# An analytic generalization of the Catalan numbers and its integral
representation

Feng Qi  Institute of Mathematics, Henan Polytechnic University, Jiaozuo
454010, Henan, China; College of Mathematics, Inner Mongolia University for
Nationalities, Tongliao 028043, Inner Mongolia, China; School of Mathematical
Sciences, Tianjin Polytechnic University, Tianjin 300387, China
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>https://qifeng618.wordpress.com, https://orcid.org/0000-0001-6239-2968 ,  Wen-
Hui Li  Department of Fundamental Courses, Zhengzhou University of Science and
Technology, Zhengzhou 450064, Henan, China<EMAIL_ADDRESS><EMAIL_ADDRESS>https://orcid.org/0000-0002-1848-8855 ,  Jian Cao
Department of Mathematics, Hangzhou Normal University, Hangzhou 311121,
Zhejiang, China<EMAIL_ADDRESS><EMAIL_ADDRESS>https://orcid.org/0000-0002-7173-0591 ,  Da-Wei Niu  Department of Science,
Henan University of Animal Husbandry and Economy, Zhengzhou 450046, Henan,
China<EMAIL_ADDRESS><EMAIL_ADDRESS>https://orcid.org/0000-0003-4033-7911
and  Jiao-Lian Zhao  School of Mathematics and Statistics, Weinan Normal
University, Weinan 714000, Shaanxi, China<EMAIL_ADDRESS><EMAIL_ADDRESS>https://orcid.org/0000-0003-3173-7075 Dedicated to people
facing and fighting COVID-19

###### Abstract.

In the paper, the authors analytically generalize the Catalan numbers in
combinatorial number theory, establish an integral representation of the
analytic generalization of the Catalan numbers by virtue of Cauchy’s integral
formula in the theory of complex functions, and point out potential directions
to further study.

###### Key words and phrases:

Catalan number; Catalan–Qi function; Catalan–Qi number; Cauchy’s integral
formula; generalization; generating function; integral representation

###### 2010 Mathematics Subject Classification:

Primary 05A15; Secondary 11B75, 11B83, 26A09, 30E20, 41A58

This paper was typeset using AmS-LaTeX

###### Contents

  1. 1 Backgrounds and motivations
  2. 2 An explicit formula for the $2$nd kind Catalan–Qi numbers
  3. 3 An integral representation for the $2$nd kind Catalan–Qi numbers
  4. 4 Potential directions to further study
    1. 4.1 Second Catalan–Qi function
    2. 4.2 Central binomial coefficients

##  1\. Backgrounds and motivations

The Catalan numbers

$C_{n}=\frac{1}{n+1}\binom{2n}{n}=\frac{4^{n}\Gamma(n+1/2)}{\sqrt{\pi}\,\Gamma(n+2)}$
(1.1)

form a sequence of integers [8, 10, 47], have combinatorial interpretations
[10, 44], have a long history [8, 24], and can be generated [44, 49] by

$G(x)=\frac{2}{1+\sqrt{1-4x}\,}=\sum_{n=0}^{\infty}C_{n}x^{n},$ (1.2)

where

$\Gamma(z)=\lim_{n\to\infty}\frac{n!n^{z}}{\prod_{k=0}^{n}(z+k)},\quad z\neq
0,-1,-2,\dotsc$

is the classical Euler gamma function [15, 30].

In the electronic preprint [35] and its formally published version [40],
starting from the second expression in terms of gamma functions in (1.1), the
Catalan numbers $C_{n}$ were analytically generalized to the Catalan–Qi
function

$C(a,b;z)=\frac{\Gamma(b)}{\Gamma(a)}\biggl{(}\frac{b}{a}\biggr{)}^{z}\frac{\Gamma(z+a)}{\Gamma(z+b)},\quad\Re(a),\Re(b)>0,\quad\Re(z)\geq
0$ (1.3)

with

$C\biggl{(}\frac{1}{2},2;n\biggr{)}=C_{n},\quad n\geq 0.$ (1.4)

Hereafter, the Catalan–Qi function $C(a,b;z)$ and its analytic generalizations
were deeply investigated in [3, 12, 13, 17, 19, 20, 22, 23, 24, 25, 26, 27,
28, 31, 34, 36, 39, 42, 43, 46, 52, 53] and closely related references
therein.

