A Unified Generalization of the Catalan, Fuss, and Fuss—Catalan Numbers

Abstract

In the paper, the authors introduce a unified generalization of the Catalan numbers, the Fuss numbers, the Fuss–Catalan numbers, and the Catalan–Qi function, and discover some properties of the unified generalization, including a product-ratio expression of the unified generalization in terms of the Catalan–Qi functions, three integral representations of the unified generalization, and the logarithmically complete monotonicity of the second order for a special case of the unified generalization.

1. Introduction

As well known from [1,2], Catalan numbers $C_n$ are used in the study of set partitions in different areas of mathematics. In particular, in combinatorial mathematics, the Catalan numbers $C_n$ form a sequence of natural numbers that occur in various counting problems, often involving recursively-defined objects. There are many counting problems in combinatorics whose solution is given by the Catalan numbers $C_n$. The book [3] contains a set of exercises which describe 66 different interpretations of the Catalan numbers.

The Catalan numbers $C_n$ can be generated by

$$\frac{2}{1+\sqrt{1-4x}}=\frac{1-\sqrt{1-4x}}{2x}=\sum _{n=0}^{\infty }{C}_n{x}^n=1+x+2x^2+5x^3+14x^4+42x^5+132x^6+\cdots .$$

One of the explicit formulas of $C_n$ for $n\ge 0$ reads that

$$C_n=\frac{1}{n+1}\left(\genfrac{}{}{0pt}{}{2n}{n}\right)=\frac{4^n\mathrm{\Gamma }(n+1/2)}{\sqrt{\pi }\mathrm{\Gamma }(n+2)},$$

where

$$\mathrm{\Gamma }\left(z\right)={\int }_0^{\infty }{t}^{z-1}{e}^{-t}\mathrm{d}t,\Re \left(z\right)>0$$

is the classical Euler gamma function. In [2,4,5] and ([1] pp. 110–111), it was mentioned that there exists an asymptotic expansion

$$C_x\triangleq \frac{4^x\mathrm{\Gamma }\left(x+\frac{1}{2}\right)}{\sqrt{\pi }\mathrm{\Gamma }(x+2)}\sim \frac{4^x}{\sqrt{\pi }}\left(\frac{1}{x^{3/2}}-\frac{9}{8}\frac{1}{x^{5/2}}+\frac{145}{128}\frac{1}{x^{7/2}}+\cdots \right)$$

(1)

for the Catalan function $C_x$. For new developments on (1), see [6] and the review paper in [7], in which there are plenty of closely related references.

A generalization of the Catalan numbers $C_n$ was defined in [8,9] by

$${}_p{d}_n=\frac{1}{n}\left(\genfrac{}{}{0pt}{}{pn}{n-1}\right)=\frac{1}{(p-1)n+1}\left(\genfrac{}{}{0pt}{}{pn}{n}\right)$$

for $n,p\ge 1$. It is obvious that $C_n={}_2{d}_n$. In ([1] pp. 375–376), the generalization ${}_{p+1}{d}_n$ of the Catalan numbers $C_n$ is denoted by $C(n,p)$ for $p\ge 0$ and is called as the generalized Catalan numbers. In ([1] pp. 377–378), the Fuss numbers

$$F(m,n)=\frac{1}{mn+1}\left(\genfrac{}{}{0pt}{}{mn+1}{n}\right)$$

were given and discussed. It is apparent that $F(2,n)=C_n$.

In combinatorial mathematics and statistics, the Fuss–Catalan numbers $A_n(p,r)$ are defined [10] as numbers of the form

$$A_n(p,r)=\frac{r}{np+r}\left(\genfrac{}{}{0pt}{}{np+r}{n}\right)=\frac{r\mathrm{\Gamma }(np+r)}{\mathrm{\Gamma }(n+1)\mathrm{\Gamma }\left(n\right(p-1)+r+1)}.$$

(2)

It is easy to see that

$$A_n(p,1)=F(p,n),A_n(2,1)=C_n,n\ge 0$$

and

$$A_{n-1}(p,p)={}_p{d}_n=C(n,p-1),n,p\ge 1.$$

There have existed some literature, such as [1,2,11,12,13,14,15,16,17,18,19,20,21], on the investigation of the Fuss–Catalan numbers $A_n(p,r)$.

In [22], an alternative and analytical generalization of the Catalan numbers $C_n$ and the Catalan function $C_x$ was introduced by

$$C(a,b;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^z\frac{\mathrm{\Gamma }(z+a)}{\mathrm{\Gamma }(z+b)},\Re \left(a\right),\Re \left(b\right)>0,\Re \left(z\right)\ge 0.$$

For uniqueness and convenience of referring to the quantity $C(a,b;z)$, we call the quantity $C(a,b;z)$ the Catalan–Qi function and, when taking $z=n\ge 0$, call $C(a,b;n)$ the Catalan–Qi numbers. It is not difficult to verify that $C\left(\frac{1}{2},2;n\right)=C_n$ and

$$C(n+1,2;(m-1)n)={\left(\frac{2}{n+1}\right)}^{(m-1)n}{}_m{d}_n={\left(\frac{2}{n+1}\right)}^{(m-1)n}C(n,m-1)$$

for $m,n\ge 1$. In [22], it was obtained that

$$\begin{array}{c}\hfill C(a,b;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^z\frac{{(z+a)}^z}{{(z+b)}^{z+b-a}}exp\left[b-a+{\int }_0^{\infty }\frac{1}{t}\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-a\right)\left(e^{-at}-e^{-bt}\right)e^{-zt}\mathrm{d}t\right]\end{array}$$

(3)

for $\Re \left(a\right),\Re \left(b\right)>0$ and $\Re \left(z\right)\ge 0$. Recently, we discovered in ([23] Theorem 1.1) relations between the Fuss–Catalan numbers $A_n(p,r)$ and the Catalan–Qi numbers $C(a,b;n)$, one of which reads that

$$A_n(p,r)=r^n\frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+r}{p},1;n\right)}{{\prod }_{k=0}^{p-2}C\left(\frac{k+r+1}{p-1},1;n\right)}$$

(4)

for integers $n\ge 0$, $p>1$, and $r>0$. In recent papers [6,22,23,24,25,26,27,28,29,30,31,32], among other things, some properties, including the general expression and a generalization of the asymptotic expansion (1), the monotonicity, logarithmic convexity, (logarithmically) complete monotonicity, minimality, Schur-convexity, product and determinantal inequalities, exponential representations, integral representations, a generating function, connections with the Bessel polynomials and the Bell polynomials of the second kind, and identities, of the Catalan numbers $C_n$, the Catalan function $C_x$, the Catalan–Qi numbers $C(a,b;n)$, the Catalan–Qi function $C(a,b;z)$, and the Fuss–Catalan numbers $A_n(p,r)$ were established.

