1 Introduction
The Bessel function of the first kind of order p given by
is a particular solution of the homogeneous Bessel differential equation
Here Γ denotes the gamma function. A solution of the homogeneous modified Bessel equation
gives the modified Bessel function of order p
Because of its importance, the Bessel function and other special functions are of continued interest to the wider scientific community. The Bessel function and its variations have gone through several generalizations, see, for example, [1‐6]. These generalized functions have also been framed as complex-valued analytic functions in the unit disk. Geometric properties of such functions have been studied, notably in [7‐13].
$$ J_{p}(x):= \sum_{k=0}^{\infty } \frac{(-1)^{k}}{k! \Gamma { ( k+p+1 ) } } \biggl( \frac{x}{2} \biggr) ^{2k+p}, \quad x \in\mathbb{R}, $$
$$\begin{aligned} x^{2} y''(x)+ x y'(x)+ \bigl(x^{2}-p^{2} \bigr)y(x)= 0. \end{aligned}$$
$$\begin{aligned} x^{2} y''(x)+ x y'(x)- \bigl(x^{2}+p^{2} \bigr)y(x)= 0 \end{aligned}$$
$$ I_{p}(x) = \sum_{k=0}^{\infty } \frac{1}{k! \Gamma { ( k +p+1 ) } } \biggl( \frac{x}{2} \biggr) ^{2k+p}. $$
Among the several generalized forms, perhaps a more complete generalization is that given by Baricz in [1]. In this case, the generalized Bessel function takes the form
for \(a \in \mathbb{N}=\{1, 2, 3, \ldots \}\), and \(b, p, c, x \in \mathbb{R}\). It is evident that the function \({}_{a}\mathtt{B}_{b, p, c}\) converges absolutely at each \(x \in \mathbb{R}\). Earlier, Galué [14] introduced a generalization of the Bessel function of the form
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x):= \sum _{k=0}^{\infty }\frac{(-c)^{k}}{k! \Gamma { ( a k +p+\frac{b+1}{2} ) } } \biggl( \frac{x}{2} \biggr) ^{2k+p} \end{aligned}$$
(1)
$$ {}_{a}J_{p}(x):= \sum_{k=0}^{\infty } \frac{(-1)^{k}}{k! \Gamma {(a k +p+1)}} \biggl( \frac{x}{2} \biggr) ^{2k+p}, \quad x \in \mathbb{R}, a \in \mathbb{N}. $$
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Apparently not much has been investigated for the extended Bessel function given by (1). Presumably such extensions would readily follow from recent results along similar used arguments, albeit involving intense computations. Still several pertinent questions remain, which include the question on how the parameter a influences the shape of the differential equation satisfied by \({}_{a}\mathtt{B}_{b, p, c}\). It is the aim of this paper to complement and to fill the void of earlier investigations on the Bessel function and its extensions.
The connection between the parameters a, b, and p in the representation formulae and recurrence relation for \({}_{a}\mathtt{B} _{b,p, c}\) are derived in Section 2. An important consequence is the derivation of an \((a+1)\)-order differential equation satisfied by the function \({}_{a}\mathtt{B}_{b,p, c}\). As applications, new functional inequalities for \({}_{a}\mathtt{B}_{b, p, -\alpha^{2}}\) are obtained, particularly in the case \(a=2\).
Section 3 is devoted to the investigation of the monotonicity properties of \({}_{a}\mathtt{B}_{b,p, c}\) for non-positive c, as well as for the normalized generalized Bessel function. Log-concavity and log-convexity properties in terms of the parameters d and p are also respectively investigated for the closely related function
As a consequence, direct and reverse Turán-type inequalities [15] are obtained.
$$ {}_{a}\mathcal{B}^{d}_{b,p, c}(x): = \sum_{k=0}^{\infty }\frac{(-c/4)^{k} \Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(d)_{k}}{k!}x^{k}. $$
2 General representation formulations and applications
This section aims to find representation formulations, including integral representations, for the generalized function \({}_{a} \mathtt{B}_{b, p, c}\) in terms of the parameters a, b, and p. A starting point is the Gauss multiplication theorem [16] for the gamma function, which states that
\(m \in \mathbb{N}\). Thus
\(l \neq -ak,-ak-1, -ak-2,\ldots \) , and \(k \in \mathbb{N}\). Here \((\alpha )_{k}\) is the Pochhammer symbol defined by \((\alpha )_{k}= \alpha (\alpha +1)_{k-1}= \Gamma (\alpha +k)/ \Gamma (\alpha )\), with \((\alpha )_{0}=1\). Substituting \(z= l/a\) and \(m=a\) gives
and thus (2) yields
Choosing \(l=p+(b+1)/2\), it is evident from (1) that
which leads to the following representation in terms of the generalized hypergeometric function (see [17]):
$$\begin{aligned} \Gamma (mz)=(2\pi )^{\frac{1-m}{2}}m^{mz-\frac{1}{2}} \prod _{j=1}^{m} \Gamma \biggl( z+ \frac{j-1}{m} \biggr) , \quad z \neq 0,- \frac{1}{m},\ldots , \end{aligned}$$
$$\begin{aligned} \Gamma (ak+l) &= \Gamma \biggl( a \biggl( k+\frac{l}{a} \biggr) \biggr) \\ &=(2\pi )^{\frac{1-a}{2}}a ^{ak+l-\frac{1}{2}} \prod_{j=1}^{a}\Gamma \biggl( k+ \frac{l+j-1}{a} \biggr) \\ &=(2\pi )^{\frac{1-a}{2}}a^{ak+l-\frac{1}{2}}\prod_{j=1}^{a} \biggl( \frac{l+j-1}{a} \biggr) _{k} \Gamma \biggl( \frac{l+j-1}{a} \biggr) , \end{aligned}$$
(2)
$$\begin{aligned} \prod_{j=1}^{a} \Gamma \biggl( \frac{l+j-1}{a} \biggr) =\frac{\Gamma (l)}{(2\pi )^{\frac{1-a}{2}}a^{l-\frac{1}{2}}}, \end{aligned}$$
$$\begin{aligned} \Gamma (ak+l) =a^{ak} \Gamma {(l)}\;\prod _{j=1} ^{a} \biggl( \frac{l+j-1}{a} \biggr) _{k}. \end{aligned}$$
$$ {}_{a}\mathtt{B}_{b, p, c}(x)= \frac{x^{p}}{2^{p} \Gamma { ( p+\frac{b+1}{2} ) }}\sum _{k=0}^{\infty }\frac{1}{ ( \frac{2p+b+1}{2a} ) _{k} ( \frac{2p+b+3}{2a} ) _{k}\cdots ( \frac{2p+b+2a-1}{2a} ) _{k} (1)_{k}} \biggl( - \frac{c x^{2}}{4 a^{a}} \biggr) ^{k}, $$
$$\begin{aligned} {}_{m} F_{n}(a_{1},a_{2}, \ldots , a_{m}; b_{1},b_{2}, \ldots , b_{n};x)= \sum_{k=0}^{\infty }\frac{ (a_{1})_{k} (a_{2})_{k} \cdots (a _{m})_{k}}{ (b_{1})_{k} (b_{2})_{k} \cdots (b_{n})_{k} k!} x^{k}. \end{aligned}$$
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Proposition 1
Let
\(a \in \mathbb{N}\), and
\(b, p, c, x \in \mathbb{R}\). Then
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x)=\frac{x^{p}}{2^{p} \Gamma { ( p+\frac{b+1}{2} ) }} {}_{0} F_{a} \biggl( -; \frac{2p+b+1}{2a}, \frac{2p+b+3}{2a} ,\ldots , \frac{2p+b+2a-1}{2a};-\frac{c x^{2}}{4 a^{a}} \biggr) . \end{aligned}$$
Another representation formula can be expressed in terms of the order \(a=1\). In the sequel, we shall simply write \(\mathtt{B}_{b, p, c} := {}_{1}\mathtt{B}_{b, p, c}\). Thus
$$ \mathtt{B}_{b, p, c}(x):= \sum_{k=0}^{\infty } \frac{(-c)^{k}}{k! \Gamma { ( k +p+\frac{b+1}{2} ) }} \biggl( \frac{x}{2} \biggr) ^{2k+p}. $$
(3)
Proposition 2
Let
\(a \in \mathbb{N}\), and
\(b, p, c, x \in \mathbb{R}\). Then
where
\(\mathtt{B}_{b, p, c}\)
is given by (3).