The Catalan numbers $C_{n}$ for $n\geq 0$ have several integral
representations which have been surveyed in [24, Section 2]. The integral
representation

$C_{n}=\frac{1}{2\pi}\int_{0}^{4}\sqrt{\frac{4-x}{x}}\,x^{n}\textup{d}x,\quad
n\geq 0$ (1.5)

was discovered in [16] and applied in [43]. An alternative integral
representation

$C_{n}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{(t+1/4)^{n+2}}\textup{d}t$
(1.6)

was derived from the integral representation

$\frac{1}{1+\sqrt{1-4x}\,}=\frac{1}{2\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{1/4+t}\frac{1}{1/4+t-x}\textup{d}t,\quad
x\in\biggl{(}-\infty,\frac{1}{4}\biggr{]}$ (1.7)

which was established in [36, Theorem 1.3] by virtue of Cauchy’s integral
formula in the theory of complex functions.

The Catalan–Qi function $C(a,b;z)$ defined by (1.3) has also several integral
representations which have been surveyed in [24, Section 2]. For example,
corresponding to integral representations in (1.5) and (1.6), integral
representations

$C(a,b;x)=\biggl{(}\frac{a}{b}\biggr{)}^{b-1}\frac{1}{B(a,b-a)}\int_{0}^{b/a}\biggl{(}\frac{b}{a}-t\biggr{)}^{b-a-1}t^{x+a-1}\textup{d}t$
(1.8)

and

$C(a,b;x)=\biggl{(}\frac{a}{b}\biggr{)}^{a}\frac{1}{B(a,b-a)}\int_{0}^{\infty}\frac{t^{b-a-1}}{(t+a/b)^{x+b}}\textup{d}t.$
(1.9)

for $b>a>0$ and $x\geq 0$ were established in [31, Theorem 4], where the
classical beta function $B(z,w)$ can be defined or expressed [19, 50] by

$B(z,w)=\int_{0}^{1}t^{z-1}(1-t)^{w-1}\textup{d}t=\int_{0}^{\infty}\frac{t^{z-1}}{(1+t)^{z+w}}\textup{d}t=\frac{\Gamma(z)\Gamma(w)}{\Gamma(z+w)}$

for $\Re(z),\Re(w)>0$. We note that, when letting $a=\frac{1}{2}$ and $b=2$,
the integral representations (1.8) and (1.9) become those in (1.5) and (1.6)
respectively.

The generating function $G(x)$ in (1.2) can be regarded as a special case
$a=\frac{1}{2}$, $b=\frac{1}{4}$, and $c=1$ of the function

$G_{a,b,c}(x)=\frac{1}{a+\sqrt{b-cx}\,},\quad a,b,c>0.$

Essentially, it is better to regard the function

$G_{a,b}(x)=\frac{1}{a+\sqrt{b-x}\,},\quad a,b>0$ (1.10)

as a generalization of the generating function $G(x)$, because

$G_{1/2,1/4}(x)=G(x),\quad G_{a,b}(x)=G_{a,b,1}(x),\quad
G_{a,b,c}(x)=\frac{G_{a/\sqrt{c}\,,b/c}(x)}{\sqrt{c}\,},$

but we can not express $G_{a,b}(x)$ in terms of $G(x)$.

Now we would like to pose the following three problems.

  1. (1)

Can one establish an explicit formula for the sequence $\mathcal{C}_{n}(a,b)$
generated by

$G_{a,b}(x)=\frac{1}{a+\sqrt{b-x}\,}=\sum_{k=0}^{\infty}\mathcal{C}_{n}(a,b)x^{n}$
(1.11)

for $a,b>0$?

  2. (2)

Can one find an integral representation for the sequence
$\mathcal{C}_{n}(a,b)$ by finding an integral representation of the generating
function $G_{a,b}(x)$ in (1.10)?

  3. (3)

Can one combinatorially interpret the sequence $\mathcal{C}_{n}(a,b)$ or some
special case of $\mathcal{C}_{n}(a,b)$ except the case $a=\frac{1}{2}$ and
$b=\frac{1}{4}$?

It is easy to see that

$\lim_{a\to
0^{+}}\mathcal{C}_{n}(a,b)=\frac{(-1)^{n}}{n!}\biggl{\langle}-\frac{1}{2}\biggr{\rangle}_{n}\frac{1}{b^{(2n+1)/2}}$
(1.12)

and

$\mathcal{C}_{n}\biggl{(}\frac{1}{2},\frac{1}{4}\biggr{)}=C_{n}$ (1.13)

for $n\geq 0$, where the notation

$\langle\alpha\rangle_{n}=\prod_{k=0}^{n-1}(\alpha-k)=\begin{cases}\alpha(\alpha-1)\dotsm(\alpha-n+1),&n\geq
1\\\ 1,&n=0\end{cases}$

for $\alpha\neq 0$ is called the falling factorial [32, 33, 37]. Comparing
(1.13) with (1.4) reveals that $C(a,b;n)\not\equiv\mathcal{C}_{n}(a,b)$,
although it is possible that

$\\{C(a,b;n):n\geq 0,a,b>0\\}=\\{\mathcal{C}_{n}(a,b):n\geq 0,a,b>0\\}$

or that there exist two $2$-tuples
$(a_{n},b_{n})\in(0,\infty)\times(0,\infty)$ and
$(\alpha_{n},\beta_{n})\in(0,\infty)\times(0,\infty)$ such that
$C(a_{n},b_{n};n)=\mathcal{C}_{n}(\alpha_{n},\beta_{n})$ for all $n\geq 0$.