In this paper, we will introduce a unified generalization of the Catalan numbers $C_n$, the generalized Catalan numbers $C(n,k)$, the Fuss numbers $F(k,n)$, the Fuss–Catalan numbers $A_n(p,r)$, and the Catalan–Qi function $C(a,b;z)$. Hereafter, we will find a product-ratio expression, similar to the product-ratio expression (4), of the unified generalization in terms of the Catalan–Qi function $C(a,b;z)$. Furthermore, based on the integral representation (3), on the Gauss multiplication formula for the gamma function, and on an integral representation for the logarithm of the gamma function $\mathrm{\Gamma }\left(z\right)$, we will derive three integral representations of the unified generalization. Finally, we will establish the logarithmically complete monotonicity of the second order for the unified generalization.

2. A Unified Generalization of the Catalan and Other Numbers

Does there exist a unified and analytic generalization of the Catalan numbers $C_n$, the Fuss numbers $F(m,n)=C(n,m)$, the Fuss–Catalan numbers $A_n(p,r)$, and the Catalan–Qi function $C(a,b;z)$? What is it concretely? In the early morning of September 15th 2015, a unified generalization was framed out eventually and successfully, and which can be described by a five-variable function

$$Q(a,b;p,q;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}{\left[\mathrm{\Gamma }(z+1)\right]}^{q-p}\frac{\mathrm{\Gamma }(pz+a)}{\mathrm{\Gamma }(qz+b)},$$

(5)

where $\Re \left(a\right),\Re \left(b\right)>0$, $\Re \left(p\right),\Re \left(q\right)>0$, and $\Re \left(z\right)\ge 0$. For uniqueness and convenience of referring to the quantity $Q(a,b;p,q;z)$, we call $Q(a,b;p,q;z)$ the Fuss–Catalan–Qi function and, when taking $z=n\ge 0$, call $Q(a,b;p,q;n)$ the Fuss–Catalan–Qi numbers.

It is easy to see that

$$\begin{array}{c}Q\left(\frac{1}{2},2;1,1;n\right)=Q(1,2;2,1;n)=C_n,Q(r,r+1;p,p-1;n)=A_n(p,r),\\ Q(p,p+1;p,p-1;n-1)={}_p{d}_n=C(n,p-1),Q(a,b;1,1;z)=C(a,b;z).\end{array}$$

(6)

Accordingly, the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$ is a unified generalization of the Catalan numbers $C_n$, the generalized Catalan numbers $C(n,m)$, the Fuss numbers $F(m,n)$, the Fuss–Catalan numbers $A_n(p,r)$, and the Catalan–Qi function $C(a,b;z)$.

It is easy to see that the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$ meets

$$\begin{array}{c}Q(b,a;p,q;z)={\left(\frac{a}{b}\right)}^{2(q-p)z}\frac{1}{Q(a,b;q,p;z)}\end{array}$$

and, when $p=q$ or $a=b$,

$$Q(a,b;q,p;z)Q(b,a;p,q;z)=1.$$

If only interchanging the role of a and b, then

$$Q(a,b;p,q;z)Q(b,a;p,q;z)=\left\{\begin{array}{c}{\displaystyle \frac{R(a,b;2p;z)}{R(b,a;2q;z)}},\hfill \\ S(a;p,q;z)S(b;p,q;z),\hfill \end{array}\right.$$

where

$$R(a,b;r;z)=\frac{\mathrm{\Gamma }(rz+a)\mathrm{\Gamma }(rz+b)}{{\left[\mathrm{\Gamma }(z+1)\right]}^r}$$

and

$$S(c;p,q;z)={\left[\mathrm{\Gamma }(z+1)\right]}^{q-p}\frac{\mathrm{\Gamma }(pz+c)}{\mathrm{\Gamma }(qz+c)}.$$

If only swapping p and q, then

$$Q(a,b;p,q;z)Q(a,b;q,p;z)=\left\{\begin{array}{c}{\displaystyle \frac{F(b;p,q;z)}{F(a;p,q;z)}},\hfill \\ G(a,b;p;z)G(a,b;q;z),\hfill \end{array}\right.$$

where

$$F(c;p,q;z)=\frac{{\left[c^z\mathrm{\Gamma }\left(c\right)\right]}^2}{\mathrm{\Gamma }(pz+c)\mathrm{\Gamma }(qz+c)}$$

and

$$G(a,b;r;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^z\frac{\mathrm{\Gamma }(rz+a)}{\mathrm{\Gamma }(rz+b)}.$$

3. A Product-Ratio Expression of the Fuss–Catalan–Qi Function

Motivated by the product-ratio expression (4), we now find out a product-ratio expression of the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$.

Theorem 1.

For $\Re \left(a\right),\Re \left(b\right)>0$ and $\Re \left(z\right)\ge 0$, when $p,q\in \mathbb{N}$, we have

$$Q(a,b;p,q;z)={\left[{\left(\frac{b}{a}\right)}^{q-p+1}\frac{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(p+a)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(q+b)}\right]}^z\frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\prod }_{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)}.$$

(7)

Proof. 

By the Gauss multiplication formula

$$\mathrm{\Gamma }\left(nz\right)=\frac{n^{nz-1/2}}{{\left(2\pi \right)}^{(n-1)/2}}\prod _{k=0}^{n-1}\mathrm{\Gamma }\left(z+\frac{k}{n}\right),n\in \mathbb{N}$$

(8)

in ([33] p. 256, 6.1.20), the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$ can be written as