$$ {}_{a} \mathtt{B}_{b, p, c}(x)=(2 \pi )^{\frac{a-1}{2}}a^{-p- \frac{b}{2}} \biggl( \frac{x}{2} \biggr) ^{p} \prod _{j=1}^{a} \biggl( \frac{x}{2 a^{a/2}} \biggr) ^{-\frac{p+j-1}{a}} \mathtt{B}_{\frac{b+1-a}{a}, \frac{p+j-1}{a}, c} \biggl( \frac{x}{ a^{a/2}} \biggr) , $$
Proof
It is clear from (2) that
Thus,
□
$$\begin{aligned} \Gamma \biggl( a k+p+\frac{b+1}{2} \biggr) = (2 \pi )^{\frac{1-a}{2}} a ^{ak+p+\frac{b}{2}} \prod_{j=1}^{a} \Gamma \biggl( k+\frac{p+ \frac{b+1}{2}+j-1}{a} \biggr) . \end{aligned}$$
(4)
$$\begin{aligned} {}_{a} \mathtt{B}_{b, p, c}(x) &= \sum _{k=0}^{\infty }\frac{(-c)^{k}}{(2 \pi )^{\frac{1-a}{2}} a^{ak+p+\frac{b}{2}} k!\; \prod_{j=1}^{a} \Gamma ( k+\frac{p+\frac{b-1}{2}+j}{a} ) } \biggl( \frac{x}{2} \biggr) ^{2k+p} \\ &=(2 \pi )^{\frac{a-1}{2}}a^{\frac{-2p-b}{2}} \biggl( \frac{x}{2} \biggr) ^{p} \prod_{j=1}^{a} \biggl( \frac{x}{2 a^{a/2}} \biggr) ^{- \frac{p+j-1}{a}} \\ & \quad {} \times \sum_{k=0}^{\infty } \frac{(-c)^{k}}{ k!\; \Gamma ( k+ \frac{p+j-1}{a}+\frac{b+1}{2a} ) } \biggl( \frac{x}{2 a^{a/2}} \biggr) ^{2k+\frac{p+j-1}{a}} \\ &=(2 \pi )^{\frac{a-1}{2}}a^{\frac{-2p-b}{2}} \biggl( \frac{x}{2} \biggr) ^{p} \prod_{j=1}^{a} \biggl( \frac{x}{2 a^{a/2}} \biggr) ^{- \frac{p+j-1}{a}}\mathtt{B}_{\frac{b+1-a}{a}, \frac{p+j-1}{a}, c} \biggl( \frac{x}{ a^{a/2}} \biggr) . \end{aligned}$$
Remark 1
As a first application, let \(a=2\). In this case, Proposition 2 yields
Now \(\mathtt{B}_{1,p, 1}(x)= J_{p}(x)\) is the classical Bessel function, while \(\mathtt{B}_{1,p,-1}(x)= I_{p}(x)\) is the modified Bessel function of the first kind of order p. Thus, for \(b=3\) and \(c=\pm 1\), it follows that
and
Thus interestingly
$$\begin{aligned} {}_{2} \mathtt{B}_{b, p, c}(x) =\sqrt{\frac{ \pi }{x 2^{b-3}}} \mathtt{B}_{\frac{b-1}{2}, \frac{p}{2}, c} \biggl( \frac{x}{ 2} \biggr) \mathtt{B}_{\frac{b-1}{2},\frac{p+1}{2}, c} \biggl( \frac{x}{ 2} \biggr) . \end{aligned}$$
$$\begin{aligned} {}_{2} \mathtt{B}_{3, p, 1}(x) &=\sqrt{\frac{ \pi }{x}}\; J_{ \frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) J_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) \end{aligned}$$
$$\begin{aligned} {}_{2} \mathtt{B}_{3, p, -1}(x) &=\sqrt{\frac{ \pi }{x}}\; I_{ \frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) I_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) . \end{aligned}$$
$$ \frac{{}_{2} \mathtt{B}_{3, 2p, 1}(x)}{{}_{2} \mathtt{B}_{3,2p+1, 1}(x)} = \frac{J_{p}(x/2)}{J_{p+1}(x/2)} \quad \text{and} \quad \frac{{} _{2} \mathtt{B}_{3, 2p, -1}(x)}{{}_{2} \mathtt{B}_{3,2p+1, -1}(x)} = \frac{I _{p}(x/2)}{I_{p+1}(x/2)}. $$
For obtaining recurrence relations, first differentiate (1) to yield
Expanding the left side of the above equation yields
$$\begin{aligned} \frac{d}{dx} \bigl( x^{-p}{ }_{a} \mathtt{B}_{b, p, c}(x) \bigr) =& \sum_{k=0}^{\infty } \frac{(-c) ^{k}(k)}{2^{p} \Gamma ( ak+p+ \frac{b+1}{2} ) \Gamma ( k+1 ) }(x/2)^{2k-1} \\ =&\sum_{k=0}^{\infty }\frac{(-c) ^{k+1}(k+1)}{2^{p} \Gamma ( ak+p+\frac{b+1}{2} ) \Gamma ( k+2 ) } (x/2)^{2k+1} \\ =&-c x^{1-a-p} \biggl( \frac{1}{2} \biggr) ^{1-a}\sum _{k=0}^{\infty }\frac{(-c)^{k} }{\Gamma ( ak+p+\frac{b+1}{2} ) \Gamma ( k+ 1 ) } \biggl( \frac{x}{2} \biggr) ^{2k+p+a} \\ =&-c x^{-p} \biggl( \frac{x}{2} \biggr) ^{1-a}{ }_{a}\mathtt{B}_{b, p+a, c}(x). \end{aligned}$$
$$ x \frac{d}{dx}{}_{a}\mathtt{B}_{b, p,c}(x) = p { }_{a}\mathtt{B}_{b, p, c}(x) -c \biggl( \frac{x}{2} \biggr) ^{1-a} x { }_{a}\mathtt{B}_{b, p+a, c}(x). $$
(5)
Yet another form for \(x{}_{a}\mathtt{B}_{b, p,c}'\) is obtained from
Expanding the left side of the above relation, it follows that
Thus (5) and (6) lead to the following recurrence relation.
$$\begin{aligned} \frac{d}{dx} \bigl( x_{{}}^{\frac{2p+b-1}{a}-p}{ }_{a} \mathtt{B}_{b, p, c}(x) \bigr) &= \sum_{k=0}^{\infty } \frac{(-c)^{k} }{2^{2k+p}\Gamma ( ak+p+\frac{b+1}{2} ) \Gamma ( k+1 ) }\frac{d}{dx}x^{2k+\frac{2p+b-1}{a}} \\ &= \sum_{k=0}^{\infty } \frac{(-c)^{k} (\frac{2}{a}) }{2^{2k+p}\Gamma ( ak+p+\frac{b+1}{2} ) \Gamma ( k+1 ) }x^{2k+\frac{2p+b-1-a}{a}} \\ &=\frac{1}{a}\;x^{\frac{2p+b-1}{a}-p}\text{ } {}_{a}B_{b, p-1,c} ( x ) . \end{aligned}$$
$$\begin{aligned} x \frac{d}{dx}{}_{a}\mathtt{B}_{b, p,c}(x) =\frac{x}{a}{}_{a} \mathtt{B}_{a, p-1,c}(x) - \biggl( \frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x). \end{aligned}$$
(6)
Proposition 3
Let
\(a \in \mathbb{N}\), and
\(b, p, c, x \in \mathbb{R}\). Then
$$ \frac{x}{a}{}_{a}\mathtt{B}_{b, p-1,c}(x)+ c \biggl( \frac{x}{2} \biggr) ^{1-a} x { }_{a} \mathtt{B}_{b, p+a, c}(x)= \biggl( \frac{2p+b-1}{a} \biggr) {}_{a}\mathtt{B}_{b, p,c}(x). $$
We next find an \((a+1)\)-order differential equation satisfied by \({ }_{a}\mathtt{B}_{b, p, c}\) from the recurrence relations (5) and (6) (see also [18]).
Theorem 1
Let the operator
D
be given by
\(D := x (d/dx)\). For each
\(k=1, \ldots , a\), the generalized Bessel function
\({}_{a}\mathtt{B}_{b, p,c}\)
satisfies the differential equation
In particular, the generalized Bessel function
\({}_{a}\mathtt{B}_{b, p,c}\)
is a solution of the differential equation
$$\begin{aligned} (D-p)\prod_{j=1}^{k} \biggl( D+\frac{2p+b+1-2j}{a}-p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) &+\frac{cx^{k+2-a}}{a^{k} 2^{1-a}} {}_{a} \mathtt{B}_{b, p-k+a,c}(x) = 0. \end{aligned}$$
(7)
$$ (D-p)\prod_{j=1}^{a} \biggl( D+\frac{2p+b+1-2j}{a}-p \biggr) y(x)+\frac{cx ^{2}}{a^{a} 2^{1-a}} y(x)= 0. $$
(8)
Proof
The proof is by induction. In terms of the differential operator D, identity (6) takes the form
Now identity (5) gives
$$\begin{aligned} \biggl( D+\frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) = \frac{x}{a}{}_{a} \mathtt{B}_{b, p-1,c}(x) . \end{aligned}$$
(9)
$$\begin{aligned} D \bigl(x {}_{a}\mathtt{B}_{b, p-1,c}(x) \bigr) &= x^{2} {}_{a}\mathtt{B}'_{b, p-1,c}(x)+ x {}_{a} \mathtt{B}_{b, p-1,c}(x) \\ &=p x{ }_{a}\mathtt{B}_{b, p-1, c}(x)-c \biggl( \frac{x}{2} \biggr) ^{1-a} x^{2} { }_{a} \mathtt{B}_{b, p-1+a, c}(x) \\ &=p a \biggl( D+\frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) -c \biggl( \frac{x}{2} \biggr) ^{1-a} x^{2} { }_{a}\mathtt{B}_{b, p-1+a, c}(x). \end{aligned}$$
Applying the operator D to both sides of (9), the latter equation leads to
whence
This establishes (7) for \(k=1\).