For our own convenience and referencing to the convention in mathematical
community, while calling $C(a,b;n)$ for $n\geq 0$ and $a,b>0$ the Cataln–Qi
numbers of the first kind, we call $\mathcal{C}_{n}(a,b)$ for $n\geq 0$ and
$a,b>0$ the Catalan–Qi numbers of the second kind.

In this paper, we will give solutions to the first two problems above:
establishing an explicit formula for the Catalan–Qi numbers of the second kind
$\mathcal{C}_{n}(a,b)$ and finding an integral representation for the
Catalan–Qi numbers of the second kind $\mathcal{C}_{n}(a,b)$ by finding an
integral representation of the generating function $G_{a,b}(x)$ in (1.10),
while leaving the third problem above to interested combinatorists.

##  2\. An explicit formula for the $2$nd kind Catalan–Qi numbers

In this section, we will establish an explicit formula for the Catalan–Qi
numbers of the second kind $\mathcal{C}_{n}(a,b)$, which gives a solution to
the first problem posed on page 1.11.

######  Theorem 2.1.

The Catalan–Qi numbers of the second kind $\mathcal{C}_{n}(a,b)$ for $n\geq 0$
and $a,b>0$ can be explicitly computed by

$\mathcal{C}_{n}(a,b)=\frac{1}{(2n)!!b^{n+1/2}}\sum_{k=0}^{n}\binom{2n-k-1}{2(n-k)}\frac{k![2(n-k)-1]!!}{\bigl{(}1+a/\sqrt{b}\,\bigr{)}^{k+1}},$
(2.1)

where the double factorial of negative odd integers $-(2\ell+1)$ is defined by

$(-2\ell-1)!!=\frac{(-1)^{\ell}}{(2\ell-1)!!}=(-1)^{\ell}\frac{(2\ell)!!}{(2\ell)!},\quad\ell\geq
0.$

###### Proof.

The Bell polynomials of the second kind
$\operatorname{B}_{n,k}(x_{1},x_{2},\dotsc,x_{n-k+1})$ for $n\geq k\geq 0$ are
defined in [5, p. 134, Theorem A] by

$\operatorname{B}_{n,k}(x_{1},x_{2},\dotsc,x_{n-k+1})=\sum_{\begin{subarray}{c}1\leq
i\leq n-k+1\\\ \ell_{i}\in\\{0\\}\cup\mathbb{N}\\\
\sum_{i=1}^{n-k+1}i\ell_{i}=n\\\
\sum_{i=1}^{n-k+1}\ell_{i}=k\end{subarray}}\frac{n!}{\prod_{i=1}^{n-k+1}\ell_{i}!}\prod_{i=1}^{n-k+1}\biggl{(}\frac{x_{i}}{i!}\biggr{)}^{\ell_{i}}.$

The famous Faà di Bruno formula can be described [5, p. 139, Theorem C] in
terms of the Bell polynomials of the second kind
$\operatorname{B}_{n,k}(x_{1},x_{2},\dotsc,x_{n-k+1})$ by

$\frac{\textup{d}^{n}}{\textup{d}x^{n}}f\circ
h(x)=\sum_{k=0}^{n}f^{(k)}(h(x))\operatorname{B}_{n,k}\bigl{(}h^{\prime}(x),h^{\prime\prime}(x),\dotsc,h^{(n-k+1)}(x)\bigr{)},$
(2.2)

where $f\circ h$ denotes the composite of the $n$-time differentiable
functions $f$ and $h$.

Let $h=h(x)=\sqrt{b-x}\,$. Then

$h^{(k)}(x)=(-1)^{k}\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{k}(b-x)^{1/2-k}\to(-1)^{k}\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{k}b^{1/2-k},\quad
x\to 0$