$$\begin{array}{c}Q(a,b;p,q;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}{\left[\mathrm{\Gamma }(z+1)\right]}^{q-p}\frac{\mathrm{\Gamma }\left(p\left(z+\frac{a}{p}\right)\right)}{\mathrm{\Gamma }\left(q\left(z+\frac{b}{q}\right)\right)}\\ =\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}{\left[\mathrm{\Gamma }(z+1)\right]}^{q-p}\frac{{\displaystyle \frac{p^{pz+a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}}{\displaystyle \prod _{k=0}^{p-1}}\mathrm{\Gamma }\left(z+\frac{k+a}{p}\right)}{{\displaystyle \frac{q^{qz+b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}}}{\displaystyle \prod _{k=0}^{q-1}}\mathrm{\Gamma }\left(z+\frac{k+b}{q}\right)}\\ =\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}\frac{{\displaystyle \frac{p^{pz+a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}}{\displaystyle \prod _{k=0}^{p-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(z+\frac{k+a}{p}\right)}{\mathrm{\Gamma }(z+1)}}}{{\displaystyle \frac{q^{qz+b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}}}{\displaystyle \prod _{k=0}^{q-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(z+\frac{k+b}{q}\right)}{\mathrm{\Gamma }(z+1)}}}\\ =\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}\frac{{\displaystyle \prod _{k=0}^{q-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(1\right)}{\mathrm{\Gamma }\left(\frac{k+b}{q}\right)}}{\left({\displaystyle \frac{q}{k+b}}\right)}^z}{{\displaystyle \prod _{k=0}^{p-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(1\right)}{\mathrm{\Gamma }\left(\frac{k+a}{p}\right)}}{\left({\displaystyle \frac{p}{k+a}}\right)}^z}\frac{{\displaystyle \frac{p^{pz+a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}}{\displaystyle \prod _{k=0}^{p-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(1\right)}{\mathrm{\Gamma }\left(\frac{k+a}{p}\right)}}{\left({\displaystyle \frac{p}{k+a}}\right)}^z{\displaystyle \frac{\mathrm{\Gamma }\left(z+\frac{k+a}{p}\right)}{\mathrm{\Gamma }(z+1)}}}{{\displaystyle \frac{q^{qz+b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}}}{\displaystyle \prod _{k=0}^{q-1}}{\displaystyle \frac{\mathrm{\Gamma }\left(1\right)}{\mathrm{\Gamma }\left(\frac{k+b}{q}\right)}}{\left({\displaystyle \frac{q}{k+b}}\right)}^z{\displaystyle \frac{\mathrm{\Gamma }\left(z+\frac{k+b}{q}\right)}{\mathrm{\Gamma }(z+1)}}}\\ =\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}\frac{{\prod }_{k=0}^{p-1}\mathrm{\Gamma }\left(\frac{a}{p}+\frac{k}{p}\right)}{{\prod }_{k=0}^{q-1}\mathrm{\Gamma }\left(\frac{b}{q}+\frac{k}{q}\right)}{\displaystyle \frac{{\prod }_{k=0}^{p-1}{(k+a)}^z}{{\prod }_{k=0}^{q-1}{(k+b)}^z}}\frac{q^{zq}}{p^{zp}}\frac{{\displaystyle \frac{p^{pz+a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}}{\displaystyle \prod _{k=0}^{p-1}}C\left(\frac{k+a}{p},1;z\right)}{{\displaystyle \frac{q^{qz+b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}}}{\displaystyle \prod _{k=0}^{q-1}}C\left(\frac{k+b}{q},1;z\right)}\\ =\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}\frac{\mathrm{\Gamma }\left(a\right){\displaystyle \frac{{\left(2\pi \right)}^{(p-1)/2}}{p^{a-1/2}}}}{\mathrm{\Gamma }\left(b\right){\displaystyle \frac{{\left(2\pi \right)}^{(q-1)/2}}{q^{b-1/2}}}}{\left[\frac{{\prod }_{k=0}^{p-1}(k+a)}{{\prod }_{k=0}^{q-1}(k+b)}\right]}^z\frac{p^{a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}\frac{{\left(2\pi \right)}^{(q-1)/2}}{q^{b-1/2}}\frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\prod }_{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)}\\ ={\left(\frac{b}{a}\right)}^{(q-p+1)z}{\left[\frac{{\prod }_{k=0}^{p-1}(k+a)}{{\prod }_{k=0}^{q-1}(k+b)}\right]}^z\frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\prod }_{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)}\\ ={\left(\frac{b}{a}\right)}^{(q-p+1)z}{\left[\frac{\mathrm{\Gamma }\left(b\right)\mathrm{\Gamma }(p+a)}{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(q+b)}\right]}^z\frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\prod }_{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)}.\end{array}$$

The identity (7) is thus proved. The proof of Theorem 1 is complete. □

Remark 1.

Before getting (7), we did not appreciate the analytic meanings of the form of the product-ratio expression (4) because before catching sight of the unified generalization (5), we did not appreciate the analytic meanings of the form of the Fuss–Catalan numbers $A_n(p,r)$ in (2).

Remark 2.

From (6) and (7), we derive the identity (4) and

$${}_p{d}_n=C(n,p-1)=p^{n-1}\frac{{\prod }_{k=0}^{p-1}C\left(1+\frac{k}{p},1;n-1\right)}{{\prod }_{k=0}^{p-2}C\left(1+\frac{k+2}{p-1},1;n-1\right)}.$$

Remark 3.

When $p=q$, the product-ratio expression (7) can be reformulated as

$$Q(a,b;q,q;z)=\frac{K(a,q,z)}{K(b,q,z)},$$

where

$$K(c,q,z)={\left[\frac{\mathrm{\Gamma }(q+c)}{\mathrm{\Gamma }(1+c)}\right]}^z\prod _{k=0}^{q-1}C\left(\frac{k+c}{q},1;z\right).$$

If taking $a=b$, then

$$Q(a,a;p,q;z)={\displaystyle \frac{{\displaystyle \frac{\mathrm{\Gamma }(pz+a)}{{\left[\mathrm{\Gamma }(z+1)\right]}^p}}}{{\displaystyle \frac{\mathrm{\Gamma }(qz+a)}{{\left[\mathrm{\Gamma }(z+1)\right]}^q}}}}=\frac{{\left[\mathrm{\Gamma }(p+a)\right]}^z{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\left[\mathrm{\Gamma }(q+a)\right]}^z{\prod }_{k=0}^{q-1}C\left(\frac{k+a}{q},1;z\right)}.$$

4. Integral Representations of the Fuss–Catalan–Qi Function

Making use of the integral representation (3) and the product-ratio expression (7), we now derive the first integral representation of the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$.

Theorem 2.

For $\Re \left(a\right),\Re \left(b\right)>0$, and $\Re \left(z\right)\ge 0$, when $p,q\in \mathbb{N}$, we have

$$\begin{array}{cc}\hfill Q(a,b;p,q;z)& ={\left(2\pi \right)}^{(q-p)/2}\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}\frac{p^{a-1/2}}{q^{b-1/2}}{(z+1)}^{(q-p)(z+1/2)+(a-b)}\hfill \\ & \times {\left[{\left(\frac{b}{a}\right)}^{q-p+1}\frac{{\prod }_{k=0}^{p-1}(a+pz+k)}{{\prod }_{k=0}^{q-1}(b+qz+k)}\right]}^{z}exp\{\frac{p-q}{2}+(b-a)\hfill \\ & +{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{a}{p}\right)\left(\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-at/p}-p{e}^{-t}\right)\hfill \\ & -\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{b}{q}\right)\left(\frac{1-e^{-t}}{1-e^{-t/q}}{e}^{-bt/q}-q{e}^{-t}\right)\hfill \\ & -\left(\frac{e^{-t/p}}{p(1-e^{-t/p})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-at/p}\hfill \\ & +\left(\frac{e^{-t/q}}{q(1-e^{-t/q})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/q}}{e}^{-bt/q}+\frac{p-q}{2}{e}^{-t}\left]\mathrm{d}t\right\}.\hfill \end{array}$$

Proof. 