$$\begin{aligned} &D \biggl( D+\frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) \\ &\quad = \frac{1}{a} D \bigl(x{}_{a} \mathtt{B}_{b, p-1,c}(x) \bigr) \\ &\quad =p \biggl( D+\frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) -\frac{c}{a 2^{1-a}} x^{3-a} { }_{a}\mathtt{B}_{b, p-1+a, c}(x), \end{aligned}$$
$$\begin{aligned} &(D-p) \biggl( D+\frac{2p+b-1}{a}- p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) + \frac{c}{a 2^{1-a}} x^{3-a} { }_{a}\mathtt{B}_{b, p-1+a, c}(x)=0. \end{aligned}$$
Assume now that (7) holds for \(k=n\). It follows from (6) that
Applying the operator D to both sides of (7) for \(k=n\), the above equation shows that
The induction formula allows us to rewrite the final term above in the form
Thus
that is,
□
$$\begin{aligned} &D \bigl(x^{n-a+2} { }_{a}\mathtt{B}_{b, p-n+a, c}(x) \bigr) \\ &\quad = x^{n-a+3} { }_{a} \mathtt{B}'_{b, p-n+a, c}(x)+(n-a+2)x^{n-a+2}{ }_{a}\mathtt{B}_{b, p-n+a, c}(x) \\ &\quad = \frac{x^{n+3-a}}{a} { }_{a}\mathtt{B}_{b, p-n-1+a, c}(x)- \biggl( \frac{2(p-n)+b-1}{a}-p \biggr) x^{n-a+2}{ }_{a}\mathtt{B} _{b, p-n+a, c}(x). \end{aligned}$$
$$\begin{aligned} &D(D-p)\prod_{j=1}^{n} \biggl( D+ \frac{2p+b+1-2j}{a}-p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) \\ &\quad =-\frac{c}{a^{n+1} 2^{1-a}} x^{n+3-a} { }_{a}\mathtt{B}_{b, p-n-1+a, c}(x) \\ & \quad \quad {}+ \frac{c}{a^{n} 2^{1-a}} \biggl( \frac{2(p-n)+b-1}{a}-p \biggr) x^{n-a+2} { }_{a}\mathtt{B}_{b, p-n+a, c}(x). \end{aligned}$$
$$\begin{aligned} &D(D-p)\prod_{j=1}^{n} \biggl( D+ \frac{2p+b+1-2j}{a}-p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) \\ &\quad =-\frac{c}{a^{n+1} 2^{1-a}} x^{n+3-a} { }_{a}\mathtt{B}_{b, p-n-1+a, c}(x) \\ &\quad \quad {}- \biggl( \frac{2(p-n)+b-1}{a}-p \biggr) \Biggl[ (D-p)\prod _{j=1} ^{n} \biggl( D+\frac{2p+b+1-2j}{a}-p \biggr) {}_{a}\mathtt{B}_{b, p,c}(x) \Biggr] . \end{aligned}$$
$$\begin{aligned} & \biggl( D+\frac{2(p-n)+b-1}{a}-p \biggr) (D-p)\prod _{j=1}^{n} \biggl( D+ \frac{2p+b+1-2j}{a}-p \biggr) {}_{a}\mathtt{B}_{b, p,c}(x) \\ &\quad =-\frac{c}{a^{n+1} 2^{1-a}} x^{n+3-a} { }_{a} \mathtt{B}_{b, p-n-1+a, c}(x), \end{aligned}$$
$$\begin{aligned} (D-p)\prod_{j=1}^{n+1} \biggl( D+ \frac{2p+b+1-2j}{a}-p \biggr) {}_{a} \mathtt{B}_{b, p,c}(x) &+ \frac{cx^{n+3-a}}{a^{n+1} 2^{1-a}} {}_{a} \mathtt{B}_{b, p-n-1+a,c}(x) =0. \end{aligned}$$
Remark 2
For \(a=1\), the differential equation (8) reduces to
This is the differential equation considered by Baricz [2] in his study on the unification of Bessel, modified Bessel, spherical Bessel, and modified spherical Bessel functions. Thus the differential equation yields the Bessel function of the first kind of order p when \(b=c=1\), and the modified Bessel function of the first kind of order p when \(b=1\) and \(c=-1\). In the case \(b=2\) and \(c=1\), there results the spherical Bessel function of order p.
$$ x^{2} y''(x) + bxy'(x) + \bigl(c x^{2}-p^{2} +(1-b)p \bigr) y(x)=0. $$
For \(a=2\), (8) reduces to
Thus its particular solution is \({}_{2}\mathtt{B}_{b, p, c}\), which from Proposition 1 can be expressed in the form
$$ x^{3} y'''(x)+(1+b-p) x^{2} y''(x)+(b-1) \biggl( \frac{b+1}{4}-p \biggr) x y'(x) + \biggl( \frac{c}{2}x^{2}-p \frac{(b-1)(b-3)}{4} \biggr) y(x)=0. $$
$$\begin{aligned} {}_{2}\mathtt{B}_{b, p, c}(x) &= \sum _{k=0}^{\infty }\frac{(-c)^{k}}{k! \Gamma { ( 2 k +p+\frac{b+1}{2} ) } } \biggl( \frac{x}{2} \biggr) ^{2k+p} \\ &=\frac{x^{p}}{2^{p} \Gamma {(p+\frac{b+1}{2})}} {}_{0} F_{2} \biggl( -; \frac{2p+b+1}{4}, \frac{2p+b+3}{4};- \frac{c x^{2}}{16} \biggr) . \end{aligned}$$
We conclude this section by establishing two integral representations for \({}_{a}\mathtt{B}_{b, p, c}\). For this purpose, first recall the integral form of the beta function \(B(x,y)\) [16, 17] given by
for \(\operatorname{Re}x >0, \operatorname{Re}y >0\). Replacing x by \((a k+1)\) and y by \((2 p+b-1)/2\) in (10), we get
where \(p> -(b-1)/2\).
$$\begin{aligned} B(x,y):=\frac{ \Gamma {(x)} \Gamma {(y)}}{\Gamma {(x+y)}}= \int_{0}^{1} t^{x-1}(1-t)^{y-1}\,dt \end{aligned}$$
(10)
$$\begin{aligned} \frac{1}{ \Gamma ( a k+ p +\frac{b+1}{2} ) }= \frac{2}{\Gamma {(a k+1)} \Gamma { ( \frac{2p+b-1}{2} ) }} \int _{0}^{1} t^{2a k+1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{2}}\,dt, \end{aligned}$$
For \(a \in \mathbb{N}\), identity (2) yields
Then the generalized Bessel function \({}_{a}\mathtt{B}_{b, p, c}\) takes the form
which establishes the following proposition.
$$ \Gamma { ( ak+1 ) }= \Gamma { \biggl( a \biggl( k+ \frac{1}{a} \biggr) \biggr) }=(2\pi )^{\frac{1-a}{2}}a^{ak+ \frac{1}{2}} \prod_{j=1}^{a} \Gamma \biggl( k+ \frac{j}{a} \biggr) . $$
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x) &= \frac{2 ( \frac{x}{2} ) ^{p}}{a ^{1/2}(2\pi )^{(1-a)/2} \Gamma { ( \frac{2p+b-1}{2} ) }} \prod_{j=1}^{a} \int_{0}^{1} t \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{2}} \sum_{k=0}^{\infty } \frac{(-c)^{k} }{k! \Gamma ( k+\frac{j}{a} ) } \biggl( \frac{x t^{a}}{2 a^{a/2}} \biggr) ^{2k}\,dt \\ &=\frac{2 ( \frac{x}{2} ) ^{p}}{a^{1/2}(2\pi )^{(1-a)/2} \Gamma { ( \frac{2p+b-1}{2} ) }} \prod_{j=1}^{a} \biggl( \frac{x }{2 a^{a/2}} \biggr) ^{1-\frac{j}{a}} \int_{0}^{1} t^{1-a+j} \bigl(1-t^{2} \bigr)^{ \frac{2p+b-3}{2}} \\ & \quad {} \times \sum_{k=0}^{\infty } \frac{(-c)^{k} }{k! \Gamma ( k+\frac{j}{a}-1+1 ) } \biggl( \frac{x t^{a}}{2 a^{a/2}} \biggr) ^{2k+\frac{j}{a}-1}\,dt \\ &=\frac{2 ( \frac{x}{2} ) ^{p}}{a^{1/2}(2\pi )^{(1-a)/2} \Gamma { ( \frac{2p+b-1}{2} ) }} \\ & \quad {} \times \prod_{j=1}^{a} \biggl( \frac{x }{2 a^{a/2}} \biggr) ^{1- \frac{j}{a}} \int_{0}^{1} t^{1-a+j} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{2}} \mathtt{B}_{1, (j/a)-1, c} \biggl( \frac{x t^{a}}{a^{a/2}} \biggr)\,dt, \end{aligned}$$
Proposition 4
Let
\(a \in \mathbb{N}\), and
\(b, p, c, x \in \mathbb{R}\). Then
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x)&=\frac{2^{1-p}x^{p}}{a^{1/2}(2\pi )^{(1-a)/2} \Gamma { ( p+\frac{b-1}{2} ) }} \\ &\quad {}\times\prod _{j=1}^{a} \biggl( \frac{x }{2 a ^{a/2}} \biggr) ^{1-\frac{j}{a}} \int_{0}^{1} t^{1-a+j} \bigl(1-t^{2} \bigr)^{ \frac{2p+b-3}{2}} \mathtt{B}_{1, (j/a)-1, c} \biggl( \frac{x t^{a}}{a ^{a/2}} \biggr)\,dt. \end{aligned}$$
Remark 3
The particular cases \(a=b=1=\pm c\) in Proposition 4 respectively lead to
for \(x>0\) and \(p>0\).
$$\begin{aligned}& {}_{1}\mathtt{B}_{1, p, 1}(x)=J_{p}(x) = \frac{2 ( \frac{x}{2} ) ^{p}}{\Gamma (p)} \int_{0}^{1} t \bigl( 1-t^{2} \bigr) ^{p-1} J_{0}(x t)\,dt, \\& {}_{1}\mathtt{B}_{1, p, -1}(x)=I_{p}(x) = \frac{2 ( \frac{x}{2} ) ^{p}}{\Gamma (p)} \int_{0}^{1} t \bigl( 1-t^{2} \bigr) ^{p-1} I_{0}(x t)\,dt, \end{aligned}$$
Another integral representation is the following.
