta+1})}\right\}-1\right|\] 
\[\leq \sum_{n=1}^{\infty}\frac{|a|^n}{n!}\frac{1}{(\zeta+1)(\zeta+2)...(\zeta+n)}<\frac{|a|}{(\zeta+1)}\sum_{n=0}^{\infty}\left(\dfrac{|a|}{\zeta+2}\right)^n\]\[=\frac{|a|}{(\zeta+1)}\dfrac{(\zeta+2)}{(\zeta+2-|a|)}\]
In view of Lemma \ref{2.2lem}, $\mathbb{J}^{\xi}_{\zeta,a}$ is starlike in $\mathscr{D}_{1/2}$, if $(\zeta+1)^2+(\zeta+1)(1-2|a|)-|a|\geq0$, but this is true in view of the hypothesis. Hence, the result is proved.\\

\noindent(ii) Using Lemma \ref{2.3lem}, we obtain
\begin{equation}
\left|(\mathbb{J}^{\xi}_{\zeta,a}(z))'-1\right|=\left|\sum_{n=1}^{\infty}\dfrac{(-a)^n(n+1)\Gamma{(\zeta+1)}z^n}{n!\Gamma{(\xi n+\zeta+1)}}\right|
\end{equation}
\[\leq \sum_{n=1}^{\infty}\dfrac{|a|^n n\Gamma{(\zeta+1)}}{n!\Gamma{(\xi n+\zeta+1)}}+\sum_{n=1}^{\infty}\dfrac{|a|^n\Gamma{(\zeta+1)}}{n!\Gamma{(\xi n+\zeta+1)}}\]
\begin{equation}\label{lavin}\leq\sum_{n=1}^{\infty}\dfrac{2|a|^n}{(\zeta+1)(\zeta+2)...(\zeta+n)}<\dfrac{2|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)}\leq 1.\end{equation}
This shows that $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is convex in $\mathscr{D}_{1/2}.$\\
\noindent
(iii) Using the Lemma \ref{2.4lem} and  equation \eqref{lavin}, we see that $\mathbb{J}^{\xi}_{\zeta,a}(z)$ is starlike  in $\mathscr{D}_{1/2},$ if\;\;\;\;\;\;\;\;\;\; $\zeta^2+\zeta(3-(1+\sqrt{5})|a|)+(2-(1+2\sqrt{5})|a|)>0,$ this proves the result. 
\end{proof}
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%%%%%%%%%%%%%%%%%%%%%%%
\begin{remark}
Setting $\xi=1,$ and $ a=1$  in Theorem \ref{2.9}, we obtain Part (a), (b) and (c) of Corollary 2.10 of \cite{prajap-2}.
\end{remark}
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\begin{theorem} Let $\xi\geq 1$ and $0\leq \eta<1$. Suppose also that {\small
\[\psi(\eta)=\dfrac{(3|a|-1)-\eta(2|a|-1)+\sqrt{\eta^2(4|a|^2+1)-2\eta(6|a|^2+|a|+1)+(9|a|^2+2|a|+1)}}{2(1-\eta)}.\] }
{\rm (i)}Let $\zeta\geq\psi(\eta)$, then $\mathbb{J}^{\xi}_{\zeta,a}(z)\in \mathscr{S}^*(\eta)$.\\
{\rm (ii)} Let $\zeta+\xi\geq\psi(\eta)$, then ${\mathbb{V}^{\xi}_{\zeta,a}(z)}\in \mathscr{K}^*(\eta)$.\\
(iii) Let $\zeta+\xi\geq\psi(\eta)$, then $\mathcal{J}^{\xi}_{\zeta,a}(z)\in \mathscr{K}^*(\eta)$.
\end{theorem}
\begin{proof}
Following the proof of Theorem \ref{first}, we find that $\mathbb{J}^{\xi}_{\zeta,a}(z)\in\mathscr{S}^*(\eta)$ii, if \[\dfrac{|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)-|a|(\zeta+2)}\leq (1-\eta),\] 
but this inequality is a direct consequence of hypothesis. Hence the result. Remaining part can be shown similarly.
\end{proof}
\begin{theorem}
Let $\xi\geq 1$ and $\zeta^*$ be the positive root of the cubic equation $\zeta^3+\zeta^2(2-3|a|)-3\zeta(1+2|a|)-(6+|a|)\geq 0$, then $\mathbb{J}^{\xi}_{\zeta,a}$ is close-to-convex with respect to $\mathbb{J}_{\zeta}$ in $\mathscr{D}$, provided $\zeta>\max\{\zeta^*, |a|-2, \sqrt{3}\}$.