Making use of the integral representation (3) leads to

$$\begin{array}{c}\prod _{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)=\prod _{k=0}^{p-1}\left[\frac{1}{\mathrm{\Gamma }\left(\frac{k+a}{p}\right)}{\left(\frac{p}{k+a}\right)}^z\frac{{\left(z+\frac{k+a}{p}\right)}^z}{{(z+1)}^{z+1-\frac{k+a}{p}}}\right]\\ \times exp\left[\sum _{k=0}^{p-1}\left(1-\frac{k+a}{p}\right)+\sum _{k=0}^{p-1}{\int }_0^{\infty }\frac{1}{t}\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{k+a}{p}\right)\left(e^{-\frac{k+a}{p}t}-e^{-t}\right)e^{-zt}\mathrm{d}t\right]\\ =\frac{1}{{\prod }_{k=0}^{p-1}\mathrm{\Gamma }\left(\frac{k+a}{p}\right)}\frac{1}{{\left[{\prod }_{k=0}^{p-1}(k+a)\right]}^z}\frac{{\prod }_{k=0}^{p-1}{(pz+k+a)}^z}{{(z+1)}^{p(z+1)}}{(z+1)}^{a+\frac{{\sum }_{k=0}^{p-1}k}{p}}\\ \times exp\left[p-a-\frac{{\sum }_{k=0}^{p-1}k}{p}+{\int }_0^{\infty }\frac{e^{-zt}}{t}\sum _{k=0}^{p-1}\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{k+a}{p}\right)\left(e^{-\frac{k+a}{p}t}-e^{-t}\right)\mathrm{d}t\right]\\ =\frac{1}{{\prod }_{k=0}^{p-1}\mathrm{\Gamma }\left(\frac{a}{p}+\frac{k}{p}\right)}{\left[\frac{\mathrm{\Gamma }\left(a\right)}{\mathrm{\Gamma }(p+a)}\right]}^z{\left[\frac{\mathrm{\Gamma }\left(p\right(z+1)+a)}{\mathrm{\Gamma }(pz+a)}\right]}^z\frac{{(z+1)}^{a+(p-1)/2}}{{(z+1)}^{p(z+1)}}\\ \times exp\left[p-a-\frac{p-1}{2}+{\int }_0^{\infty }\frac{e^{-zt}}{t}\sum _{k=0}^{p-1}\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{k+a}{p}\right)\left(e^{-\frac{k+a}{p}t}-e^{-t}\right)\mathrm{d}t\right]\\ =\frac{p^{a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}\mathrm{\Gamma }\left(a\right)}{\left[\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }\left(p\right)}{\mathrm{\Gamma }(p+a)}\right]}^z{\left[\frac{\mathrm{\Gamma }\left(p\right(z+1)+a)}{\mathrm{\Gamma }(pz+a)\mathrm{\Gamma }\left(p\right)}\right]}^z\frac{1}{{(z+1)}^{p(z+1/2)-a+1/2}}\\ \times exp\{\frac{p}{2}-a+\frac{1}{2}+{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{a}{p}\right)\left(\sum _{k=0}^{p-1}{e}^{-\frac{k+a}{p}t}-p{e}^{-t}\right)\\ -\sum _{k=0}^{p-1}\frac{k}{p}{e}^{-\frac{k+a}{p}t}+\frac{p-1}{2}{e}^{-t}\left]\mathrm{d}t\right\}\\ =\frac{p^{a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}\mathrm{\Gamma }\left(a\right)}{\left[\frac{B(a,p)}{B(a+pz,p)}\right]}^z\frac{1}{{(z+1)}^{p(z+1/2)-a+1/2}}\\ \times exp\{\frac{p}{2}-a+\frac{1}{2}+{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{a}{p}\right)\left(\sum _{k=0}^{p-1}{e}^{-\frac{k+a}{p}t}-p{e}^{-t}\right)\\ -\sum _{k=0}^{p-1}\frac{k}{p}{e}^{-\frac{k+a}{p}t}+\frac{p-1}{2}{e}^{-t}\left]\mathrm{d}t\right\}\\ =\frac{p^{a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}\mathrm{\Gamma }\left(a\right)}{\left[\frac{B(a,p)}{B(a+pz,p)}\right]}^z\frac{1}{{(z+1)}^{p(z+1/2)-a+1/2}}\\ \times exp\{\frac{p}{2}-a+\frac{1}{2}+{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{a}{p}\right)\left(e^{-at/p}\frac{1-e^{-t}}{1-e^{-t/p}}-p{e}^{-t}\right)\\ -\frac{p\left(e^{-t/p}-1\right)e^{-t}+(1-e^{-t})e^{-t/p}}{p{(1-e^{-t/p})}^2}{e}^{-at/p}+\frac{p-1}{2}{e}^{-t}\left]\mathrm{d}t\right\},\end{array}$$

where $B(p,q)=\frac{\mathrm{\Gamma }\left(p\right)\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(p+q)}$ is the classical beta function. Similarly, we also have

$$\begin{array}{c}\prod _{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)=\frac{q^{b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}\mathrm{\Gamma }\left(b\right)}{\left[\frac{B(b,q)}{B(b+qz,q)}\right]}^z\frac{1}{{(z+1)}^{q(z+1/2)-b+1/2}}\\ \times exp\{\frac{q}{2}-b+\frac{1}{2}+{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{b}{q}\right)\left(e^{-bt/q}\frac{1-e^{-t}}{1-e^{-t/q}}-q{e}^{-t}\right)\\ -\frac{q\left(e^{-t/q}-1\right)e^{-t}+(1-e^{-t})e^{-t/q}}{q{(1-e^{-t/q})}^2}{e}^{-bt/q}+\frac{q-1}{2}{e}^{-t}\left]\mathrm{d}t\right\}.\end{array}$$

Consequently, we obtain

$$\begin{array}{cc}\hfill \frac{{\prod }_{k=0}^{p-1}C\left(\frac{k+a}{p},1;z\right)}{{\prod }_{k=0}^{q-1}C\left(\frac{k+b}{q},1;z\right)}& ={\left(2\pi \right)}^{(q-p)/2}{(z+1)}^{(q-p)(z+1/2)+(a-b)}\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}\frac{p^{a-1/2}}{q^{b-1/2}}\hfill \\ & \times {\left[\frac{B(a,p)}{B(a+pz,p)}\frac{B(b+qz,q)}{B(b,q)}\right]}^{z}exp\{\frac{p-q}{2}+(b-a)\hfill \\ & +{\int }_0^{\infty }\frac{e^{-zt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{a}{p}\right)\left(e^{-at/p}\frac{1-e^{-t}}{1-e^{-t/p}}-p{e}^{-t}\right)\hfill \\ & -\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{b}{q}\right)\left(e^{-bt/q}\frac{1-e^{-t}}{1-e^{-t/q}}-q{e}^{-t}\right)\hfill \\ & -\frac{p\left(e^{-t/p}-1\right)e^{-t}+(1-e^{-t})e^{-t/p}}{p{(1-e^{-t/p})}^2}{e}^{-at/p}\hfill \\ & +\frac{q\left(e^{-t/q}-1\right)e^{-t}+(1-e^{-t})e^{-t/q}}{q{(1-e^{-t/q})}^2}{e}^{-bt/q}+\frac{p-q}{2}{e}^{-t}\left]\mathrm{d}t\right\}.\hfill \end{array}$$

Substituting this into (5) and simplifying yields the integral representation in Theorem 2. The proof of Theorem 2 is complete. □

By the Gauss multiplication formula (8) and an integral representation for the logarithm of the gamma function $\mathrm{\Gamma }\left(z\right)$, we can acquire the second integral representation of the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$, which is seemingly simpler than the one in Theorem 2.