Proposition 5
Let
\(a \in \mathbb{N}\), and
\(b, p, c, x \in \mathbb{R}\). Then
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x) & =\frac{2^{1-p} x^{p}}{\sqrt{\pi } (2 \pi )^{{\frac{1-a}{2}}}a^{p+\frac{b}{2}}} \prod _{j=1}^{a}\frac{1}{\Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ & \quad {} \times \int_{0}^{1}\sum_{k=0}^{\infty } \frac{(-c)^{k}}{(2k)!} \biggl( \frac{xt}{ a^{a/2}} \biggr) ^{2k} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}}\,dt. \end{aligned}$$
Proof
Replace x by \((k+1/2)\) and y by \(((2p+2j+b-1-a)/2a)\) in (10). Then
where \(p>(a-1-b)/2\). On the other hand, (4) yields
Thus (11) and (12) shows that the generalized Bessel function \({}_{a}\mathtt{B}_{b, p, c}\) takes the form
$$ \frac{1}{ \Gamma ( k+\frac{2 p+2 j+b-1}{2 a} ) }= \frac{2}{\Gamma { ( k+\frac{1}{2} ) } \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \int_{0}^{1} t^{2k} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}}\,dt, $$
(11)
$$\begin{aligned} \Gamma { \biggl( ak+p+\frac{b+1}{2} \biggr) } &= \Gamma { \biggl( a \biggl( k+\frac{2p+b+1}{2a} \biggr) \biggr) } \\ &=(2\pi )^{\frac{1-a}{2}}a^{ak+\frac{2p+b}{2}}\prod_{j=1}^{a} \Gamma \biggl( k+\frac{2p+b+2j-1}{2a} \biggr) . \end{aligned}$$
(12)
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, c}(x) &=\sum _{k=0}^{\infty }(-c)^{k} \frac{ ( \frac{x}{2} ) ^{2k+p}}{\Gamma {(k+1)} \Gamma ( k+ \frac{1}{2} ) }\frac{2}{(2\pi )^{\frac{1-a}{2}}a^{ak+p+ \frac{b}{2}}} \\ & \quad {} \times \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \int _{0}^{1} t^{2k} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}}\,dt \\ & =\frac{2 ( \frac{x}{2} ) ^{p}}{(2\pi )^{{\frac{1-a}{2}}}a ^{p+\frac{b}{2}}} \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \int _{0}^{1}\sum_{k=0}^{\infty } \frac{(-c)^{k}(xt)^{2k}(1-t^{2})^{ \frac{2p+2j+b-1-3a}{2a}}}{\Gamma ( k+1 ) \Gamma ( k+ \frac{1}{2} ) (2 a^{a/2})^{2k}}\,dt. \end{aligned}$$
(13)
Now the Legendre duplication formula (see [16, 17])
shows that
This reduces (13) to the desired representation and completes the proof. □
$$ \Gamma {(z)} \Gamma { \biggl( z+ \frac{1}{2} \biggr) }= 2^{1-2z} \; \sqrt{\pi } \Gamma {(2z)} $$
$$\begin{aligned} \Gamma {(k+1)}\; \Gamma { \biggl( k+\frac{1}{2} \biggr) }=k \Gamma {(k)} \Gamma { \biggl( k+\frac{1}{2} \biggr) }= 2^{1-2k} k\sqrt{ \pi } \Gamma {(2k)} = \frac{\sqrt{\pi } (2k)!}{2^{2k}}. \end{aligned}$$
Remark 4
For another application, choose \(c=\pm \alpha^{2}\). Then Proposition 5 leads to
and
Substituting \(t=\cos \theta \) yields
and
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, \alpha^{2}}(x) & = \frac{2^{1-p}x^{p}}{\sqrt{ \pi } (2\pi )^{{\frac{1-a}{2}}}a^{p+\frac{b}{2}}} \prod_{j=1}^{a} \frac{1}{\Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ & \quad {} \times \int_{0}^{1}\sum_{k=0}^{\infty } \frac{(-1)^{k}}{(2k)!} \biggl( \frac{ \alpha xt}{ a^{a/2}} \biggr) ^{2k} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}}\,dt \\ & =\frac{{2^{1-p}x^{p}}}{\sqrt{\pi } (2\pi )^{{\frac{1-a}{2}}}a ^{p+\frac{b}{2}}} \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \int _{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}} \cos \biggl( \frac{\alpha xt}{ a^{a/2}} \biggr)\,dt, \end{aligned}$$
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, -\alpha^{2}}(x) & =\frac{{2^{1-p}x^{p}}}{\sqrt{ \pi } (2\pi )^{{\frac{1-a}{2}}}a^{p+\frac{b}{2}}} \prod _{j=1}^{a}\frac{1}{\Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ & \quad {} \times \int_{0}^{1}\sum_{k=0}^{\infty } \frac{1}{(2k)!} \biggl( \frac{ \alpha xt}{ a^{a/2}} \biggr) ^{2k} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}}\,dt \\ & =\frac{{2^{1-p}x^{p}}}{\sqrt{\pi } (2\pi )^{{\frac{1-a}{2}}}a ^{p+\frac{b}{2}}} \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \int _{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+2j+b-1-3a}{2a}} \cosh \biggl( \frac{\alpha xt}{ a^{a/2}} \biggr)\,dt. \end{aligned}$$
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, \alpha^{2}}(x) &= \frac{-2^{1-p}x^{p}}{\sqrt{ \pi } (2\pi )^{{\frac{1-a}{2}}}a^{\frac{2p+b}{2}}} \prod_{j=1}^{a}\frac{1}{\Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ & \quad \times \int_{\frac{\pi }{2}}^{0} \bigl(1-\cos^{2} \theta \bigr)^{ \frac{2p+2j+b-1-3a}{2a}} \cos \biggl( \frac{\alpha x \cos \theta }{ a ^{\frac{a}{2}}} \biggr) \sin \theta \,d\theta \\ &=\frac{2^{1-p}x^{p}}{\sqrt{\pi } (2\pi )^{{\frac{1-a}{2}}}a^{ \frac{2p+b}{2}}} \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ & \quad {}\times \int_{0}^{\frac{\pi }{2}} (\sin \theta )^{ \frac{2p+2j+b-1-3a}{a}+1} \cos \biggl( \frac{\alpha x \cos (\theta )}{ a ^{\frac{a}{2}}} \biggr)\,d\theta \\ & =\frac{2^{1-p}x^{p}}{\sqrt{\pi } (2\pi )^{{\frac{1-a}{2}}}a^{ \frac{2p+b}{2}}} \prod_{j=1}^{a} \frac{1}{ \Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ &\quad {}\times \int _{0}^{\frac{\pi }{2}} (\sin \theta )^{\frac{2p+2j+b-1-2a}{a}} \cos \biggl( \frac{\alpha x \cos \theta }{ a^{\frac{a}{2}}} \biggr)\,d\theta , \end{aligned}$$
$$\begin{aligned} {}_{a}\mathtt{B}_{b, p, -\alpha^{2}}(x) & =\frac{2^{1-p}x^{p}}{\sqrt{ \pi } (2\pi )^{{\frac{1-a}{2}}}a^{\frac{2p+b}{2}}} \prod _{j=1}^{a}\frac{1}{\Gamma { ( \frac{2p+2j+b-1-a}{2a} ) }} \\ &\quad {}\times \int_{0}^{ \frac{\pi }{2}} (\sin \theta )^{\frac{2p+2j+b-1-2a}{a}} \cosh \biggl( \frac{ \alpha x \cos \theta }{ a^{\frac{a}{2}}} \biggr)\,d\theta . \end{aligned}$$
The particular case \(a=b=1=\alpha \) in the above representations gives respectively the integral representation for the classical Bessel and modified Bessel functions of order p:
and
These integrations for \(J_{p}\) and \(I_{p}\) can also be found in [16, 9.1.20, p. 360] and [16, 9.6.18, p. 376].
$$\begin{aligned} {}_{1}\mathtt{B}_{1, p, 1}(x)=J_{p}(x) &= \frac{2^{1-p}x^{p}}{\sqrt{ \pi } \Gamma { ( p +\frac{1}{2} ) }} \int_{0}^{\frac{ \pi }{2}} (\sin \theta )^{2p} \cos ( x \cos \theta )\,d\theta \\ &=\frac{ ( \frac{x}{2} ) ^{p}}{\sqrt{\pi } \Gamma { ( p +\frac{1}{2} ) }} \int_{0}^{\pi } (\sin \theta )^{2p} \cos ( x \cos \theta )\,d\theta \\ &=\frac{2^{1-p}x^{p}}{\sqrt{\pi } \Gamma { ( p + \frac{1}{2} ) }} \int_{0}^{1} \bigl(1-t^{2} \bigr)^{p-\frac{1}{2}} \cos ( x t )\,dt, \end{aligned}$$
$$\begin{aligned} {}_{1}\mathtt{B}_{1, p, -1}(x)=I_{p}(x) & = \frac{ ( \frac{x}{2} ) ^{p}}{\sqrt{\pi } \Gamma { ( p +\frac{1}{2} ) }} \int_{0}^{\pi } (\sin \theta )^{2p} \cosh ( x \cos \theta )\,d\theta \\ & =\frac{2^{1-p}x^{p}}{\sqrt{\pi } \Gamma { ( p + \frac{1}{2} ) }} \int_{0}^{1} \bigl(1-t^{2} \bigr)^{p-\frac{1}{2}} \cosh ( x t )\,dt. \end{aligned}$$
3 Monotonicity and consequences
Investigations into the monotonicity properties of the generalized function \({}_{a}\mathtt{B}_{b,p,c}\) hinges on the following result of Biernacki and Krzyż [19].
Lemma 1
([19])
Suppose
\(f(x)=\sum_{k=0}^{\infty }a_{k} x^{k}\)
and
\(g(x)=\sum_{k=0}^{\infty }b_{k} x^{k}\), where
\(a_{k} \in \mathbb{R}\)
and
\(b_{k} > 0\)
for all k. Further suppose that both series converge on
\(\vert x \vert < r\). If the sequence
\(\{a_{k}/b_{k}\}_{k\geq 0}\)
is increasing (or decreasing), then the function
\(x \mapsto f(x)/g(x)\)
is also increasing (or decreasing) on
\((0,r)\).
Evidently, the above lemma also holds true when both f and g are even functions, or both odd.
Theorem 2
Let
\(c \leq 0\).
(a)
If
\(q \geq p > -(b+1)/2\)
and
\(a \leq d\), then
\(x \mapsto ( 2^{p}x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x) ) / ( 2^{q}x^{-q} {}_{d}\mathtt{B}_{b,q,c}(x) ) \)
is increasing on
\((0, \infty )\).
(b)
The function
\(p\mapsto {}_{a}\mathtt{B}_{b,p+a,c}(x)/{} _{a}\mathtt{B}_{b,p,c}(x)\)
is decreasing on
\((-(b+1)/2, \infty )\)
for each fixed
\(x>0\).