for $k\geq 0$ and, in light of the formula (2.2),

$\displaystyle\frac{\textup{d}^{n}G_{a,b}(x)}{\textup{d}x^{n}}=\sum_{k=0}^{n}\frac{\textup{d}^{k}}{\textup{d}h^{k}}\biggl{(}\frac{1}{1+h}\biggr{)}\operatorname{B}_{n,k}\bigl{(}h^{\prime}(x),h^{\prime\prime}(x),\dotsc,h^{(n-k+1)}(x)\bigr{)}$
$\displaystyle=\sum_{k=0}^{n}(-1)^{k}\frac{k!}{[1+h(x)]^{k+1}}\operatorname{B}_{n,k}\bigl{(}h^{\prime}(x),h^{\prime\prime}(x),\dotsc,h^{(n-k+1)}(x)\bigr{)}$
$\displaystyle\to\sum_{k=0}^{n}(-1)^{k}\frac{k!}{[1+h(0)]^{k+1}}\operatorname{B}_{n,k}\biggl{(}-\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{1}b^{-1/2},\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{2}b^{-3/2},$
$\displaystyle\dotsc,(-1)^{n-k+1}\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{n-k+1}b^{1/2-(n-k+1)}\biggr{)},\quad
x\to 0$
$\displaystyle=\sum_{k=0}^{n}(-1)^{k}\frac{k!}{\bigl{(}a+\sqrt{b}\,\bigr{)}^{k+1}}(-1)^{n}b^{k/2-n}\operatorname{B}_{n,k}\biggl{(}\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{1},\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{2},\dotsc,\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{n-k+1}\biggr{)}$
$\displaystyle=\frac{1}{2^{n}b^{n+1/2}}\sum_{k=0}^{n}k![2(n-k)-1]!!\binom{2n-k-1}{2(n-k)}\biggl{(}\frac{\sqrt{b}\,}{a+\sqrt{b}\,}\biggr{)}^{k+1},$

where we used the formulas

$\operatorname{B}_{n,k}\bigl{(}abx_{1},ab^{2}x_{2},\dotsc,ab^{n-k+1}x_{n-k+1}\bigr{)}=a^{k}b^{n}\operatorname{B}_{n,k}(x_{1},x_{2},\dotsc,x_{n-k+1})$

and

$\operatorname{B}_{n,k}\biggl{(}\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{1},\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{2},\dotsc,\biggl{\langle}\frac{1}{2}\biggr{\rangle}_{n-k+1}\biggr{)}\\\
=(-1)^{n+k}[2(n-k)-1]!!\biggl{(}\frac{1}{2}\biggr{)}^{n}\binom{2n-k-1}{2(n-k)}$
(2.3)

for $n\geq k\geq 0$, see [5, p. 135] and the formula (3.6) in the first two
lines on [32, p. 168] respectively. By the way, the formula (2.3) is connected
with [29, Remark 1], [33, Section 1.3], [38, Theorem 4], and closely related
references therein.

The equation (1.11) means that

$n!\mathcal{C}_{n}(a,b)=\lim_{x\to
0}\frac{\textup{d}^{n}G_{a,b}(x)}{\textup{d}x^{n}}.$

Consequently, we obtain the explicit formula

$\mathcal{C}_{n}(a,b)=\frac{1}{(2n)!!b^{n+1/2}}\sum_{k=0}^{n}k![2(n-k)-1]!!\binom{2n-k-1}{2(n-k)}\biggl{(}\frac{\sqrt{b}\,}{a+\sqrt{b}\,}\biggr{)}^{k+1},$

which can be rearranged as (2.1). The proof of Theorem 2.1 is complete. ∎

######  Corollary 2.1 ([38, Theorem 1.3]).

The Catalan number $C_{n}$ for $n\geq 0$ can be explicitly computed by

$C_{n}=\frac{1}{n!}\sum_{\ell=0}^{n}\binom{n+\ell-1}{2\ell}2^{\ell}(n-\ell)!(2\ell-1)!!.$
(2.4)

###### Proof.

This follows from utilizing the relation (1.13) and applying $a=\frac{1}{2}$
and $b=\frac{1}{4}$ in (2.1). The proof of Corollary 2.1 is complete. ∎

######  Remark 2.1.

Taking $a\to 0^{+}$ on both sides of the formula (2.1) and employing the
equation (1.12) result in

$\sum_{k=0}^{n}\binom{n+\ell-1}{2\ell}(n-\ell)!(2\ell-1)!!=(2n-1)!!.$

Combining (2.4) with the first equality in (1.1) gives

$\sum_{\ell=0}^{n}\binom{n+\ell-1}{2\ell}(n-\ell)!(2\ell-1)!!2^{\ell}=\frac{n!}{n+1}\binom{2n}{n}.$

Stimulated by these two identities and the formula (2.1) in Theorem 2.1, we
would like to ask a question: can one use a simple quantity to express the sum

$\sum_{\ell=0}^{n}\binom{n+\ell-1}{2\ell}(n-\ell)!(2\ell-1)!!t^{\ell}$

for $t\in\mathbb{R}\setminus\\{0,1,2\\}$?