Theorem 3.

For $a,b>0$ and $x\ge 0$, when $p,q\in \mathbb{N}$, we have

$$\begin{array}{cc}Q(a,b;p,q;x)=\hfill & {\left(2\pi \right)}^{(q-p)/2}{e}^{(p-q)/2+b-a}\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)x}\hfill \\ & \times {(x+1)}^{(q-p)(x+1/2)}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}\frac{{\prod }_{k=0}^{p-1}{\left(x+\frac{a+k}{p}\right)}^{x+(a+k)/p-1/2}}{{\prod }_{k=0}^{q-1}{\left(x+\frac{b+k}{q}\right)}^{x+(b+k)/q-1/2}}\hfill \\ & \times exp\left\{{\int }_0^{\infty }\beta \left(t\right)\left[(q-p)e^{-t}+\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-at/p}-\frac{1-e^{-t}}{1-e^{-t/q}}{e}^{-bt/q}\right]e^{-xt}\mathrm{d}t\right\},\hfill \end{array}$$

(9)

where

$$\beta \left(t\right)=\frac{1}{t}\left(\frac{1}{e^t-1}-\frac{1}{t}+\frac{1}{2}\right).$$

(10)

Proof. 

By Formula (8) and

$$\ln \mathrm{\Gamma }\left(z\right)=\ln \left(\sqrt{2\pi }{z}^{z-1/2}{e}^{-z}\right)+{\int }_0^{\infty }\beta \left(t\right)e^{-zt}\mathrm{d}t$$

(11)

in ([34] (3.22)), we have

$$\begin{array}{c}\ln \left({\left[\mathrm{\Gamma }\right(x+1\left)\right]}^{q-p}\frac{\mathrm{\Gamma }(px+a)}{\mathrm{\Gamma }(qx+b)}\right)=(q-p)\ln \mathrm{\Gamma }(x+1)+\ln \frac{\mathrm{\Gamma }\left(p\left(x+\frac{a}{p}\right)\right)}{\mathrm{\Gamma }\left(q\left(x+\frac{b}{q}\right)\right)}\hfill \\ =(q-p)\ln \mathrm{\Gamma }(x+1)+\ln \frac{\frac{p^{px+a-1/2}}{{\left(2\pi \right)}^{(p-1)/2}}{\prod }_{k=0}^{p-1}\mathrm{\Gamma }\left(x+\frac{a+k}{p}\right)}{\frac{q^{qx+b-1/2}}{{\left(2\pi \right)}^{(q-1)/2}}{\prod }_{k=0}^{q-1}\mathrm{\Gamma }\left(x+\frac{b+k}{q}\right)}\hfill \\ =(q-p)\ln \mathrm{\Gamma }(x+1)+\ln \left[{\left(2\pi \right)}^{(q-p)/2}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}\right]\hfill \\ +\sum _{k=0}^{p-1}\ln \mathrm{\Gamma }\left(x+\frac{a+k}{p}\right)-\sum _{k=0}^{q-1}\ln \mathrm{\Gamma }\left(x+\frac{b+k}{q}\right)\hfill \\ =(q-p)\ln \left[\sqrt{2\pi }{(x+1)}^{x+1/2}{e}^{-(x+1)}\right]+(q-p){\int }_0^{\infty }\beta \left(t\right)e^{-(x+1)t}\mathrm{d}t\hfill \\ +\ln \left[{\left(2\pi \right)}^{(q-p)/2}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}\right]-\sum _{k=0}^{q-1}{\int }_0^{\infty }\beta \left(t\right)e^{-[x+(b+k)/q]t}\mathrm{d}t\hfill \\ +\sum _{k=0}^{p-1}\ln \left[\sqrt{2\pi }{\left(x+\frac{a+k}{p}\right)}^{x+(a+k)/p-1/2}{e}^{-[x+(a+k)/p]}\right]\hfill \\ -\sum _{k=0}^{q-1}\ln \left[\sqrt{2\pi }{\left(x+\frac{b+k}{q}\right)}^{x+(b+k)/q-1/2}{e}^{-[x+(b+k)/q]}\right]+\sum _{k=0}^{p-1}{\int }_0^{\infty }\beta \left(t\right)e^{-[x+(a+k)/p]t}\mathrm{d}t\hfill \\ =\ln \left[{\left(2\pi \right)}^{(q-p)/2}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}{(x+1)}^{(q-p)(x+1/2)}\frac{{\prod }_{k=0}^{p-1}{\left(x+\frac{a+k}{p}\right)}^{x+(a+k)/p-1/2}}{{\prod }_{k=0}^{q-1}{\left(x+\frac{b+k}{q}\right)}^{x+(b+k)/q-1/2}}\right]\hfill \\ +(p-q)(x+1)-\sum _{k=0}^{p-1}\left(x+\frac{a+k}{p}\right)+\sum _{k=0}^{q-1}\left(x+\frac{b+k}{q}\right)\hfill \\ +(q-p){\int }_0^{\infty }\beta \left(t\right)e^{-(x+1)t}\mathrm{d}t+{\int }_0^{\infty }\beta \left(t\right)e^{-(x+a/p)t}\sum _{k=0}^{p-1}{e}^{-kt/p}\mathrm{d}t\hfill \\ -{\int }_0^{\infty }\beta \left(t\right)e^{-(x+b/q)t}\sum _{k=0}^{q-1}{e}^{-kt/q}\mathrm{d}t\hfill \\ =\ln \left[{\left(2\pi \right)}^{(q-p)/2}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}{(x+1)}^{(q-p)(x+1/2)}\frac{{\prod }_{k=0}^{p-1}{\left(x+\frac{a+k}{p}\right)}^{x+(a+k)/p-1/2}}{{\prod }_{k=0}^{q-1}{\left(x+\frac{b+k}{q}\right)}^{x+(b+k)/q-1/2}}\right]\hfill \\ +\frac{p-q}{2}+b-a+(q-p){\int }_0^{\infty }\beta \left(t\right)e^{-(x+1)t}\mathrm{d}t\hfill \\ +{\int }_0^{\infty }\beta \left(t\right)e^{-(x+a/p)t}\frac{1-e^{-t}}{1-e^{-t/p}}\mathrm{d}t-{\int }_0^{\infty }\beta \left(t\right)e^{-(x+b/q)t}\frac{1-e^{-t}}{1-e^{-t/q}}\mathrm{d}t\hfill \\ =\ln \left[{\left(2\pi \right)}^{(q-p)/2}\frac{p^{px+a-1/2}}{q^{qx+b-1/2}}{(x+1)}^{(q-p)(x+1/2)}\frac{{\prod }_{k=0}^{p-1}{\left(x+\frac{a+k}{p}\right)}^{x+(a+k)/p-1/2}}{{\prod }_{k=0}^{q-1}{\left(x+\frac{b+k}{q}\right)}^{x+(b+k)/q-1/2}}\right]\hfill \\ +\frac{p-q}{2}+b-a+{\int }_0^{\infty }\beta \left(t\right)\left[(q-p)e^{-t}+\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-at/p}-\frac{1-e^{-t}}{1-e^{-t/q}}{e}^{-bt/q}\right]e^{-xt}\mathrm{d}t.\hfill \end{array}$$

Substituting this into (5) leads to the integral representation (9). □

Only by the integral representation (11), we can establish the third integral representation of the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$, which is seemingly simpler than the previous ones.