(c)
The function
\(x\mapsto x{}_{a}\mathtt{B}_{b,p,c}'(x)/{} _{a}\mathtt{B}_{b,p,c}(x)\)
is increasing on
\((0, \infty )\)
for each fixed
\(p > -{(b+1)}/{2}\).
Proof
(a) From (1) it is evident that
where
Write \(w_{k}= \alpha_{k,p,a}/\alpha_{k,q,d}\); since \(d \geq a\) and \(q \geq p\), it follows that
The result now readily follows from Lemma 1.
$$\begin{aligned} \frac{ x^{q-p}{}_{a}\mathtt{B}_{b,p,c}(x)}{2^{q-p}{}_{b}B_{b,q,c}(x)}=\frac{ \sum_{k=0}^{\infty }\alpha_{k, p,a} ( \frac{x}{2} ) ^{2k}}{ \sum_{k=0}^{\infty }\alpha_{k, q,d} ( \frac{x}{2} ) ^{2k}}, \end{aligned}$$
$$ \alpha_{k, p,a}=\frac{(-c)^{k}}{k! \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \quad \text{and} \quad \alpha_{k, q,d}=\frac{(-c)^{k}}{k! \Gamma { ( dk+q+\frac{b+1}{2} ) }}. $$
$$\begin{aligned} \frac{w_{k+1}}{w_{k}} &=\frac{ \Gamma { ( ak+p+ \frac{b+1}{2} ) } \Gamma { ( dk+d+q+\frac{b+1}{2} ) }}{ \Gamma { ( dk+q+\frac{b+1}{2} ) } \Gamma { ( ak+a+p+\frac{b+1}{2} ) }}=\frac{ ( d k +q+\frac{b+1}{2} ) _{d}}{ ( a k+p+\frac{b+1}{2} ) _{a}} \geq 1. \end{aligned}$$
(b) Let \(q \geq p>-(b+1)/2\). It follows from part (a) that
on \((0,\infty )\). Thus
It now follows from (5) that
whence \({}_{a} B_{b,p+a,c}/{}_{a} B_{b,p,c}\) is decreasing for \(p>-(b+1)/2\).
$$\begin{aligned} \frac{d}{dx} \biggl( \frac{2^{p} x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x)}{ {2^{q} x^{-q}{}_{a}\mathtt{B}_{b,q,c}(x)}} \biggr) \geq 0 \end{aligned}$$
$$\begin{aligned} \bigl( x^{-p}{}_{a}\mathtt{B}_{b,p,c}(x) \bigr) ' \bigl( x^{-q}{}_{a} \mathtt{B}_{b,q,c}(x) \bigr) - \bigl( x^{-p}{}_{a} \mathtt{B}_{b,p,c}(x) \bigr) \bigl( x^{-q}{}_{a} \mathtt{B}_{b,q,c}(x) \bigr) '\geq 0. \end{aligned}$$
$$\begin{aligned} (-c)\;x^{-p-q} \biggl( \frac{x}{2} \biggr) ^{1-a} \bigl( {}_{a}\mathtt{B} _{b,p+a,c}(x)\; {}_{a} \mathtt{B}_{b,q,c}(x)-{}_{a}\mathtt{B}_{b,p,c}(x) \;{}_{a}\mathtt{B}_{b,q+a,c}(x) \bigr) \geq 0, \end{aligned}$$
(c) Let \(\beta_{k,p,a}:=(2k+p)\alpha_{k, p,a}\). Then the quotient \(x {}_{a}\mathtt{B}_{b,p,c}'/{}_{a}\mathtt{B}_{b,p,c}\) can be written as
Clearly, the sequence \(\{\beta_{k,p,a}/\alpha_{k,p,a}\}_{k \geq 0}=\{2k+p \}_{k\geq 0}\) is increasing, and hence Lemma 1 shows that the function \(x \mapsto x {}_{a}\mathtt{B}_{b,p,c}'/{}_{a}\mathtt{B} _{b,p,c}\) is increasing on \(( 0, \infty ) \). □
$$\begin{aligned} \frac{x {}_{a}B'_{b,p,c}(x)}{{}_{a}\mathtt{B}_{b,p,c}(x)}= \frac{ \sum_{k=0}^{\infty }{\beta_{k,p,a} ( \frac{x}{2} ) ^{2k}}}{ \sum_{k=0}^{\infty }{\alpha_{k,p,a} ( \frac{x}{2} ) ^{2k}}}. \end{aligned}$$
Next consider the normalized function
Also let \({}_{1}\Phi_{1}\) be the confluent hypergeometric function
The next result discusses the monotonicity property of rational functions involving \({}_{a}\mathcal{B}_{b,p,c}\).
$$\begin{aligned} {}_{a}\mathcal{B}_{b,p,c}(x)&=2^{p}x^{-p} \Gamma { \biggl( p+\frac{b+1}{2} \biggr) } {}_{a}\mathtt{B}_{b, p,c}(x) \\ &= \Gamma { \biggl( p+\frac{b+1}{2} \biggr) }\sum_{k=0}^{\infty }\frac{(-c)^{k}}{k! \Gamma { ( ak+p+\frac{b+1}{2} ) }} \biggl( \frac{x}{2} \biggr) ^{2k}. \end{aligned}$$
(14)
$$ {}_{1}\Phi_{1}(\alpha ; \beta ; x)= \sum _{k=0}^{\infty }\frac{ ( \alpha )_{k} x^{k}}{(\beta )_{k} k!}. $$
Theorem 3
Let
\(c \leq 0\).
(a)
If
\(\alpha \geq \beta >0\), then the function
\(x \mapsto {}_{a}\mathcal{B}_{b,p,c}(x)/{}_{1}\Phi_{1}(\alpha ; \beta ; -c x^{2}/4)\)
is decreasing on
\(\mathbb{R}\)
for each fixed
\(p > -{(b+1)}/ {2}\).
(b)
If
\(0< \beta \leq (2p+b+1)/(2a)\), then the function
\(x \mapsto {}_{a}\mathcal{B}_{b,p,c}(x)/F_{a}(\beta ; -c x^{2}/4)\)
is decreasing on
\(\mathbb{R}\)
for each fixed
\(p > -{(b+1)}/{2}\), where
$$ F_{a}(\beta , x):= {}_{0}F_{a} \biggl( -; \beta , \beta +\frac{1}{a}, \ldots , \beta +\frac{a-1}{a}; x \biggr) . $$
Proof
(a) It follows from (14) that
with
Set \(w_{k}:= \delta_{1}(k)/\delta_{2}(k)\). Since \(\alpha \geq \beta \),
Thus \(\{w_{k}\}_{k}\) is decreasing, and the result follows from Lemma 1.
$$ \frac{{}_{a}\mathcal{B}_{b, p, c}(x)}{{}_{1}\Phi_{1} ( \alpha ; \beta ; -c\frac{x^{2}}{4} ) }=\frac{\sum_{k=0}^{\infty }\delta _{1}(k) ( \frac{x}{2} ) ^{2k}}{\sum_{k=0}^{\infty }\delta _{2}(k) ( \frac{x}{2} ) ^{2k}} $$
$$ \delta_{1}(k)=\frac{(-c)^{k}\Gamma ( p+\frac{b+1}{2} ) }{k! \Gamma ( ak+p+\frac{b+1}{2} ) } \quad \text{and} \quad \delta _{2}(k)=\frac{(-c)^{k} (\alpha )_{k}}{k! (\beta )_{k}}. $$
$$ \frac{w_{k+1}}{w_{k}}= \frac{(\beta +k)\Gamma ( ak+p+ \frac{b+1}{2} ) }{(\alpha +k)\Gamma ( ak+a+p+\frac{b+1}{2} ) } \leq 1. $$
(b) From Proposition 1, a representation of \({}_{a} \mathcal{B}_{b, p, c}\) by the generalized hypergeometric function is
where
Let
Then
With \(\tau_{k}:= \sigma_{1}(k)/\sigma_{2}(k)\), a computation shows that
Now
for each fixed j, \(1 \leq j \leq a\), provided \(0 < \beta \leq (2p+b+1)/(2a)\). Hence \(\tau_{k+1} \leq \tau_{k}\), and the result follows from Lemma 1. □
$$ {}_{a}\mathcal{B}_{b, p, c}(x)=\sum _{k=0}^{\infty }\sigma_{1}(k) \biggl( \frac{x}{2} \biggr) ^{2k}, $$
$$ \sigma_{1}(k):=\frac{(-c)^{k} }{a^{ak}k! \prod_{j=1}^{a} ( \frac{2p+b+2j-1}{2a} ) _{k}}. $$
$$ \sigma_{2}(k):=\frac{(-c)^{k} }{k! \prod_{j=1}^{a} ( \beta + \frac{j-1}{a} ) _{k}}. $$
$$ \frac{{}_{a}\mathcal{B}_{b, p, c}(x)}{F_{a} ( \beta ,-c\frac{x ^{2}}{4} ) } = \frac{\sum_{k=0}^{\infty }\sigma_{1}(k) ( \frac{x}{2} ) ^{2k}}{\sum_{k=0}^{\infty }\sigma_{2}(k) ( \frac{x}{2} ) ^{2k}}. $$
$$ \frac{\tau_{k+1}}{\tau_{k}}= \frac{1}{a^{a}} \prod_{j=1}^{a} \frac{ ( \frac{2p+b+2j-1}{2a} ) _{k}}{ ( \frac{2p+b+2j-1}{2a} ) _{k+1}} \frac{ ( \beta +\frac{j-1}{a} ) _{k+1}}{ ( \beta + \frac{j-1}{a} ) _{k}} =\frac{1}{a^{a}} \prod _{j=1}^{a} \frac{ \beta +\frac{j-1}{a}+k}{\frac{2p+b+2j-1}{2a}+k}. $$
$$ \frac{2p+b+2j-1}{2a} \geq \beta +\frac{j-1}{a} $$
Another function of interest is that given by
Note that \({}_{a}\mathcal{B}^{1}_{b,p,c}(x)={}_{a}\mathcal{B}_{b,p,c}( \sqrt{x})\), where \({}_{a}\mathcal{B}_{b,p,c}\) is given by (14). The following result by Karp and Sitnik [20, Theorem 1] is required to deduce the log-concavity of \({}_{a}\mathcal{B}^{d}_{b,p, c}\) in terms of the parameter d.