##  3\. An integral representation for the $2$nd kind Catalan–Qi numbers

In this section, we will find an integral representation for the Catalan–Qi
numbers of the second kind $\mathcal{C}_{n}(a,b)$ by finding an integral
representation of the generating function $G_{a,b}(x)$ in (1.10), which gives
a solution to the second problem posed on page 1.11.

######  Theorem 3.1.

The principal branch of the generating function $G_{a,b}(z)$ for $a,b>0$ can
be represented by

$G_{a,b}(z)=\frac{1}{a+\sqrt{b-z}\,}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{b+t-z}\textup{d}t,\quad
z\in\mathbb{C}\setminus[b,\infty).$ (3.1)

Consequently, the Catalan–Qi numbers of the second kind $\mathcal{C}_{n}(a,b)$
for $a,b>0$ can be represented by

$\mathcal{C}_{n}(a,b)=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{n+1}}\textup{d}t,\quad
n\geq 0.$ (3.2)

###### Proof.

Let

$F(z)=\frac{1}{a+\exp\frac{\ln(-z)}{2}},\quad
z\in\mathbb{C}\setminus[0,\infty),\quad\arg z\in(0,2\pi),$

where $i=\sqrt{-1}\,$ is the imaginary unit and $\arg z$ stands for the
principal value of the argument of $z$. By virtue of Cauchy’s integral formula
[6, p. 113] in the theory of complex functions, for any fixed point
$z_{0}=x_{0}+iy_{0}\in\mathbb{C}\setminus[0,\infty)$, we have

$F(z_{0})=\frac{1}{2\pi i}\int_{L}\frac{F(\xi)}{\xi-z_{0}}\textup{d}\xi,$

where $L$ is a positively oriented contour $L(r,R)$ in
$\mathbb{C}\setminus[0,\infty)$, as showed in Figure 1,

Figure 1. The positively oriented contour $L(r,R)$ in
$\mathbb{C}\setminus[0,\infty)$

satisfying

  1. (1)

$0<r<|z_{0}|<R$;

  2. (2)

$L(r,R)$ consists of the half circle $z=re^{i\theta}$ for
$\theta\in\bigl{[}\frac{\pi}{2},\frac{3\pi}{2}\bigr{]}$;

  3. (3)

$L(r,R)$ consists of the half lines $z=x\pm ir$ for $x\in(0,R(r)]$, where
$R(r)=\sqrt{R^{2}-r^{2}}\,$;

  4. (4)

$L(r,R)$ consists of the circular arc $z=Re^{i\theta}$ for

$\theta\in\biggl{(}\arctan\frac{r}{R(r)},2\pi-\arctan\frac{r}{R(r)}\biggr{)};$

  5. (5)

the half lines $z=x\pm ir$ for $x\in(0,R(r)]$ cut the circle $|z|=R$ at the
points $R(r)\pm ir$ and $R(r)\to R$ as $r\to 0^{+}$.

The integral on the circular arc $z=Re^{i\theta}$ equals

$\displaystyle\quad\frac{1}{2\pi
i}\int_{\arcsin[r/R(r)]}^{2\pi-\arcsin[r/R(r)]}\frac{Rie^{i\theta}}{\bigl{(}Re^{i\theta}-z_{0}\bigr{)}\bigl{[}a+\exp\frac{\ln(-Re^{i\theta})}{2}\bigr{]}}\textup{d}\theta$
$\displaystyle=\frac{1}{2\pi}\int_{\arcsin[r/R(r)]}^{2\pi-\arcsin[r/R(r)]}\frac{1}{\bigl{(}1-\frac{z_{0}}{Re^{i\theta}}\bigr{)}\bigl{[}a+\exp\frac{\ln(-Re^{i\theta})}{2}\bigr{]}}\textup{d}\theta$
$\displaystyle=\frac{1}{2\pi}\int_{\arcsin[r/R(r)]}^{2\pi-\arcsin[r/R(r)]}\frac{1}{\bigl{(}1-\frac{z_{0}}{Re^{i\theta}}\bigr{)}\bigl{[}a+\exp\frac{\ln
R+i\arg(-Re^{i\theta})}{2}\bigr{]}}\textup{d}\theta$ $\displaystyle\to 0$

uniformly as $R\to\infty$.