Theorem 4.

For $a,b,p,q>0$ and $x\ge 0$, we have

$$\begin{array}{c}Q(a,b;p,q;x)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(2\pi \right)}^{(q-p)/2}{e}^{p-q+b-a}{\left(\frac{b}{a}\right)}^{(q-p+1)x}{(x+1)}^{(q-p)(x+1/2)}\frac{{(px+a)}^{px+a-1/2}}{{(qx+b)}^{qx+b-1/2}}\hfill \\ \hfill \times exp\left\{{\int }_0^{\infty }\beta \left(t\right)\left[(q-p)e^{-(x+1)t}-e^{-(qx+b)t}+e^{-(px+a)t}\right]\mathrm{d}t\right\},\end{array}$$

(12)

where $\beta \left(t\right)$ is defined by (10).

Proof. 

By virtue of (11), we obtain

$$\begin{array}{c}\ln \left({\left[\mathrm{\Gamma }\right(x+1\left)\right]}^{q-p}\frac{\mathrm{\Gamma }(px+a)}{\mathrm{\Gamma }(qx+b)}\right)=(q-p)\ln \mathrm{\Gamma }(x+1)+\ln \mathrm{\Gamma }(px+a)-\ln \mathrm{\Gamma }(qx+b)\\ =(q-p)\ln \left[\sqrt{2\pi }{(x+1)}^{x+1/2}{e}^{-(x+1)}\right]+(q-p){\int }_0^{\infty }\beta \left(t\right)e^{-(x+1)t}\mathrm{d}t\\ +\ln \left[\sqrt{2\pi }{(px+a)}^{px+a-1/2}{e}^{-(px+a)}\right]+{\int }_0^{\infty }\beta \left(t\right)e^{-(px+a)t}\mathrm{d}t\\ -\ln \left[\sqrt{2\pi }{(qx+b)}^{qx+b-1/2}{e}^{-(qx+b)}\right]-{\int }_0^{\infty }\beta \left(t\right)e^{-(qx+b)t}\mathrm{d}t\\ =p-q+b-a+\ln \left[{\left(2\pi \right)}^{(q-p)/2}{(x+1)}^{(q-p)(x+1/2)}\frac{{(px+a)}^{px+a-1/2}}{{(qx+b)}^{qx+b-1/2}}\right]\\ +{\int }_0^{\infty }\beta \left(t\right)\left[(q-p)e^{-(x+1)t}+e^{-(px+a)t}-e^{-(qx+b)t}\right]\mathrm{d}t.\end{array}$$

Substituting this into (5) leads to the integral representation (12). The proof of Theorem 4 is complete. □

Remark 4.

From (6) and the integral representation in Theorem 2, we obtain

$$\begin{array}{c}{A}_n(p,r)=\frac{1}{\sqrt{2\pi }}r\frac{p^{r-1/2}}{{(p-1)}^{r+1/2}}\frac{1}{{(n+1)}^{n+3/2}}{\left[\frac{{\prod }_{k=0}^{p-1}(k+r+pv)}{{\prod }_{k=0}^{p-2}(k+r+(p-1)v+1)}\right]}^n\hfill \\ \times exp\{\frac{3}{2}+{\int }_0^{\infty }\frac{e^{-vt}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{r}{p}\right)\left(e^{-\frac{r}{p}t}\frac{1-e^{-t}}{1-e^{-t/p}}-p{e}^{-t}\right)\hfill \\ -\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{r+1}{p-1}\right)\left(e^{-\frac{r+1}{p-1}t}\frac{1-e^{-t}}{1-e^{-t/(p-1)}}-(p-1)e^{-t}\right)\hfill \\ -\left(\frac{e^{-t/p}}{p(1-e^{-t/p})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-\frac{r}{p}t}\hfill \\ +\left(\frac{e^{-t/(p-1)}}{(p-1)(1-e^{-t/(p-1)})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/(p-1)}}{e}^{-\frac{r+1}{p-1}t}+\frac{1}{2}{e}^{-t}\left]\mathrm{d}t\right\}\hfill \end{array}$$

and

$$\begin{array}{c}C(n,p-1)=\frac{1}{\sqrt{2\pi }}{\left(\frac{p}{p-1}\right)}^{p+1/2}\frac{1}{{(n+1)}^{n+1/2}}{\left[\frac{{\prod }_{k=1}^{p-1}(k+pn)}{{\prod }_{k=0}^{p-2}(k+(p-1)n+2)}\right]}^n\hfill \\ \times exp\{\frac{3}{2}+{\int }_0^{\infty }\frac{e^{-(n-1)t}}{t}[\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-1\right)\left(\frac{1-e^{-t}}{1-e^{-t/p}}-p\right)e^{-t}\hfill \\ -\left(\frac{1}{1-e^{-t}}-\frac{1}{t}-\frac{p+1}{p-1}\right)\left(e^{-\frac{p+1}{p-1}t}\frac{1-e^{-t}}{1-e^{-t/(p-1)}}-(p-1)e^{-t}\right)\hfill \\ -\left(\frac{e^{-t/p}}{p(1-e^{-t/p})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/p}}{e}^{-t}\hfill \\ +\left(\frac{e^{-t/(p-1)}}{(p-1)(1-e^{-t/(p-1)})}-\frac{e^{-t}}{1-e^{-t}}\right)\frac{1-e^{-t}}{1-e^{-t/(p-1)}}{e}^{-\frac{p+1}{p-1}t}+\frac{1}{2}{e}^{-t}\left]\mathrm{d}t\right\}.\hfill \end{array}$$

Remark 5.

When $p=q$, the integral representation in Theorem 2 is reduced to

$$\begin{array}{c}Q(a,b;q,q;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{q}^{a-b}{(z+1)}^{a-b}{\left[\frac{b}{a}\prod _{k=0}^{q-1}\frac{a+qz+k}{b+qz+k}\right]}^z\hfill \\ \times exp\{b-a+{\int }_0^{\infty }\frac{e^{-zt}}{t}[(a-b)e^{-t}+\left(\frac{b}{q}{e}^{-bt/q}-\frac{a}{q}{e}^{-at/q}\right)\frac{1-e^{-t}}{1-e^{-t/q}}\hfill \\ +\left(\frac{1+e^{-t}}{1-e^{-t}}-\frac{e^{-t/q}}{q(1-e^{-t/q})}-\frac{1}{t}\right)\frac{1-e^{-t}}{1-e^{-t/q}}\left(e^{-at/q}-e^{-bt/q}\right)\left]\mathrm{d}t\right\}.\hfill \end{array}$$

Remark 6.