\end{theorem}
\begin{proof}
Using definition, we need to show $\exists h\in\mathcal{S^*}$, such that
\[\Re \left(\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{h(z)}\right)>0\;\;\;(z\in\mathscr{D}),\]
this can be easily shown by proving
\[\left|\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{h(z)}-1\right|<1,\;\;\;z\in\mathscr{D}.\]
If $z\in\mathscr{D}$, then a computation gives
\[\left|(\mathbb{J}^{\xi}_{\zeta,a}(z))'-\dfrac{\mathbb{J}_{\xi}(z)}{z}\right|\leq\sum_{n=1}^{\infty}\dfrac{\Gamma{(\zeta+1)}}{n!}\left|\dfrac{(-a)^n(n+1)}{\Gamma{(\xi n+\zeta+1)}}-\dfrac{(-1)^n}{\Gamma{(n+\zeta+1)}}\right|\]
\[\leq\sum_{n=1}^{\infty}\dfrac{\Gamma{(\zeta+1)}}{n!}\left|\dfrac{(a)^n(n+1)}{\Gamma{(\xi n+\zeta+1)}}+\dfrac{1}{\Gamma{(n+\zeta+1)}}\right|\]
\[\leq\sum_{n=1}^{\infty}\dfrac{\Gamma{(\zeta+1)}}{n!}\left[\dfrac{|a|^n(n+1)+1}{\Gamma{(n+\zeta+1)}}\right]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\leq\sum_{n=1}^{\infty}\dfrac{2|a|^n}{(\zeta+1)(\zeta+2)...(\zeta+n)}+\sum_{n=1}^{\infty}\dfrac{1}{(\zeta+1)(\zeta+2)...(\zeta+n)}\]
\begin{equation}\label{jsr1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;<\dfrac{1}{(\zeta+1)}\sum_{n=0}^{\infty}\dfrac{(2|a|^{n+1}+1)}{(\zeta+2)^n}=\dfrac{(\zeta+2)[|a|(2\zeta+1)+\zeta+2]}{(\zeta+2-|a|)(\zeta+1)^2}\end{equation}
and
\[\left|\dfrac{\mathbb{J}_{\zeta}(z)}{z}\right|\geq1-\sum_{n=1}^{\infty}\dfrac{1}{n!(\zeta+1)(\zeta+2)...(\zeta+n)}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\]
\begin{equation}\label{jsr2}\;\;\;\;\;\;>1-\dfrac{1}{\zeta+1}\sum_{n=0}^{\infty}\left(\frac{1}{\zeta+2}\right)^n=\dfrac{\zeta^2+\zeta-1}{(\zeta+1)^2}.\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\end{equation}
From \eqref{jsr1} and \eqref{jsr2}
\[\left|\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{\mathbb{J}_{\zeta}(z)}-1\right|=\left|\dfrac{(\mathbb{J}^{\xi}_{\zeta,a}(z))'-\dfrac{\mathbb{J}_{\xi}(z)}{z}}{\dfrac{\mathbb{J}_{\zeta}(z)}{z}}\right|\]
\[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;<\dfrac{(\zeta+2)}{(\zeta+2-|a|)(\zeta^2+\zeta-1)}[|a|(2\zeta+1)+\zeta+2]\leq1,\;\;\;z\in\mathscr{D}.\]
This shows that $\Re\left(z(\mathbb{J}^{\xi}_{\zeta,a}(z))'/\mathbb{J}_{\zeta}(z)\right)>0$ and hence $\mathbb{J}^{\xi}_{\zeta,a}$ is close-to-convex in $\mathscr{D}$. Starlikeness of $\mathbb{J}_{\zeta}$ can be deduced from Theorem \ref{first} for $a=1$ and it comes out $\xi\ge 1$ and $\zeta\ge \sqrt{3}$. 
\end{proof}
\noindent
For a non-zero complex number $\delta$, we define an integral operator $\mathcal{F}_{\delta}:\mathscr{D}\rightarrow\mathbb{C}$, by
\begin{equation}
\mathcal{F}_{\delta}(z)=\left\{\delta\int_{0}^z t^{\delta-2}\mathbb{J}^{\xi}_{\zeta,a}(t)dt\right\}^{\frac{1}{\delta}},\;\;\;z\in\mathscr{D}.
