In this section, we state and prove the reduction formula for the q-integral \(I_{n,q}\) stated in the introduction. We start with the following result.
Proposition 3.1
The
q-integral
\(I_{n,q}\)
can be represented in terms of basic hypergeometric series
\({_{3}\phi_{2}}\), that is,
$$ I_{n,q}= S_{n} {_{3}\phi_{2}} \bigl(q^{-n}, q^{-\beta-\delta-\alpha-\gamma +n-1}, q^{-\beta}; q^{-\beta-\delta}, q^{-\alpha-\beta}; q, q\bigr), $$
where
$$ S_{n}= q^{\frac{n}{2}(2\beta+5-n)} \frac{\Gamma_{q}(\beta+\delta +1)\Gamma _{q}(\alpha+\gamma+1)}{\Gamma_{q}(\alpha+\gamma+\beta+\delta-n+2)} \frac {(q^{-\gamma};q)_{n} (q^{-\alpha-\beta};q)_{n}}{(q^{-\alpha-\gamma };q)_{n} (q^{1-n+\gamma};q)_{n}}. $$
Proof
Calculating \(D_{q^{-1}}^{n} [x^{\gamma}(q^{\beta+1}x;q)_{\delta}]\) by using a \(q^{-1}\)-type Leibiniz rule (see [6], p.27) gives
$$ I_{n,q}= \int_{0}^{1} x^{\alpha} (qx;q)_{\beta} \sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix} _{q^{-1}} \bigl(D_{q^{-1}}^{n-k} f\bigr) \bigl(q^{-k}x\bigr) D_{q^{-1}}^{k} g(x) \,d_{q}x, $$
where
$$f(x)= x^{\gamma}\quad\mbox{and}\quad g(x)= \bigl(q^{\beta+1}x,q \bigr)_{\delta}. $$
Note that
$$\begin{aligned}& \bigl\{ D_{q^{-1}}^{n-k} (\cdot)^{\gamma}\bigr\} \bigl(q^{-k}x\bigr)= \frac {(-q)^{n-k}(q^{-\gamma};q)_{n-k}}{(1-q)^{n-k}} q^{-k^{2}+nk-\gamma k} x^{\gamma-n+k}, \\& D_{q^{-1}}^{k}\bigl(q^{\beta+1}x,q\bigr)_{\delta}= q^{(\beta+\delta+1)k} \frac {q^{-\frac{1}{2}k(k-1)}}{(1-q)^{k}}\bigl(q^{\beta+1}x,q\bigr)_{\delta -k} \bigl(q^{-\delta },q\bigr)_{k} . \end{aligned}$$
This implies
$$\begin{aligned} I_{n,q}={}& \sum _{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix} _{q^{-1}} \frac{(-q)^{n-k}(q^{-\gamma};q)_{n-k}}{(1-q)^{n}} \bigl(q^{-\delta};q \bigr)_{k} q^{-\frac{1}{2}k(k-1)} \\ &{}\times q^{(\beta +\delta +1)k-(k-n+\gamma)k} \int_{0}^{1} x^{\alpha+\gamma-n+k} (qx;q)_{\beta} \bigl(q^{\beta+1}x;q\bigr)_{\delta-k} \,d_{q}x. \end{aligned}$$
(3.1)
Denoting the q-integral in the left-hand-side of (3.1) by \(J_{n}\), we obtain
$$\begin{aligned} J_{n} &= \int_{0}^{1} x^{\alpha+\gamma-n+k} (qx;q)_{\beta+\delta-k} \,d_{q}x\\ &= B_{q}(\alpha+\gamma-n+k+1,\beta+\delta-k+1)\\ &=\frac{\Gamma_{q}(\alpha+\gamma-n+k+1)\Gamma_{q}(\beta+\delta -k+1)}{\Gamma _{q}(\alpha+\gamma+ \beta+\delta-n+2)}. \end{aligned}$$
Using [6], Eq. (I.8) and Eq. (I.35), we get