By the integral representation (12) and the second relation in (6), we find

$$\begin{array}{c}{A}_n(p,r)=\frac{e^2}{\sqrt{2\pi }}r{(n+1)}^{-(n+1/2)}\frac{{(pn+r)}^{pn+r-1/2}}{{[(p-1)n+r+1]}^{(p-1)n+r+1/2}}\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\times exp\left\{{\int }_0^{\infty }\beta \left(t\right)\left[e^{-(pn+r)t}-e^{-\left(\right(p-1)n+r+1)t}-e^{-(n+1)t}\right]\mathrm{d}t\right\}.\end{array}$$

Remark 7.

The function $Q(a,b;p,q;z)$ defined by (5) can be rewritten as

$$Q(a,b;p,q;z)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^{(q-p+1)z}\frac{G(b,q,z)}{G(a,p,z)},$$

where

$$G(c,r,z)=\frac{{\left[\mathrm{\Gamma }(z+1)\right]}^r}{\mathrm{\Gamma }(rz+c)}.$$

Taking the logarithm of $G(c,r,z)$ and differentiating gives

$$\ln G(c,r,z)=r\ln \mathrm{\Gamma }(z+1)-\ln \mathrm{\Gamma }(rz+c)$$

and

$$\frac{\mathrm{d}}{\mathrm{d}z}\left[\ln G(c,r,z)\right]=r[\psi (z+1)-\psi (rz+c)].$$

Therefore, we obtain

$$\begin{array}{cc}\hfill \frac{\mathrm{d}}{\mathrm{d}z}\left[\ln \frac{G(b,q,z)}{G(a,p,z)}\right]& =q\left[\psi \right(z+1)-\psi (qz+b\left)\right]-p\left[\psi \right(z+1)-\psi (pz+a\left)\right]\hfill \\ & =(q-p)\psi (z+1)-\left[q\psi \right(qz+b)-p\psi (pz+a\left)\right]\hfill \end{array}$$

and

$$\frac{{\mathrm{d}}^{k+1}}{\mathrm{d}{z}^{k+1}}\left[\ln \frac{G(b,q,z)}{G(a,p,z)}\right]=(q-p){\psi }^{\left(k\right)}(z+1)-q^{k+1}{\psi }^{\left(k\right)}(qz+b)+p^{k+1}{\psi }^{\left(k\right)}(pz+a)$$

for $k\in \left\{0\right\}\cup \mathbb{N}$. Further making use of (15) arrives at

$${(-1)}^{k+1}\frac{{\mathrm{d}}^{k+1}}{\mathrm{d}{z}^{k+1}}\left[\ln \frac{G(b,q,z)}{G(a,p,z)}\right]={\int }_0^{\infty }\frac{t^k}{1-e^{-t}}\left[(q-p)e^{-(z+1)t}-q^{k+1}{e}^{-(qz+b)t}+p^{k+1}{e}^{-(pz+a)t}\right]\mathrm{d}t$$

for $k\in \mathbb{N}$. Consequently, for $k\in \mathbb{N}$, we have

$${(-1)}^{k+1}\frac{{\mathrm{d}}^{k+1}\ln Q(a,b;p,q;z)}{\mathrm{d}{z}^{k+1}}={\int }_0^{\infty }\frac{t^k}{1-e^{-t}}\left[(q-p)e^{-(z+1)t}-q^{k+1}{e}^{-(qz+b)t}+p^{k+1}{e}^{-(pz+a)t}\right]\mathrm{d}t.$$

5. Properties of the Fuss–Catalan–Qi Function

Recall from ([35] Chapter XIII), ([36] Chapter 1), and ([37] Chapter IV) that an infinitely differentiable function f is said to be completely monotonic on an interval I if it satisfies $0\le {(-1)}^k{f}^{\left(k\right)}\left(x\right)<\infty$ on I for all $k\ge 0$.

Recall from [38,39] that an infinitely differentiable and positive function f is said to be logarithmically completely monotonic on an interval I if $0\le {(-1)}^k{\left[\ln f\left(x\right)\right]}^{\left(k\right)}<\infty$ holds on I for all $k\in \mathbb{N}$. For more information on logarithmically completely monotonic functions, please refer to [40,41,42,43] and plenty of references therein.

Recall from [38] that if $f^{\left(k\right)}\left(x\right)$ for some nonnegative integer k is completely monotonic on an interval I but $f^{(k-1)}\left(x\right)$ is not completely monotonic on I, then $f\left(x\right)$ is called a completely monotonic function of the k-th order on an interval I.

Stimulated by the above definitions and main results in [44], we now introduce the concept of logarithmically completely monotonic functions of the k-th order.

Definition 1.

For a positive function $f\left(x\right)$ on an interval I, if ${\left[\ln f\left(x\right)\right]}^{\left(k\right)}$ for some nonnegative integer k is completely monotonic on an interval I but ${\left[\ln f\left(x\right)\right]}^{(k-1)}$ is not completely monotonic on I, then we call $f\left(x\right)$ a logarithmically completely monotonic function of the k-th order on I.

In terms of the terminology of logarithmically completely monotonic functions of the k-th order, we can state the main results of this section as the following theorem.

Theorem 5.

The function

$$Q(a,b;q,q;x)=\frac{\mathrm{\Gamma }\left(b\right)}{\mathrm{\Gamma }\left(a\right)}{\left(\frac{b}{a}\right)}^x\frac{\mathrm{\Gamma }(qx+a)}{\mathrm{\Gamma }(qx+b)},a,b,p>0,x\ge 0$$

satisfies the following conclusions:

1. 

if $a<b$ and $q\le \frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the function $Q(a,b;q,q;x)$ is increasing on $[0,\infty )$;

2. 

if $a>b$ and $q\le \frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the function $Q(a,b;q,q;x)$ is decreasing on $[0,\infty )$;

3. 

if $a<b$ and $q>\frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the function $Q(a,b;q,q;x)$ has a unique minimum on $(0,\infty )$;

4. 

if $a>b$ and $q>\frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the function $Q(a,b;q,q;x)$ has a unique maximum on $(0,\infty )$;

5. 

if and only if $a\lessgtr b$, the function ${\left[Q(a,b;q,q;x)\right]}^{\pm }$ is logarithmically completely monotonic of the second order on $[0,\infty )$; in particular, if and only if $a\lessgtr b$, the function ${\left[Q(a,b;q,q;x)\right]}^{\pm 1}$ is logarithmically convex on $[0,\infty )$.

Proof. 