$$ {}_{a}\mathcal{B}^{d}_{b,p, c}(x):= \sum_{k=0}^{\infty }\frac{(-c/4)^{k}\Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(d)_{k}}{k!}x^{k} =\sum_{k=0}^{\infty } \frac{(-c/4)^{k}(d)_{k}}{\Gamma{ ( k+1 ) } ( p+\frac{b+1}{2} ) _{ak} k!}x ^{k}. $$
(15)
Lemma 2
([20])
Let
where
\(f_{k}>0\) (and is independent of
d). Suppose
\(e>d>0\), \(\delta >0\). Then the function
has positive power series coefficients
\(\phi_{m} >0\)
so that
\(d \mapsto f(d,x)\)
is strictly log-concave for
\(x>0\)
whenever the sequence
\(\{ f_{k}/f_{k-1} \} \)
is decreasing. On the other hand, \(\phi_{d, e, \delta }(x)\)
has negative power series coefficients
\(\phi_{m} <0\)
so that
\(d \mapsto f(d,x)\)
is strictly log-convex for
\(x>0\)
whenever the sequence
\(\{ f_{k}/f_{k-1} \} \)
is increasing.
$$ f(d,x)=\sum_{k=0}^{\infty }f_{k} \frac{(d)_{k}}{k!} x^{k}, $$
$$ \phi_{d, e, \delta }(x)=f(d+\delta ,x)f(e,x)-f(e+\delta ,x)f(d,x)= \sum _{m=2}^{\infty }\phi_{m} x^{m} $$
Theorem 4
Let
\(c \leq 0\)
and
\(d>0\).
(a)
The function
given by (15) is decreasing and log-convex on
\((-(b+1)/2, \infty )\)
for each fixed
\(x>0\)
and
\(d>0\).
$$ p \mapsto {}_{a}\mathcal{B}^{d}_{b,p, c}(x) $$
(b)
The function
is increasing on
\((-(b+1)/2, \infty )\), that is, for
\(q \geq p >-(b+1)/2\), the inequality
holds for each fixed
\(x>0\)
and
\(d>0\).
$$ p \mapsto {}_{a}\mathcal{B}_{b,p+1, c}^{d}(x)/{}_{a} \mathcal{B}_{b,p, c}^{d}(x) $$
$$ {}_{a}\mathcal{B}_{b, q+1, c}^{d}(x){}_{a} \mathcal{B}_{b,p, c}^{d}(x) \geq {}_{a} \mathcal{B}_{b, q, c}^{d}(x){}_{a}\mathcal{B}_{b,p+1, c} ^{d}(x) $$
(16)
(c)
The function
\(d \mapsto {}_{a}\mathcal{B}^{d}_{b,p, c}(x)\)
is log-concave on
\((0, \infty )\)
for each fixed
\(x>0\)
and
\(p >(2a-b-1)/2\).
Proof
(a) Let \(q\geq p > -(b+1)/2\). Then \((q+(b+1)/2)_{ak} > (p+(b+1)/2)_{ak}\) for all \(k \in \{0, 1, 2, \ldots \}\). Thus
Since
we deduce that
for each fixed \(x >0\) and \(d>0\). Therefore \(p \mapsto {}_{a} \mathcal{B}_{b,p, c}^{d}\) is decreasing for \(p > -(b+1)/2\).
$$\begin{aligned} \gamma_{k}(q, d):=\frac{(-c)^{k}(d)_{k}}{4^{k} \Gamma { ( k+1 ) } ( q+\frac{b+1}{2} ) _{ak}k!} \leq \frac{(-c)^{k} (d)_{k}}{4^{k}\Gamma { ( k+1 ) } ( p+\frac{b+1}{2} ) _{ak}k!}:= \gamma_{k}(p, d). \end{aligned}$$
$$ {}_{a}\mathcal{B}_{b,q, c}^{d}(x)=\sum _{k=0}^{\infty } \gamma_{k}(q, d) x^{k}, $$
$$\begin{aligned} {}_{a}\mathcal{B}_{b,q, c}^{d}(x) \leq {}_{a}\mathcal{B}_{b,p, c}^{d}(x) \end{aligned}$$
To show log-convexity of \({}_{a}\mathcal{B}_{b,p, c}^{d}\), it suffices to show that \(p \mapsto \gamma_{k}(p, d)\) is log-convex for all \(k \in \{0,1, 2, 3, \ldots \}\) and fixed \(d>0\). Then the result follows from the fact that sums of log-convex functions are also log-convex.
Let Ψ be the digamma function given by \(\Psi (p)= \Gamma '{(p)}/ \Gamma {(p)}\). Then evidently
Note that [16, p. 260] \(\Psi '\) has the explicit form
This implies that
for all \(k \in \{0, 1, 2, \ldots \}\) and \(p > -(b+1)/2\). Thus \(p \mapsto \gamma_{k}( p, d)\) is log-convex on \((-(b+1)/2, \infty )\), and consequently \({}_{a}\mathcal{B}_{b,p, c}^{d}\) is log-convex for each fixed \(x>0\).
$$\begin{aligned} \frac{\partial^{2}}{\partial {p^{2}}} \bigl(\log \bigl(\gamma_{k}( p,d) \bigr) \bigr)= \Psi ' \biggl( p+\frac{b+1}{2} \biggr) -\Psi ' \biggl( ak+p+\frac{b+1}{2} \biggr) . \end{aligned}$$
$$\begin{aligned} \Psi '(t)=\sum_{n=0}^{\infty } \frac{1}{(t+n)^{2}}, \quad t \in \mathbb{R}\setminus \{0,-1,-2,\ldots \}. \end{aligned}$$
$$\begin{aligned} \frac{\partial^{2}}{\partial {p^{2}}} \bigl(\log \bigl(\gamma_{k}(p, d) \bigr) \bigr)= \sum_{n=0}^{\infty }\frac{ak(ak+2p+(b+1)+2n)}{ ( p+\frac{b+1}{2}+n ) ^{2} ( ak+p+\frac{b+1}{2}+n ) ^{2}} \geq 0 \end{aligned}$$
(b) It is clear that (16) is equivalent to showing
which holds whenever
for all \(i, j \in \mathbb{N}\).
$$\begin{aligned} \Biggl( \sum_{k=0}^{\infty } \gamma_{k}(q+1, d)x^{k} \Biggr) \Biggl( \sum _{k=0}^{\infty }\gamma_{k}(p, d)x^{k} \Biggr) \geq \Biggl( \sum_{k=0} ^{\infty } \gamma_{k}(q, d)x^{k} \Biggr) \Biggl( \sum _{k=0}^{\infty } \gamma_{k}(p+1, d)x^{k} \Biggr) , \end{aligned}$$
$$\begin{aligned} \gamma_{i}(q+1, d)\gamma_{j}(p, d)+ \gamma_{j}(q+1, d) \gamma_{i}(p, d) \geq \gamma_{j}(q, d)\gamma_{i}(p+1, d)+\gamma_{i}(q, d) \gamma_{j}(p+1, d) \end{aligned}$$
(17)
Let
and
Then
Similarly,
$$ \lambda_{1}= \Gamma { \biggl( ai+q+\frac{b+3}{2} \biggr) } \Gamma { \biggl( aj+p+\frac{b+3}{2} \biggr) } $$
$$ \lambda_{2}= \Gamma { \biggl( aj+q+\frac{b+3}{2}\biggr) } \Gamma { \biggl( ai+p+\frac{b+3}{2} \biggr) }. $$
$$\begin{aligned} &\gamma_{i}(q+1, d)\gamma_{j}(p, d)+\gamma_{j}(q+1, d) \gamma_{i}(p, d) \\ &\quad = \frac{(-c/4)^{i+j} \Gamma ( p+\frac{b+1}{2} ) \mathrm{ \Gamma } ( q+\frac{b+3}{2} ) (d)_{i} (d)_{j}}{ \Gamma ( i+1 ) \Gamma ( j+1 ) i! j!} \biggl[ \frac{aj+p+ \frac{b+1}{2}}{\lambda_{1}} +\frac{ai+q+\frac{b+1}{2}}{\lambda_{2}} \biggr] . \end{aligned}$$
(18)
$$\begin{aligned} &\gamma_{j}(q, d)\gamma_{i}(p+1, d)+\gamma_{i}(q, d) \gamma_{j}(p+1, d) \\ &\quad = \frac{ ( -c/4 ) ^{i+j} \Gamma ( q+ \frac{b+1}{2} ) \Gamma ( p+\frac{b+3}{2} ) (d)_{i} (d)_{j}}{ \Gamma ( i+1 ) \Gamma ( j+1 ) i! j!} \biggl[ \frac{ai+q+\frac{b+1}{2}}{ \lambda_{1}} +\frac{aj+p+\frac{b+1}{2}}{\lambda_{2}} \biggr] . \end{aligned}$$
(19)
Next, for \(i \geq j\), relations (18) and (19) show that inequality (17) is equivalent to
Since \(q\geq p\), this can be further simplified to showing
The latter inequality clearly holds true whenever \(\lambda_{1} \geq \lambda_{2}\).