The integral on the half circle $z=re^{i\theta}$ for
$\theta\in\bigl{[}\frac{\pi}{2},\frac{3\pi}{2}\bigr{]}$ is

$\displaystyle\quad-\frac{1}{2\pi
i}\int_{\pi/2}^{3\pi/2}\frac{rie^{i\theta}}{\bigl{(}re^{i\theta}-z_{0}\bigr{)}\bigl{[}a+\exp\frac{\ln(-re^{i\theta})}{2}\bigr{]}}\textup{d}\theta$
$\displaystyle=-\frac{1}{2\pi}\int_{\pi/2}^{3\pi/2}\frac{re^{i\theta}}{re^{i\theta}-z_{0}}\frac{1}{a+\exp\frac{\ln(-re^{i\theta})}{2}}\textup{d}\theta$
$\displaystyle=-\frac{1}{2\pi}\int_{\pi/2}^{3\pi/2}\frac{re^{i\theta}}{re^{i\theta}-z_{0}}\frac{1}{a+\exp\frac{\ln
r+i\arg(-re^{i\theta})}{2}}\textup{d}\theta$
$\displaystyle=-\frac{1}{2\pi}\int_{\pi/2}^{3\pi/2}\frac{re^{i\theta}}{re^{i\theta}-z_{0}}\frac{1}{a+\sqrt{r}\,\exp\frac{i\arg(-re^{i\theta})}{2}}\textup{d}\theta$
$\displaystyle\to 0$

uniformly as $r\to 0^{+}$.

Since

$\displaystyle F(x+ir)$ $\displaystyle=\frac{1}{a+\exp\frac{\ln(-x-ri)}{2}}$
$\displaystyle=\frac{1}{a+\exp\frac{\ln\sqrt{x^{2}+r^{2}}\,+i[\arctan(r/x)-\pi]}{2}}$
$\displaystyle=\frac{1}{a+\sqrt[4]{x^{2}+r^{2}}\,\bigl{[}\cos\frac{\arctan(r/x)-\pi}{2}+i\sin\frac{\arctan(r/x)-\pi}{2}\bigr{]}}$
$\displaystyle=\frac{1}{a+\sqrt[4]{x^{2}+r^{2}}\,\bigl{[}\sin\frac{\arctan(r/x)}{2}-i\cos\frac{\arctan(r/x)}{2}\bigr{]}}$
$\displaystyle=\frac{1}{a+\sqrt[4]{x^{2}+r^{2}}\,\sin\frac{\arctan(r/x)}{2}-i\sqrt[4]{x^{2}+r^{2}}\,\cos\frac{\arctan(r/x)}{2}}$
$\displaystyle=\frac{a+\sqrt[4]{x^{2}+r^{2}}\,\sin\frac{\arctan(r/x)}{2}+i\sqrt[4]{x^{2}+r^{2}}\,\cos\frac{\arctan(r/x)}{2}}{\bigl{[}a+\sqrt[4]{x^{2}+r^{2}}\,\sin\frac{\arctan(r/x)}{2}\bigr{]}^{2}+\bigl{[}\sqrt[4]{x^{2}+r^{2}}\,\cos\frac{\arctan(r/x)}{2}\bigr{]}^{2}}$
$\displaystyle\to\frac{a+i\sqrt{x}\,}{a^{2}+x}$

as $r\to 0^{+}$ and $\overline{F(z)}=F(\bar{z})$, the integral on the half
lines $z=x\pm ir$ for $x\in(0,R(r)]$ is equal to

$\displaystyle\quad\frac{1}{2\pi
i}\Biggl{[}\int_{0}^{R(r)}\frac{F(x+ir)}{x+ir-
z_{0}}\textup{d}x+\int_{R(r)}^{0}\frac{F(x-ir)}{x-ir-
z_{0}}\textup{d}x\Biggr{]}$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{R(r)}\frac{(x-ir-z_{0})F(x+ir)-(x+ir-z_{0})F(x-ir)}{(x+ir-
z_{0})(x-ir-z_{0})}\textup{d}x$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{R(r)}\frac{(x-z_{0})[F(x+ir)-F(x-ir)]-ir[F(x+ir)+F(x-ir)]}{(x+ir-
z_{0})(x-ir-z_{0})}\textup{d}x$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{R(r)}\frac{(x-z_{0})\bigl{[}F(x+ir)-F\bigl{(}\overline{x+ir}\bigr{)}\bigr{]}-ir\bigl{[}F(x+ir)+F\bigl{(}\overline{x+ir}\bigr{)}\bigr{]}}{(x+ir-
z_{0})(x-ir-z_{0})}\textup{d}x$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{R(r)}\frac{(x-z_{0})\bigl{[}F(x+ir)-\overline{F(x+ir)}\,\bigr{]}-ir\bigl{[}F(x+ir)+\overline{F(x+ir)}\,\bigr{]}}{(x+ir-
z_{0})(x-ir-z_{0})}\textup{d}x$ $\displaystyle=\frac{1}{2\pi
i}\int_{0}^{R(r)}\frac{(x-z_{0})[2i\Im(F(x+ir))]-ir[2\Re(F(x+ir))]}{(x+ir-
z_{0})(x-ir-z_{0})}\textup{d}x$ $\displaystyle\to\frac{1}{2\pi
i}\int_{0}^{\infty}\frac{2i}{x-z_{0}}\frac{\sqrt{x}\,}{a^{2}+x}\textup{d}x,\quad
r\to 0^{+},\quad R\to\infty$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{x}}{(a^{2}+x)(x-z_{0})}\textup{d}x.$

