# An extension for Rice’s integral and applications

## Abstract

In this paper, we generalize Rice’s integral and give some applications. Some old and new binomial identities are obtained.

## 1 Introduction

For complex z and $$m, n \in\mathbb{N}_{0}=\mathbb{N} \cup\{0\}$$, we define that

$$(z)_{0} = 1,\qquad (z)_{n}=z(z+1)\cdots(z+n-1)$$
(1)

and

$$(z+m)_{n-m}= \left \{ \begin{array}{@{}l@{\quad}l} (m+z)(m+z+1)\cdots(z+n-1), & \mbox{when } 0 \leq m < n,\\ 1, & \mbox{when } m \geq n. \end{array} \right .$$
(2)

The standard Bell polynomials (see [1, p.133]). The modified Bell polynomials $$L_{m}(x_{1}, x_{2},\ldots)$$ are defined by

$$\exp \Biggl(\sum_{k= 1}^{\infty}x_{k} \frac{t^{k}}{k} \Biggr)=1+\sum_{m=1}^{\infty}L_{m}(x_{1},x_{2},\ldots) \frac{t^{m}}{m!}.$$
(3)

We have

$$L_{m}(x_{1},x_{2},\ldots)=\sum _{m_{1}+2m_{2}+3m_{3}+\cdots=m}\frac {m!}{m_{1}!m_{2}!m_{3}! \cdots} \biggl(\frac{x_{1}}{1} \biggr)^{m_{1}} \biggl(\frac{x_{2}}{2} \biggr)^{m_{2}} \biggl( \frac{x_{3}}{3} \biggr)^{m_{3}}\cdots.$$
(4)

Recalling Rice’s integral (see [2] and [3]), suppose that $$f(z)$$ is a rational function analytic on $$[0,+\infty[$$, and $$n \in\mathbb{N}_{0}$$, then we have

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}f(k)=\frac{(-1)^{n}}{2\pi{i}}\int_{\mathcal{C}} \frac{n!}{z(z-1)\cdots(z-n)}f(z)\,dz,$$
(5)

where $$\mathcal{C}$$ is a positively oriented, encloses the simple poles at the integers $$0, 1, 2,\ldots,n$$ and no others.

All technicalities and many examples of Rice’s integral are involved in [2] and [3].

Well-known is the following binomial identity: For $$n \in\mathbb{N}=\{ 1,2,\ldots\}$$ and $$\theta> 0$$, we have

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\frac{\theta}{\theta +k}=\prod_{k=1}^{n} \frac{k}{\theta+k}.$$
(6)

Recently, Jonathon [4] gave a simple and interesting probabilistic proof of the above binomial identity, and he further extended this binomial identity based on the probabilistic method. We state his result as follows. For all $$r, n \in\mathbb{N}$$; $$\theta> 0$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta }{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(1+\sum_{j=1}^{r-1} \sum _{1\leq k_{1}\leq\cdots\leq k_{j}\leq n }\frac{\theta^{j}}{(\theta +k_{1})(\theta+k_{2})\cdots(\theta+k_{j})} \Biggr). \end{aligned}
(7)

More recently, Bayad et al. [5, 6] established new identities for some new polynomials associated with the λ-array type polynomials. Simsek [6] gave an approach to combinatorial sums involving binomial coefficients and the Catalan number from the analysis of the Bernstein basis functions. There is much literature about the binomial identities, we refer the readers to [717].

In the present paper, we further extend Rice’s integral by using Cauchy’s integral formula. As some applications, some old and new binomial identities are obtained.

## 2 Rice’s integral

In this section, we generalize Rice’s integral using Cauchy’s integral formula.

### Theorem 1

Let $$f(z)$$ be a rational function analytic on $$[0,+\infty[$$ and $$m, n \in\mathbb{N}_{0}$$. Then we have

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} f(k)=\frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int _{\mathcal{C}}\frac{(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)}f(z)\,dz,$$
(8)

where $$\mathcal{C}$$ is a positively oriented, encloses the poles $$0, 1, 2,\ldots,n$$ and no others.