$$\begin{aligned} & \biggl( q+\frac{b+1}{2} \biggr) \biggl[ \biggl( aj+p+\frac{b+1}{2} \biggr) \lambda_{2} + \biggl( ai+q+\frac{b+1}{2} \biggr) \lambda_{1} \biggr] \\ &\quad \geq \biggl( p+\frac{b+1}{2} \biggr) \biggl[ \biggl( aj+p+ \frac{b+1}{2} \biggr) \lambda_{1}+ \biggl( ai+q+\frac{b+1}{2} \biggr) \lambda_{2} \biggr] . \end{aligned}$$
$$\begin{aligned} a ( \lambda_{1}-\lambda_{2} ) (i-j) \biggl( p+\frac{b+1}{2} \biggr) \geq 0. \end{aligned}$$
To see that this is indeed the case, for \(q \geq p\), let
Since \(x \mapsto \Gamma {(ax+y)}\) is log-convex, it follows that \(\phi '(x) \geq 0\). Thus \(\phi (i) \geq \phi (j)\) for \(i \geq j\), and consequently \(\lambda_{1} \geq \lambda_{2}\). This validates inequality (16).
$$\begin{aligned} \phi (x):= \frac{ \Gamma { ( ax+ q +\frac{b+3}{2} ) }}{\mathrm{ \Gamma }{ ( ax+ p +\frac{b+3}{2} ) }}. \end{aligned}$$
(c) To apply Lemma 2, let
We shall show that the sequence \(b_{k} = \{ f_{k}/f_{k-1} \} \) is decreasing. A calculation gives
and so we need to show that the function \(\xi :(0,\infty )\to \mathbb{R}\) given by
is decreasing for \(p>(2a-b-1)/2\). Logarithmic differentiation gives
Since the digamma function is known to be increasing on \((0, \infty )\) for \(p>(2a-b-1)/2\) and \(x>0\), it follows that
Thus ξ is indeed decreasing, and Lemma 2 shows that the function
is log-concave. □
$$ f_{k}:=\frac{(-c/4)^{k} {\Gamma ( p+\frac{b+1}{2} ) }}{ \Gamma (k+1) \Gamma ( ak+p+\frac{b+1}{2} ) }. $$
$$ b_{k}= -\frac{c}{4} \frac{\Gamma ( ak+p+\frac{b+1}{2}-a ) }{k \; \Gamma ( ak+p+\frac{b+1}{2} ) }, $$
$$ \xi (x)= \frac{\Gamma ( ax+p+\frac{b+1}{2}-a ) }{x\; \Gamma ( ax+p+\frac{b+1}{2} ) } $$
$$\begin{aligned} \frac{\xi '(x)}{\xi (x)}= a \Psi \biggl( ax+p+\frac{b+1}{2}-a \biggr) - a \Psi \biggl( ax+p+\frac{b+1}{2} \biggr) -\frac{1}{x}. \end{aligned}$$
$$\begin{aligned} \frac{\xi '(x)}{\xi (x)} < 0. \end{aligned}$$
$$ d\mapsto {}_{a}\mathcal{B}_{b,p, c}^{d}(x) =\sum _{k=0}^{\infty } f _{k} \frac{(d)_{k}}{k!} x^{k} $$
The results of parts (a) and (b) in Theorem 4 in the case \(d=1\) were also obtained by Baricz [1, Theorem 3, Theorem 4].
Remark 5
Theorem 4 has interesting consequences, among which is the Tur\(\acute{\text{a}}\)n-type inequality for the function \({}_{a}\mathcal{B}_{b,p, c}^{d}\) given by (15). From the definition of log-convexity, it follows from Theorem 4(a) that
where \(\alpha \in [0,1]\), \(p_{1}, p_{2} > -(b+1)/2\), and \(x >0\). Choosing \(\alpha =1/2\), \(p_{1}=p-\nu \) and \(p_{2}=p+\nu \), the above inequality yields
$$\begin{aligned} {}_{a}\mathcal{B}_{b, \alpha p_{1}+(1-\alpha )p_{2}, c}^{d}(x) \leq \bigl( {}_{a}\mathcal{B}_{b,p_{1}, c}^{d}(x) \bigr) ^{\alpha } \bigl( {} _{a}\mathcal{B}_{b,p_{2}, c}^{d}(x) \bigr) ^{1-\alpha }, \end{aligned}$$
$$\begin{aligned} \bigl({}_{a}\mathcal{B}_{b,p,c}^{d}(x) \bigr)^{2} - {}_{a}\mathcal{B} _{b,p+\nu , c}^{d}(x) {}_{a}\mathcal{B}_{b,p-\nu }^{d}(x)\leq 0. \end{aligned}$$
(20)
On the other hand, the log-concavity of \(d \mapsto {}_{a}\mathcal{B} _{b,p, c}^{d}\) implies that
for \(t \in [0, 1]\), \(d_{2} > d_{1} >0\), \(p> (2a-b-1)/2\) and \(x>0\). Choosing \(t=1/2\), \(d_{1}=d-\mu \), and \(d_{2}=d+\mu \), \(\mu \in \mathbb{R}\), the inequality reduces to
Thus (20) and (21) lead to direct and reverse Tur\(\acute{\text{a}}\)n-type inequalities
$$\begin{aligned} {}_{a}\mathcal{B}_{b, p, c}^{td_{1}+(1-t)d_{2}}(x) \geq \bigl( {}_{a} \mathcal{B}_{b,p, c}^{d_{1}}(x) \bigr) ^{t} \bigl( {}_{a}\mathcal{B} _{b,p, c}^{d_{2}}(x) \bigr) ^{1-t} \end{aligned}$$
$$\begin{aligned} \bigl( {}_{a}\mathcal{B}_{b, p, c}^{d}(x) \bigr) ^{2} \geq {}_{a} \mathcal{B}_{b,p, c}^{d+\mu }(x) {}_{a}\mathcal{B}_{b,p, c}^{d-\mu }(x). \end{aligned}$$
(21)
$$\begin{aligned} {}_{a}\mathcal{B}_{b,p, c}^{d+\mu }(x) {}_{a} \mathcal{B}_{b,p, c}^{d- \mu }(x)\leq \bigl( {}_{a} \mathcal{B}_{b, p, c}^{d}(x) \bigr) ^{2} \leq {}_{a}\mathcal{B}_{b,p+\nu , c}^{d}(x) {}_{a} \mathcal{B}_{b,p- \nu }^{d}(x). \end{aligned}$$
Remark 6
For \(d=2\), it follows from (15) that
where \({}_{a}\mathcal{B}_{b,p, c}\) is given by (14). With \(d=1\) and \(\mu =1\) in (21), then \({}_{a}\mathcal{B} _{b,p, c}^{1}(x)={}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \). Thus (22) shows that
$$\begin{aligned} {}_{a}\mathcal{B}^{2}_{b,p, c}(x) &=\sum_{k=0}^{\infty }\frac{(-c/4)^{k}\mathrm{ \Gamma }{ ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(2)_{k}}{k!}x^{k} \\ &=\sum_{k=0}^{\infty }\frac{(-c/4)^{k} \Gamma { ( p+ \frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \mathrm{ \Gamma }{ ( ak+p+\frac{b+1}{2} ) }} \frac{(k+1)!}{k!}x^{k} \\ &=x \sum_{k=1}^{\infty } \frac{(-c/4)^{k} \Gamma { ( p+ \frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \mathrm{ \Gamma }{ ( ak+p+\frac{b+1}{2} ) }} kx^{k-1} \\ &\quad {}+ \sum_{k=0}^{ \infty } \frac{(-c/4)^{k} \Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+\frac{b+1}{2} ) }} x^{k} \\ &= x \frac{d}{dx} \bigl( {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) + {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) , \end{aligned}$$
(22)
$$\begin{aligned} \bigl( {}_{a}\mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) ^{2} \geq x \frac{d}{dx} \bigl( {}_{a} \mathcal{B}_{b,p, c} ( \sqrt{x} ) \bigr) + {}_{a} \mathcal{B}_{b,p, c} ( \sqrt{x} ) . \end{aligned}$$
Remark 7
Inequality (16) leads to a generalization of the Turán-type inequality
$$ \bigl( {}_{a}\mathcal{B}_{b, p+1,c}^{d}(x) \bigr) ^{2}\leq {}_{a} \mathcal{B}_{b, p,c}^{d}(x) \;{}_{a}\mathcal{B}_{b, p+2,c}^{d}(x). $$
Inequality (16) also yields
Thus
$$\begin{aligned} \frac{{}_{a}\mathcal{B}_{b, q,c}^{d}(x)}{{}_{a}\mathcal{B}_{b,p,c} ^{d}(x)}\leq \frac{{}_{a}\mathcal{B}_{b, q+1,c}^{d}(x)}{{}_{a} \mathcal{B}_{b,p+1,c}^{d}(x)} \leq \frac{{}_{a}\mathcal{B}_{b, q+2,c} ^{d}(x)}{{}_{a}\mathcal{B}_{b,p+2,c}^{d}(x)}\leq \cdots \leq \frac{ {}_{a}\mathcal{B}_{b, q+a,c}^{d}(x)}{{}_{a}\mathcal{B}_{b,p+a,c}^{d}(x)}. \end{aligned}$$
$$ \frac{{}_{a}\mathcal{B}_{b, q,c}^{d}(x)}{{}_{a}\mathcal{B}_{b, p,c} ^{d}(x)}\leq \frac{{}_{a}\mathcal{B}_{b, q+a,c}^{d}(x)}{{}_{a} \mathcal{B}_{b, p+a,c}^{d}(x)}. $$
The next result gives a dominant function for \({}_{a}\mathtt{B}_{b,p, -\alpha^{2}}\).