$$ J_{n}= \frac{\Gamma_{q}(\alpha+\gamma+1)\Gamma_{q}(\beta+\delta +1)}{\Gamma _{q}(\alpha+\gamma+ \beta+\delta-n+2)} \frac {(1-q)^{n}q^{k^{2}-nk+\frac {1}{2}n(n+1)}q^{-(\beta+\delta)k-(n-k)(\alpha+\gamma)}}{(-1)^{n} (q^{-\beta-\delta};q)_{k} (q^{-\alpha-\gamma};q)_{n-k}}. $$
(3.2)
Using [6], Eq. (I.11), we get
$$ \frac{(q^{-\gamma};q)_{n-k}}{(q^{-\alpha-\gamma};q)_{n-k}}= \frac {(q^{-\gamma};q)_{n}}{(q^{-\alpha-\gamma} ;q)_{n}} \frac {(q^{1-n+\alpha +\gamma};q)_{k}}{(q^{1-n+\gamma};q)_{k}} q^{-\alpha k}. $$
(3.3)
Substituting (3.2) into (3.1), using (3.3), yields
$$\begin{aligned} I_{n,q}={}& \frac{\Gamma_{q}(\alpha+\gamma+1)\Gamma_{q}(\beta+\delta +1)}{\Gamma _{q}(\alpha+\gamma+ \beta+\delta-n+2)} q^{n(1-\gamma-\alpha )+\frac {1}{2}n(n+1)} \frac{ (q^{-\gamma};q)_{n}}{(q^{-\alpha-\gamma};q)_{n}} \\ &{}\times\sum_{k=0}^{n} \begin{bmatrix} n \\ k \end{bmatrix} _{q^{-1}}(-1)^{k} q^{-\frac{1}{2}k(k-1)} \frac{(q^{1-n+\alpha +\gamma};q)_{k} (q^{-\delta};q)_{k}}{(q^{1-n+\gamma};q)_{k} (q^{-\delta -\beta};q)_{k}}. \end{aligned}$$
(3.4)
Now using [6], Eq. (I.42) and Eq. (I.47), we get
$$\begin{aligned} I_{n,q} =& K_{n} \sum_{k=0}^{n} q^{k} \frac{(q^{-n},q^{-\delta },q^{1-n+\alpha+\gamma};q)_{k}}{(q^{-\beta-\delta},q,q^{1-n+\gamma};q)_{k}}\\ =& K_{n} {_{3} \phi_{2}}\bigl(q^{-n}, q^{1-n+\alpha+\gamma}, q^{-\delta }; q^{-\beta-\delta}, q^{1-n+\gamma}; q, q\bigr), \end{aligned}$$
where
$$ K_{n}= \frac{\Gamma_{q}(\beta+\delta+1)\Gamma_{q}(\alpha+\gamma +1)}{\Gamma _{q}(\alpha+\gamma+\beta+\delta-n+2)}q^{n(1-\gamma-\alpha)+ \frac{1}{2}n(n+1)}\frac{(q^{-\gamma};q)_{n}}{(q^{-\alpha-\gamma};q)_{n}}. $$
Using the transformation (2.9) yields the required result and completes the proof. □
Corollary 3.2
If
\(\gamma=0\)
then
\(I_{n,q}\)
vanishes for all values of
n
where
\(n-\beta-\delta\)
and
β
are nonnegative integers.
Proof
Since
$$ {_{3}\phi_{2}}\bigl(a,b_{1}q^{m_{1}},b_{2}q^{m_{2}};b_{1},b_{2};q,a^{-1}q^{-(m_{1}+m_{2})} \bigr)= 0, $$
(3.5)
where \(m_{1}\), \(m_{2}\) are arbitrary nonnegative integers (see [6]), the proof follows directly from Proposition 3.1 and (3.5). □
Watson remarked \(I_{n}\) vanishes for odd values of n in two special cases, (i) \(\alpha=\gamma\) and \(\beta=\delta\) and (ii) \(\alpha=\beta \) and \(\gamma=\delta\).
Now, we can derive the reduction formula for \(I_{n,q}\).