### Proof

By using Cauchy’s integral formula, via the residues calculus, we have

\begin{aligned} &\frac{(-1)^{n}n!}{2\pi{i}(n-m)!} \int_{\mathcal{C}}\frac {(z+m+1)_{n-m}}{z(z-1)\cdots(z-n)}f(z)\,dz\\ &\quad= \frac{(-1)^{n}n!}{(n-m)!}\sum_{k=1}^{n+1} \mathop{\operatorname {Res}}\limits_{z=k-1}\frac {(z+m+1)_{n-m}}{z(z-1)\cdots(z-n)}f(z) \\ &\quad=\frac{(-1)^{n}n!}{(n-m)!}\sum_{k=1}^{n+1}\lim _{z\rightarrow k-1}\frac{(z+m+1)_{n-m}}{z(z-1)\cdots(z-k+2)(z-k)\cdots(z-n)}f(z) \\ &\quad=\frac{(-1)^{n}n!}{(n-m)!} \sum_{k=1}^{n+1} \frac {(m+k)_{n-m}}{(k-1)\cdots2\cdot1(-1)(-2)\cdots(-n+k-1)}f(k-1) \\ &\quad=\frac{n!}{(n-m)!}\sum_{k=0}^{n}(-1)^{k} \frac {(m+k+1)_{n-m}}{k!(n-k)!}f(k) \\ &\quad=\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k}f(k). \end{aligned}

The proof is complete. □

### Corollary 2

Let $$f(z)$$ be a rational function analytic on $$[0,+\infty[$$ and $$n \in\mathbb{N}_{0}$$. Then we have

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} f(k)=\frac{(-1)^{n}}{2\pi{i}}\int _{\mathcal{C}}\frac {(z+1)_{n}}{z(z-1)\cdots(z-n)}f(z)\,dz,$$
(9)

where $$\mathcal{C}$$ is a positively oriented, encloses the poles $$0, 1, 2,\ldots,n$$ and no others.

### Proof

Taking $$m=0$$ in Theorem 1, Corollary 2 follows. □

### Remark 3

Taking $$m=n$$ in Theorem 1 and noting that $$(z+m+1)_{0}= 1$$, we get Rice’s integral (5) immediately. Obviously, Theorem 1 is an extension of Rice’s integral (5).

## 3 Applications

In this section, we give some applications of Theorem 1 and Corollary 2. We can obtain many old and new binomial identities.

Below we give an extension of binomial identity (7) and deduce some new binomial identities.

### Theorem 4

Let $$r, l, n \in\mathbb{N}$$, $$m \in\mathbb{N}_{0}$$, $$0 \leq m \leq n$$; $$\theta> 0$$, $$\beta\neq0$$.

When $$\theta\notin\{m+1, m+2,\ldots, n\}$$ and $$r>0$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr) \\ &\qquad{}\times\Biggl( \biggl(\frac{\beta }{\theta} \biggr)^{r}+ \sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta +s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{\theta(k_{1}-\theta)\cdots (k_{i}-\theta)(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}
(10)

When $$\theta\in\{m+1, m+2,\ldots, n\}$$ and $$r>1$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \\ &\qquad{}\times \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr) \Biggl(\frac {\beta^{r}}{\theta^{r-1}}+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta +s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-m-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta ^{r}}{\theta(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}
(11)

### Proof

Taking $$f(z)= (\frac{\beta}{\theta+z} )^{r}$$ in (8), and $$\mathcal{C}$$ is a positively oriented, encloses the poles $$0, 1, 2,\ldots,n$$ and no others, by means of the residues calculus, we obtain the following.