Consequently, it follows that

$\frac{1}{a+\exp\frac{\ln(-z_{0})}{2}}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{x}}{(a^{2}+x)(x-z_{0})}\textup{d}x$
(3.3)

for any $z_{0}\in\mathbb{C}\setminus[0,\infty)$ and $\arg z_{0}\in(0,2\pi)$.
Due to the point $z_{0}$ in (3.3) being arbitrary, the integral formula (3.3)
can be rearranged as

$F(z)=\frac{1}{a+\exp\frac{\ln(-z)}{2}}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{x}}{(a^{2}+x)(x-z)}\textup{d}x$
(3.4)

for $z\in\mathbb{C}\setminus[0,\infty)$ and $\arg z\in(0,2\pi)$.

Let

$f(z)=\frac{1}{a+\exp\frac{\ln(b-z)}{2}},\quad
z\in\mathbb{C}\setminus[b,\infty),\quad\arg(z-b)\in(0,2\pi).$

Then $f(z)=F(z-b)$. Therefore, from (3.4), it follows that

$f(z)=\frac{1}{a+\exp\frac{\ln(b-z)}{2}}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{x}}{(a^{2}+x)(b+x-z)}\textup{d}x$

for $z\in\mathbb{C}\setminus[b,\infty)$ and $\arg(z-b)\in(0,2\pi)$. The
integral representation (3.1) is thus proved.

Differentiating $n\geq 0$ times with respect to $z$ on both sides of (3.1) and
taking the limit $z\to 0$ yield

$\displaystyle G_{a,b}^{(n)}(z)$
$\displaystyle=\frac{\textup{d}^{n}}{\textup{d}z^{n}}\biggl{(}\frac{1}{a+\sqrt{b-z}\,}\biggr{)}$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{\textup{d}^{n}}{\textup{d}z^{n}}\biggl{(}\frac{1}{b+t-z}\biggr{)}\textup{d}t$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{n!}{(b+t-z)^{n+1}}\textup{d}t$
$\displaystyle\to\frac{n!}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{n+1}}\textup{d}t,\quad
z\to 0.$

As a result, by virtue of (1.11), we have

$\mathcal{C}_{n}(a,b)=\frac{G_{a,b}^{(n)}(0)}{n!}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{n+1}}\textup{d}t.$

The integral representation (3.2) for the Catalan–Qi numbers of the second
kind $\mathcal{C}_{n}(a,b)$ is thus proved. The proof of Theorem 3.1 is
complete. ∎

######  Remark 3.1.

When taking $z=x\in(-\infty,b)$, the integral formula (3.1) becomes

$\frac{1}{a+\sqrt{b-x}\,}=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{b+t-x}\textup{d}t,\quad
a,b>0.$ (3.5)

When taking $x\to b^{-}$, the integral in (3.5) converges. Consequently, the
integral representation (3.5) is valid on $(-\infty,b]$.

######  Remark 3.2.

When taking $a=\frac{1}{2}$ and $b=\frac{1}{4}$, integral representations
(3.1) and (3.2) reduce to (1.7) and (1.6) respectively.

######  Remark 3.3.

Combining the explicit formula (2.1) with the integral representation (3.2)
and simplifying lead to

$\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{n+1}}\textup{d}t=\frac{\pi}{(2n)!!b^{n+1/2}}\sum_{k=0}^{n}\binom{2n-k-1}{2(n-k)}\frac{k![2(n-k)-1]!!}{\bigl{(}1+a/\sqrt{b}\,\bigr{)}^{k+1}}$

for $a,b>0$ and $n\geq 0$.

##  4\. Potential directions to further study

In this section, we will try to point out two potential directions to further
study.

###  4.1. Second Catalan–Qi function

Motivated by (3.2), we can consider the function

$\mathcal{C}(a,b;z)=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{z+1}}\textup{d}t,\quad
a,b>0,\quad\Re(z)\geq 0$ (4.1)

and call it the second Catalan–Qi function, while calling $C(a,b;z)$ in (1.3)
the first Catalan–Qi function.