Theorem 3.3
The reduction formula satisfies a three term recurrence relations of
\(I_{n,q}\). More precisely, the following holds:
If
$$W_{n}= {_{3}\phi_{2}}\bigl(q^{-n}, q^{-\beta-\delta-\alpha-\gamma+n-1}, q^{-\beta }; q^{-\beta-\delta}, q^{-\alpha-\beta}; q, q \bigr), $$
then
$$ L_{n} W_{n+1} - Q_{n} W_{n-1}+ M_{n} W_{n}=0, $$
(3.6)
where
$$\begin{aligned}& L_{n}=\bigl(q^{-n}-q^{\theta_{1}} \bigr) \bigl(q^{-n}-q^{\theta_{2}}\bigr) \bigl(1-q^{\theta _{3}+n} \bigr) \bigl(q^{\theta_{3}+n}-q^{1-n}\bigr),\\& Q_{n}= \bigl(1-q^{-n}\bigr) \bigl(q^{\theta_{3}+n}-q^{\theta_{1}} \bigr) \bigl(q^{\theta _{3}+n}-q^{\theta _{2}}\bigr) \bigl(q^{-n}-q^{\theta_{3}+n+1} \bigr),\\& \begin{aligned}[b] M_{n}={}& q^{\theta_{2}} \bigl(1-q^{-n}\bigr)\bigl[q^{-n}\bigl(q^{\theta_{1}}-q^{\theta_{3}+n} \bigr)+ q^{2(\theta_{3}+n)+1}\bigr]+ q^{3(\theta_{3}+n)}\bigl(q^{1-n}-q^{-\beta} \bigr) \\ &{}-[2]q^{\theta_{3}+n}\bigl(q^{\theta_{1}+\theta_{2}}+q^{1+\theta_{3}}\bigr)+ q^{1+\theta _{3}-n} \bigl([2]-q^{-n}-q^{\theta_{2}} \bigr) + q^{1-2n}\bigl(q^{-n-\beta}-q^{\theta_{2}}\bigr)\\ &{}+q\bigl(q^{\theta_{3}+n}-q^{-n}\bigr) \bigl(q^{1+\theta_{3}-\beta}+ q^{\theta _{1}}\bigl(q^{\theta _{3}+n}+q^{-n}\bigr)\bigr)+ q^{\theta_{1}+\theta_{2}}\bigl(q^{1-n}+q^{2(\theta_{3}+n)}\bigr)\\ &{}+[2]q^{\theta_{3}}\bigl(q^{\theta_{1}}-q^{-\beta}\bigr) \bigl(q^{-n}-q^{\theta_{3}+n}\bigr), \end{aligned} \end{aligned}$$
and
\(\theta_{1}=-\alpha-\beta\), \(\theta_{2}=-\delta-\beta\), \(\theta _{3}=-\alpha-\beta-\delta-\gamma-1\).
Proof
This result follows by applying Proposition 3.1 and using equation (2.18) with
$$b=q^{-\beta-\delta-\alpha-\gamma+n-1},\qquad c=q^{-\beta},\qquad d=q^{-\beta -\delta}\quad \mbox{and}\quad e=q^{-\alpha-\beta}. $$
□
Recall that the little q-Jacobi polynomials, see [16], are defined by
$$ P_{n}(x;a,b;q)= {_{2}\phi_{1}} \bigl(q^{-n}, abq^{n+1}; aq; q, qx\bigr), $$
(3.7)
and the formula
$$P_{n}(x;c,d;q)= \sum_{k=0}^{n} a_{k,n}P_{k} (x;a,b;q) $$
holds with
$$ \begin{aligned} & a_{k,n}= C_{k} {_{3} \phi _{2}}\bigl(q^{k-n},cdq^{n+k+1},aq^{k+1};cq^{k+1},abq^{2k+2};q,q \bigr),\\ &C_{k}= (-1)^{k} q \frac{(q^{-n},aq,cdq^{n+1};q)_{k}}{(q,cq,abq^{k+1};q)_{k}}. \end{aligned} $$
(3.8)
Remark 3.4
In (3.7), if we take \(a=q^{-\beta-1}\), \(b=q^{-\alpha-1}\), \(c=q^{-\beta-\delta-1}\) and \(d=q^{-\alpha-\gamma-1}\), we get \(a_{0,n} =\frac{1}{S_{n}} I_{n,q}\). Thus, the little q-Jacobi polynomials and the q-integrals \(I_{n,q}\) are related in the following way:
$$ P_{n}(x;c,d;q) = \frac{1}{S_{n}} I_{n,q}+ \sum _{k=1}^{n} a_{k,n} P_{k} (x;a,b;q). $$