• When $$\theta\notin\{m+1, m+2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!} \int_{\mathcal{C}}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r}\,dz \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!} \mathop{\operatorname {Res}}\limits_{z=-\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(\theta+z)^{r-1}\bigr]\frac{(z+1+m)\cdots (z+n)\beta^{r}}{z(z-1)\cdots(z-n)} \\ &\quad= \frac{n!}{(n-m)!}\bigl[z^{r-1}\bigr] \Biggl(\prod _{k=1}^{n}\frac{1}{\theta +k} \Biggr) \Biggl(\prod _{k=m+1}^{n}(k-\theta) \Biggr) \frac{(1+\frac {z}{m+1-\theta})\cdots(1+\frac{z}{n-\theta})\beta^{r}}{\theta (1-\frac{z}{\theta})(1-\frac{z}{\theta+1})\cdots(1-\frac {z}{\theta+n})} \\ &\quad= \Biggl(\prod_{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr)\bigl[z^{r -l-1}\bigr] \Biggl(1+\sum _{i=1}^{n-m}\sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \frac{z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots (\theta+s_{l})} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr)\bigl[z^{r -l-1}\bigr] \Biggl(1+\sum _{i=1}^{n-m}\sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \frac{z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{\theta^{l+1}}+\sum_{j=1}^{l} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\beta ^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \Biggr)\\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1}^{n} \frac{k-\theta}{k-m} \Biggr) \Biggl( \biggl(\frac{\beta }{\theta} \biggr)^{r}+ \sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta +s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{\theta(k_{1}-\theta)\cdots (k_{i}-\theta)(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}

• When $$\theta\in\{m+1, m+2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{\theta+k} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r}\,dz \\ &\quad=\frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=-\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{\theta +z} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=-\theta}\frac{(z+1+m)\cdots (z+\theta-1)(z+\theta+1)\cdots(z+n-\theta)}{z(z-1)\cdots(z-n)} \frac{\beta^{r}}{(\theta+z)^{r-1}} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(\theta+z)^{r-2}\bigr]\frac{(z+1+m)\cdots (z+\theta-1)(z+\theta+1)\cdots(z+n-\theta)\beta^{r}}{z(z-1)\cdots (z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r-2}\bigr]\frac{(z+1+m-\theta)\cdots (z-1)(z+1)\cdots(z+n-\theta)\beta^{r}}{(z-\theta)(z-\theta -1)\cdots(z-n-\theta)} \\ &\quad= \frac{n!}{(n-m)!}\bigl[z^{r-2}\bigr] \Biggl(\prod _{k=1}^{n}\frac{1}{\theta +k} \Biggr) \Biggl(\prod _{k=m+1,k\neq\theta}^{n}(k-\theta) \Biggr)\\ &\qquad{}\times \frac{(1+\frac{z}{m+1-\theta})\cdots(1-z)(1+z)\cdots(1+\frac {z}{n-\theta})\beta^{r}}{\theta(1-\frac{z}{\theta})(1-\frac {z}{\theta+1})\cdots(1-\frac{z}{\theta+n})} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\bigl[z^{r -l-2}\bigr] \\ &\qquad{}\times\Biggl(1+\sum _{i=1}^{n-m-1}\mathop{\sum_{k_{1},\ldots ,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \frac {z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots (\theta+s_{l})} \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\bigl[z^{r -l-2}\bigr] \\ &\qquad{}\times\Biggl(1+\sum _{i=1}^{n-m-1}\mathop{\sum_{k_{1},\ldots ,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \frac {z^{i}}{(k_{1}-\theta)\cdots(k_{i}-\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{\theta^{l+1}}+\sum_{j=1}^{l} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\beta ^{r}}{\theta(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \Biggr) \\ &\quad= \frac{1}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta+k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr)\\ &\qquad{}\times \Biggl(\frac {\beta^{r}}{\theta^{r-1}}+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{\theta(\theta +s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-m-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta ^{r}}{\theta(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}

This proof is complete. □

Setting $$r=1$$ and $$\theta=\beta$$ in (10) of Theorem 4 and $$r=2$$ and $$\theta=\beta$$ in (11) of Theorem 4, we have the following new binomial identity.

### Corollary 5

Let $$n \in\mathbb{N}$$, $$m \in\mathbb{N}_{0}$$, $$0 \leq m \leq n$$; $$\theta> 0$$.