Theorem 5
Let
\(p\geq -(b+1)/2\)
and
\(x \geq 0\). Then
$$ {}_{a}\mathtt{B}_{b,p, -\alpha^{2}}(x) \leq \frac{1}{\Gamma ( p+ \frac{b+1}{2} ) } \biggl( \frac{x}{2} \biggr) ^{p}\exp \biggl( \frac{ \alpha^{2} x^{2}}{4 ( p+\frac{b+1}{2} ) ^{a}} \biggr) . $$
Proof
Clearly the estimate trivially holds for \(x=0\). Let \(x>0\). It is readily established by mathematical induction that \(\Gamma (m+x) \geq x^{m} \Gamma (x)\) for \(m \in \mathbb{N}\) and \(x\geq 0\). Then
and thus
□
$$ \Gamma \biggl( a k + p+\frac{b+1}{2} \biggr) \geq \biggl( p+ \frac{b+1}{2} \biggr) ^{a k} \Gamma \biggl( p+\frac{b+1}{2} \biggr) , $$
$$\begin{aligned} {}_{a}\mathtt{B}_{b,p, -\alpha^{2}}(x) &\leq \frac{1}{\Gamma ( p+ \frac{b+1}{2} ) }\sum _{k=0}^{\infty }\frac{\alpha^{2k}}{ ( p+ \frac{b+1}{2} ) ^{ak} k!} \biggl( \frac{x}{2} \biggr) ^{2k+p} \\ &=\frac{1}{\Gamma ( p+\frac{b+1}{2} ) } \biggl( \frac{x}{2} \biggr) ^{p}\exp \biggl( \frac{\alpha^{2} x^{2}}{4 ( p+\frac{b+1}{2} ) } \biggr) ^{a} . \end{aligned}$$
For \(\alpha =\pm 1\), \(b=a=1\), Theorem 5 leads to a dominant for the modified Bessel function
obtained by Baricz in [21].
$$ I_{p}(x) \leq \frac{1}{\Gamma ( p+1 ) } \biggl( \frac{x}{2} \biggr) ^{p}e^{\frac{ x^{2}}{4 ( p+1 ) }} $$
The final result uses the Chebyshev integral inequality [22, p. 40]: Suppose f and g are two integrable functions monotonic in the same sense (either both decreasing or both increasing). Let \(q: (a, b) \to \mathbb{R}\) be a positive integrable function. Then
The inequality in (23) is reversed if f and g are monotonic of the opposite sense.
$$\begin{aligned} \biggl( \int_{a}^{b} q(t) f(t)\,dt \biggr) \biggl( \int_{a}^{b} q(t) g(t)\,dt \biggr) \leq \biggl( \int_{a}^{b} q(t)\,dt \biggr) \biggl( \int_{a}^{b} q(t) f(t) g(t)\,dt \biggr) . \end{aligned}$$
(23)
Theorem 6
Let
\(p > -(b-1)/2\), \(\alpha \in \mathbb{R}\backslash \{0\}\), and
\(x \in (0, \pi /\vert \alpha \vert )\). Then
$$\begin{aligned}& {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) \geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma (\frac{2p+b-1}{4})\Gamma ( \frac{2p+b+1}{4})}\; \biggl( \mathtt{J}_{\frac{2p+b-2}{4}} \biggl( \frac{ \alpha x}{2} \biggr) \biggr) ^{2}, \\& {}_{2}\mathtt{B}_{b, p, -\alpha^{2}}(x) \geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma (\frac{2p+b-1}{4})\Gamma ( \frac{2p+b+1}{4})}\; \biggl( \mathtt{I}_{\frac{2p+b-2}{4}} \biggl( \frac{ \alpha x}{2} \biggr) \biggr) ^{2}. \end{aligned}$$
Proof
Putting \(a=2\) in Remark 4, the integral form for \({}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x)\) is
To establish the subordinant for \({}_{2}\mathtt{B}_{b, p, \alpha^{2}}\), let
Then
$$\begin{aligned} {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) & = \frac{x^{p}}{2^{2p+ \frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-5}{4}} \cos \biggl( \frac{\alpha xt}{2} \biggr)\,dt \biggr) \\ & \quad {} \times \biggl( \int_{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{ \alpha xt}{2} \biggr)\,dt \biggr) . \end{aligned}$$
$$ q(t) = \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{\alpha x t}{2} \biggr) ; \quad \quad f(t)= g(t) := \bigl(1-t^{2} \bigr)^{-\frac{1}{4}}, \quad 0 < t < 1. $$
$$\begin{aligned} \int_{0}^{1} q(t)f(t)\,dt= \int_{0}^{1} q(t)g(t)\,dt &= \int_{0}^{1} \bigl(1-t ^{2} \bigr)^{\frac{2p+b-4}{4}} \cos \biggl( \frac{\alpha x t}{2} \biggr)\,dt. \end{aligned}$$
It is known that for \(\operatorname{Re} \nu >-1/2\), the classical Bessel function \(\mathtt{J}_{\nu }\) has the integral representation
Replacing y by \((\alpha x)/2\) and ν by \((2p+b-2)/4\), we obtain
Since f and g both are increasing on \((0, 1)\), it is evident from (23) that
$$\begin{aligned} \mathtt{J}_{\nu }(y)= \frac{2^{1-\nu }y^{\nu }}{\sqrt{\pi } \mathrm{ \Gamma }{ ( \nu +\frac{1}{2} ) }} \int_{0}^{1} \bigl(1-t^{2} \bigr)^{ \nu -\frac{1}{2}} \cos ( y t )\,dt. \end{aligned}$$
$$\begin{aligned} \biggl( \int_{0}^{1}q(t) f(t)\,dt \biggr) \biggl( \int_{0}^{1}q(t) g(t)\,dt \biggr) = 2^{2p+b-4} \pi (\alpha x)^{1-p-\frac{b}{2}} \biggl( \Gamma \biggl( \frac{2p+b}{4} \biggr) \biggr) ^{2} \mathtt{J}^{2}_{ \frac{2p+b-2}{4}} \biggl( \frac{\alpha x}{2} \biggr) . \end{aligned}$$
$$\begin{aligned} {}_{2}\mathtt{B}_{b, p, \alpha^{2}}(x) & =\frac{x^{p}}{2^{2p+ \frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-5}{4}} \cos \biggl( \frac{\alpha xt}{2} \biggr)\,dt \biggr) \\ & \quad {} \times \biggl( \int_{0}^{1} \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cos \biggl( \frac{ \alpha xt}{2} \biggr)\,dt \biggr) \\ &=\frac{x^{p}}{2^{2p+\frac{b-3}{2}} { \Gamma { ( \frac{2p+b-1}{4} ) }} { \Gamma { ( \frac{2p+b+1}{4} ) }}} \biggl( \int_{0} ^{1}p(t)f(t) g(t)\,dt \biggr) \biggl( \int_{0}^{1}p(t)\,dt \biggr) \\ &\geq \frac{ \pi \alpha^{1-p-\frac{b}{2}}x^{1-\frac{b}{2}} ( \Gamma ( \frac{2p+b}{4} ) ) ^{2}}{2^{\frac{5-b}{2}} \Gamma ( \frac{2p+b-1}{4})\Gamma (\frac{2p+b+1}{4})}\; \mathtt{J}^{2}_{ \frac{2p+b-2}{4}} \biggl( \frac{\alpha x}{2} \biggr) . \end{aligned}$$
The subordinant for \({}_{2}\mathtt{B}_{b, p, - \alpha^{2}}\) is readily established in a similar manner by choosing
□
$$ q(t) = \bigl(1-t^{2} \bigr)^{\frac{2p+b-3}{4}} \cosh \biggl( \frac{\alpha x t}{2} \biggr) ; \quad \quad f(t)= g(t) := \bigl(1-t^{2} \bigr)^{-\frac{1}{4}}, \quad 0 < t < 1. $$
Remark 8
As a final application, choose \(b=3\) and \(\alpha =1\) in Theorem 6. Then
Remark 1 now shows that
$$ {}_{2}\mathtt{B}_{3, p, 1}(x) \geq \frac{ \pi x^{-\frac{1}{2}} ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4}) \Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{J}_{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$
$$ J_{\frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) J_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) \geq \frac{ \sqrt{\pi } ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{J} _{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$
Similarly,
and thus
$$ {}_{2}\mathtt{B}_{3, p, -1}(x) \geq \frac{ \pi x^{-\frac{1}{2}} ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma ( \frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{I}_{ \frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}, $$
$$ I_{\frac{p}{2}} \biggl( \frac{x}{ 2} \biggr) I_{\frac{p+1}{2}} \biggl( \frac{x}{ 2} \biggr) \geq \frac{ \sqrt{\pi } ( \Gamma ( \frac{2p+3}{4} ) ) ^{2}}{2 \Gamma (\frac{2p+2}{4})\Gamma (\frac{2p+4}{4})}\; \biggl( \mathtt{I} _{\frac{2p+1}{4}} \biggl( \frac{ x}{2} \biggr) \biggr) ^{2}. $$
4 Concluding remarks
This paper derived representation relations and functional inequalities for the extended Bessel function
in terms of the parameters a, b, and p. An important consequence is the \((a+1)\)-order differential equation satisfied by the function \({}_{a}\mathtt{B}_{b,p, c}\). Monotonicity properties of \({}_{a} \mathtt{B}_{b,p, c}\) are obtained for non-positive c. Connections with earlier works on the Bessel function and its generalizations are also made.
$$ {}_{a}\mathtt{B}_{b, p, c}(x):= \sum _{k=0}^{\infty }\frac{(-c)^{k}}{k! \Gamma { ( a k +p+\frac{b+1}{2} ) } } \biggl( \frac{x}{2} \biggr) ^{2k+p} $$
Additionally, this paper also studied log-concavity and log-convexity properties in terms of the parameters d and p for the closely related function
As a consequence, direct and reverse Turán-type inequalities are obtained.
$$ {}_{a}\mathcal{B}^{d}_{b,p, c}(x): =\sum _{k=0}^{\infty }\frac{(-c/4)^{k}\Gamma { ( p+\frac{b+1}{2} ) }}{ \Gamma { ( k+1 ) } \Gamma { ( ak+p+ \frac{b+1}{2} ) }} \frac{(d)_{k}}{k!}x^{k}. $$
Acknowledgements
The first author gratefully acknowledges support from FRGS research grant 203.PMATHS.6711568, and the second author acknowledges support from USM Research University Individual grant (RUI) 1001/PMATHS/8011038.
Competing interests
The authors declare that they have no competing interests.
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