Taking the logarithm on both sides of Equation (5) and differentiating with respect to x yields

$$\frac{\mathrm{d}\left[\ln Q\right(a,b;p,q;x\left)\right]}{\mathrm{d}x}=(q-p+1)\ln \frac{b}{a}+(q-p)\psi (x+1)+p\psi (px+a)-q\psi (qx+b)$$

and

$$\frac{{\mathrm{d}}^2\left[\ln Q(a,b;p,q;x)\right]}{\mathrm{d}{x}^2}=(q-p){\psi }^{\prime }(x+1)+p^2{\psi }^{\prime }(px+a)-q^2{\psi }^{\prime }(qx+b).$$

It is not difficult to see that

$$\underset{x\to 0^{+}}{lim}\frac{\mathrm{d}\left[\ln Q\right(a,b;p,q;x\left)\right]}{\mathrm{d}x}=(q-p+1)\ln \frac{b}{a}+(q-p)\psi \left(1\right)+p\psi \left(a\right)-q\psi \left(b\right)$$

(13)

and

$$\begin{array}{c}\underset{x\to \infty }{lim}\frac{\mathrm{d}\left[\ln Q\right(a,b;p,q;x\left)\right]}{\mathrm{d}x}=(q-p+1)\ln \frac{b}{a}+\underset{x\to \infty }{lim}\left\{(q-p)\right[\ln (x+1)\\ -\frac{1}{2(x+1)}-{\displaystyle \sum _{n=1}^{\infty }}\frac{B_{2n}}{2n{(x+1)}^{2n}}]+p[\ln (px+a)-\frac{1}{2(px+a)}\\ -{\displaystyle \sum _{n=1}^{\infty }}\frac{B_{2n}}{2n{(px+a)}^{2n}}]-q\left[\ln (qx+b)-\frac{1}{2(qx+b)}-{\displaystyle \sum _{n=1}^{\infty }}\frac{B_{2n}}{2n{(qx+b)}^{2n}}\right]\}\\ \begin{array}{c}=(q-p+1)\ln \frac{b}{a}+\underset{x\to \infty }{lim}[(q-p)\ln (x+1)+p\ln (px+a)-q\ln (qx+b)]\hfill \\ +\underset{x\to \infty }{lim}\{(p-q)\left[\frac{1}{2(x+1)}+\sum _{n=1}^{\infty }\frac{B_{2n}}{2n{(x+1)}^{2n}}\right]\hfill \\ -p\left[\frac{1}{2(px+a)}+\sum _{n=1}^{\infty }\frac{B_{2n}}{2n{(px+a)}^{2n}}\right]+q\left[\frac{1}{2(qx+b)}+\sum _{n=1}^{\infty }\frac{B_{2n}}{2n{(qx+b)}^{2n}}\right]\}\hfill \end{array}\\ \begin{array}{c}=(q-p+1)\ln \frac{b}{a}+\underset{x\to \infty }{lim}[(q-p)\ln (x+1)+p\ln (px+a)-q\ln (qx+b)]\hfill \\ =(q-p+1)\ln \frac{b}{a}+p\ln p-q\ln q\hfill \\ =\ln \left[{\left(\frac{b}{a}\right)}^{q-p+1}\frac{p^p}{q^q}\right],\hfill \end{array}\end{array}$$

(14)

where the asymptotic expansion

$$\psi \left(z\right)\sim \ln z-\frac{1}{2z}-\sum _{n=1}^{\infty }\frac{B_{2n}}{2n{z}^{2n}}$$

as $z\to \infty$ in $|argz|<\pi$ (see [33] p. 259, 6.3.18) was used, and $B_k$ stands for the Bernoulli numbers that are defined by

$$\frac{x}{e^x-1}=\sum _{k=0}^{\infty }\frac{B_k{x}^k}{k!}.$$

When $p=q$, making use of

$${\psi }^{\left(k\right)}\left(z\right)={(-1)}^{k+1}{\int }_0^{\infty }\frac{t^k}{1-e^{-t}}{e}^{-zt}\mathrm{d}t,\Re \left(z\right)>0,k\in \mathbb{N}$$

(15)

in ([33] p. 260, 6.4.1) leads to

$$\frac{{\mathrm{d}}^2\left[\ln Q(a,b;q,q;x)\right]}{\mathrm{d}{x}^2}=q^2[{\psi }^{\prime }(qx+a)-{\psi }^{\prime }(qx+b)]=q^2{\int }_0^{\infty }\frac{t}{1-e^{-t}}{e}^{-qxt}\left(e^{-at}-e^{-bt}\right)\mathrm{d}t,$$

which means that the derivative $\pm \frac{{\mathrm{d}}^2\left[\ln Q(a,b;q,q;x)\right]}{\mathrm{d}{x}^2}$ is completely monotonic on $[0,\infty )$ if and only if $a\lessgtr b$. Hence, the first derivative $\pm \frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}$ is increasing on $[0,\infty )$ if and only if $a\lessgtr b$. Meanwhile, the limits (13) and (14) become

$$\underset{x\to 0^{+}}{lim}\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}=\ln \frac{b}{a}+q[\psi \left(a\right)-\psi \left(b\right)]=[\psi \left(a\right)-\psi \left(b\right)]\left[q-\frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}\right]$$

and

$$\underset{x\to \infty }{lim}\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}=\ln \frac{b}{a}.$$

As a result,

• if $a<b$ and $q\le \frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the first derivative $\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}$ is non-negative on $[0,\infty )$;
• if $a<b$ and $q>\frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the first derivative $\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}$ has a unique zero, which is a minimum point of $\ln Q(a,b;q,q;x)$, on $(0,\infty )$;
• if $a>b$ and $q\le \frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the first derivative $\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}$ is non-positive on $[0,\infty )$;
• if $a>b$ and $q>\frac{\ln b-\ln a}{\psi \left(b\right)-\psi \left(a\right)}$, the first derivative $\frac{\mathrm{d}\left[\ln Q\right(a,b;q,q;x\left)\right]}{\mathrm{d}x}$ has a unique zero, which is a maximum point of $\ln Q(a,b;q,q;x)$, on $(0,\infty )$.

Therefore, the conclusions on $Q(a,b;q,q;x)$ are thus proved. The proof of Theorem 5 is complete. □

6. Remarks

Finally, we list additional several remarks.

Remark 8.

Combining Theorem 5 and the last relation in (6), we obtain that the Catalan–Qi function $C(a,b;x)$ is a logarithmically completely monotonic function of the second order.

Remark 9.

Similar to the introduction of the Catalan–Qi function $C(a,b;z)$ in [22], we had better leave the combinatorial interpretation of the Fuss–Catalan–Qi function $Q(a,b;p,q;z)$ to combinatorialists and number theorists.

Remark 10.

In recent years, there were many results on the Catalan numbers $C_n$ and the Catalan–Qi numbers $C(a,b;n)$ in [45,46,47,48,49,50,51,52,53] and the closely related references therein.

Remark 11.

This paper is a corrected and revised version of the preprint [54].

7. Conclusions

In this paper, we introduce a unified generalization of the Catalan numbers, the Fuss numbers, the Fuss–Catalan numbers, and the Catalan–Qi function, and discover some properties of the unified generalization, including a product-ratio expression of the unified generalization in terms of the Catalan–Qi functions, three integral representations of the unified generalization, and the logarithmically complete monotonicity of the second order for a special case of the unified generalization.