We can study the second Catalan–Qi function $\mathcal{C}(a,b;z)$ as a function
of three variables $a,b$, and $z$. It is easy to see that

$\displaystyle\frac{\textup{d}^{n}\mathcal{C}(a,b;z)}{\textup{d}b^{n}}$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{\textup{d}^{n}}{\textup{d}b^{n}}\biggl{[}\frac{1}{(b+t)^{z+1}}\biggr{]}\textup{d}t$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{\langle-(z+1)\rangle_{n}}{(b+t)^{z+n+1}}\textup{d}t$
$\displaystyle=(-1)^{n}\frac{(z+1)_{n}}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{z+n+1}}\textup{d}t,$

where the rising factorial $(z)_{n}$ is defined [32, 37] by

$(z)_{n}=\prod_{\ell=0}^{n-1}(z+\ell)=\begin{cases}z(z+1)\dotsm(z+n-1),&n\geq
1\\\ 1,&n=0\end{cases}$

which is also known as the Pochhammer symbol or shifted factorial in the
theory of special functions [15, 50]. This means that the second Catalan–Qi
function $\mathcal{C}(a,b;z)$ is a completely monotonic function with respect
to $b\in(0,\infty)$. Utilizing complete monotonicity [18, 21, 45], we can
derive many new analytic properties of the second Catalan–Qi function
$\mathcal{C}(a,b;z)$.

In a word, employing the integral representation (4.1), we believe that we can
discover some new properties of the second Catalan–Qi function
$\mathcal{C}(a,b;z)$, of the Catalan–Qi numbers of the second kind
$\mathcal{C}_{n}(a,b)$, and the Catalan numbers $C_{n}$. For the sake of
length limit of this paper, we would not like to further study in details.

###  4.2. Central binomial coefficients

It is known that

$\frac{1}{\sqrt{1-4x}\,}=\sum_{n=0}^{\infty}\binom{2n}{n}x^{n}=1+2x+6x^{2}+20x^{3}+70x^{4}+252x^{5}+\dotsm,$
(4.2)

where $\binom{2n}{n}$ is called central binomial coefficient. It has been an
attracted point for mathematicians to study central binomial coefficients. For
example, we can rewritten [41, Lemma 3] as

$\sum_{k=0}^{n-1}\binom{2k}{k}\frac{1}{(k+1)4^{k}}=2\biggl{[}1-\frac{1}{4^{n}}\binom{2n}{n}\biggr{]}$

and

$\sum_{k=0}^{n-1}\binom{2k}{k}\frac{4^{n-k}}{n-k}=2\binom{2n}{n}\sum_{k=1}^{n}\frac{1}{2k-1}.$

For more information on results at this point, please refer to [1, 2, 4, 7, 9,
11, 14, 24, 41, 48, 51] and closely related references therein.

Combining (4.2) with (1.11) and (3.2) arrives at

$\displaystyle\binom{2n}{n}$ $\displaystyle=\frac{1}{n!}\lim_{x\to
0}\frac{\textup{d}^{n}}{\textup{d}x^{n}}\biggl{(}\frac{1}{\sqrt{1-4x}\,}\biggr{)}$
$\displaystyle=\frac{1}{n!}\lim_{x\to
0}\frac{\textup{d}^{n}}{\textup{d}x^{n}}\left(\frac{1}{2}\lim_{\begin{subarray}{c}a\to
0^{+}\\\ b\to 1/4\end{subarray}}\frac{1}{a+\sqrt{b-x}\,}\right)$
$\displaystyle=\frac{1}{n!}\frac{1}{2}\lim_{\begin{subarray}{c}a\to 0^{+}\\\
b\to 1/4\end{subarray}}\lim_{x\to
0}\frac{\textup{d}^{n}}{\textup{d}x^{n}}\biggl{(}\frac{1}{a+\sqrt{b-x}\,}\biggr{)}$
$\displaystyle=\frac{1}{2}\lim_{\begin{subarray}{c}a\to 0^{+}\\\ b\to
1/4\end{subarray}}\mathcal{C}_{n}(a,b)$
$\displaystyle=\frac{1}{2}\lim_{\begin{subarray}{c}a\to 0^{+}\\\ b\to
1/4\end{subarray}}\frac{1}{\pi}\int_{0}^{\infty}\frac{\sqrt{t}\,}{a^{2}+t}\frac{1}{(b+t)^{n+1}}\textup{d}t$
$\displaystyle=\frac{1}{2\pi}\int_{0}^{\infty}\frac{1}{\sqrt{t}\,}\frac{1}{(1/4+t)^{n+1}}\textup{d}t$
$\displaystyle=\frac{1}{\pi}\int_{0}^{\infty}\frac{1}{(1/4+s^{2})^{n+1}}\textup{d}s$
$\displaystyle=\frac{2^{2n+1}}{\pi}\int_{0}^{\infty}\frac{1}{(1+t^{2})^{n+1}}\textup{d}t.$

This integral representation should provides an effective tool for further
studying central binomial coefficients $\binom{2n}{n}$.

### Competing interests

The authors declare that they have no competing interests.

### Author’s contributions

All authors contribute equally to the manuscript.

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