\end{equation}
Note that $\mathcal{F}_{\delta}(z)\in\mathscr{A}$. In the next theorem, we find  conditions so that $\mathcal{F}_{\delta}$ is univalent in $\mathscr{D}$.
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\begin{theorem}\label{operator}
Let $\xi>-1$, $\zeta>-1$, $\kappa=\dfrac{|a|(\zeta+2)}{(\zeta+1)(\zeta+2-|a|)-|a|(\zeta+2)}$ and $L\in \mathbb{R}^{+}$  such that $\left|\mathbb{J}^{\xi}_{\zeta,a}(z)\right|\leq L$ in $\mathscr{D}$, then following results holds\\
\noindent{\rm(i)} If $\kappa+|\delta-1|+L/\delta\leq 1$, then $\mathcal{F}_{\delta}$ is univalent in $\mathscr{D}.$\\
\noindent{\rm(ii)} If $d\in\mathbb{C}$ with $|d|\le 1,\;d\neq-1$ and $|d|+\kappa/|\delta|\leq 1,$ then $\mathcal{F}_{\delta}$ is univalent in $\mathscr{D}.$ 
\end{theorem}
\begin{proof}
(i) A simple calculation gives us
\begin{equation}
\dfrac{z\mathcal{F}_{\delta}''(z)}{\mathcal{F}_{\delta}'(z)}=\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{\mathbb{J}_{\zeta}(z)}+\frac{z^{\delta-1}}{\delta}\mathbb{J}^{\xi}_{\zeta,a}(z)+\delta-2,\;\;\;z\in\mathscr{D}.
\end{equation}
Since $\mathbb{J}^{\xi}_{\zeta,a}\in\mathscr{A}$, so using Schwarz Lemma, we obtain $\left|\mathbb{J}^{\xi}_{\zeta,a}(z)\right|\leq L|z|$ in $\mathscr{D}$.\\
Now using \eqref{lo2.4} and the triangle inequality $\left(|z_1+z_2|\le |z_1|+|z_2|\right)$, we obtain
\[(1-|z|^2)\left|\dfrac{z\mathcal{F}_{\delta}''(z)}{\mathcal{F}_{\delta}'(z)}\right|\leq(1-|z|^2)\left\{|\delta-1|+\left|\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{\mathbb{J}^{\xi}_{\zeta,a}(z)}-1\right|+\dfrac{|z|^{\Re(\delta)}}{|\delta|}\left|\frac{\mathbb{J}^{\xi}_{\zeta,a}(z)}{z}\right|\right\}\]
\[<(1-|z|^2)\left\{\zeta+|\delta-1|+\frac{L}{|\delta|}\right\}\leq 1.\;\;\;\]
This implies that $\mathcal{F}_{\delta}$ satisfy Becker's criterion for univalence, hence $\mathcal{F}_{\delta}$ is univalent in $\mathscr{D}$.\\
(ii) Let us consider the function
\[\mathcal{G}(z)=\int_{0}^z\dfrac{\mathbb{J}^{\xi}_{\zeta,a}(t)}{t}dt,\;\;\;z\in\mathscr{D}.\] 
Observe that, $\mathcal{G}\in\mathscr{A}$. Using \eqref{lo2.4} and the triangle inequality, we get
\[\left|d|z|^{2\delta}+(1-|z|^{2\delta})\frac{z\mathcal{G}''(z)}{\delta\mathcal{G}'(z)}\right|\leq\left|d|z|^{2\delta}+(1-|z|^{2\delta})\dfrac{1}{\delta}\left(\dfrac{z(\mathbb{J}^{\xi}_{\zeta,a}(z))'}{\mathbb{J}_{\zeta}(z)}-1\right)\right|\]
\[\le |d|+\frac{\zeta}{|\delta|}\le 1.\;(\rm using\;the \;hypothesis\;of\;Theorem\; \ref{operator})\]
This in view of Lemma \ref{2.6}, implies that $\mathcal{F}_{\delta}(z)$ defined by 
\begin{equation}
\mathcal{F}_{\delta}(z)=\left\{\delta\int_{0}^z t^{\delta-1}\mathcal{G}'(t)dt\right\}^{{1}/{\delta}}=\left\{\delta\int_{0}^z t^{\delta-2}\mathbb{J}^{\xi}_{\zeta,a}(t)dt\right\}^{{1}/{\delta}}\;(z\in\mathscr{D}).
\end{equation}
is univalent in $\mathscr{D}$.
\end{proof}
\noindent
{\it\bf Conflicts of interest}. The author declare that there is no conflict of interest regarding the publication of this paper.
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