When $$\theta\notin\{m+1, m+2,\ldots, n\}$$, we have

\begin{aligned} \sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \frac{\theta}{\theta+k} = \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(\prod _{k=m+1}^{n}\frac{k-\theta}{k-m} \Biggr). \end{aligned}
(12)

When $$\theta\in\{m+1, m+2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2} \\ &\quad=\frac{\theta}{(n-m)!} \Biggl(\prod_{k=1}^{n} \frac{k}{\theta +k} \Biggr) \Biggl(\prod_{k=m+1,k\neq\theta}^{n}(k- \theta) \Biggr). \end{aligned}
(13)

### Corollary 6

(see [4])

For $$r, n \in\mathbb{N}$$; $$\theta> 0$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta }{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(1+\sum_{j=1}^{r-1}\sum _{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta +s_{j})} \Biggr). \end{aligned}
(14)

### Proof

Take $$m=n$$, $$\beta= \theta$$ in (10) of Theorem 4, it is a must that $$\theta\notin\{m+1, m+2,\ldots, n\}$$, and define that $$\prod_{k=n+1}^{n}=1$$, $$\sum_{i=1}^{0}=0$$, then Corollary 6 follows. □

Setting $$r=0$$ in (14), defining the empty sum $$\sum_{j=1}^{-1}=-1$$, we have the following familiar formula.

### Corollary 7

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}=0.$$
(15)

Setting $$r=1$$ in (14), defining the empty sum $$\sum_{j=1}^{0}=0$$, we have the following.

### Corollary 8

(see [4])

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\frac{\theta}{\theta+k} =\prod_{k=1}^{n} \frac{k}{\theta+k}.$$
(16)

Setting $$m=2$$ in (14), we have the following.

### Corollary 9

(see [4])

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2} = \Biggl(\prod _{k=1}^{n}\frac{k}{\theta+k} \Biggr) \Biggl(1+\sum_{k=1}^{n}\frac{\theta}{\theta+k} \Biggr).$$
(17)

### Corollary 10

Let $$r, l, n \in\mathbb{N}$$ and $$\theta> 0$$.

When $$\theta\notin\{1, 2,\ldots, n\}$$,

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{k-\theta}{\theta+k} \Biggr) \Biggl(1+\sum_{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac{\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots<k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta^{r-1}}{(k_{1}-\theta)\cdots(k_{i}-\theta )(\theta+s_{1})\cdots(\theta+s_{l})} \Biggr). \end{aligned}
(18)

When $$\theta\in\{1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{r} \\ &\quad= \Biggl(\prod_{k=1}^{n} \frac{1}{\theta+k} \Biggr) \Biggl(\prod_{k=1,k \neq\theta}^{n}(k- \theta) \Biggr) \Biggl(\theta+\sum_{j=1}^{r-2} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(\theta+s_{1})(\theta+s_{2})\cdots(\theta+s_{j})} \\ &\qquad{} +\sum_{i+l=r-2}\sum_{i=1}^{n-1} \mathop{\sum_{k_{1},\ldots,k_{i}\neq\theta}}_{ 1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta ^{r-1}}{(k_{1}-\theta)\cdots(k_{i}-\theta)(\theta+s_{1})\cdots (\theta+s_{l})} \Biggr). \end{aligned}
(19)

### Proof

Take $$m=0$$, $$\beta= \theta$$ in Theorem 4, then Corollary 10 follows. □

Setting $$r=0$$ in (18) and defining the suitable empty sum, we have the following familiar formula.

### Corollary 11

([7])

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k}= (-1)^{n}.$$
(20)

Setting $$r=1$$ and $$\theta=\beta$$ in (18) and defining the empty sum $$\sum_{j=1}^{0}=0$$, we have the following new binomial identity.

### Corollary 12

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k}\frac{\theta}{\theta+k}= \Biggl(\prod _{k=1}^{n}\frac {k-\theta}{\theta+k} \Biggr).$$
(21)

Setting $$r=2$$ and $$\theta=\beta$$ in (18) and defining the empty sum $$\sum_{j=1}^{0}=0$$, we have the following new binomial identity.

### Corollary 13

For all $$n \in\mathbb{N}$$ and $$\theta> 0$$,

$$\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{\theta+k} \biggr)^{2}= \Biggl(\prod_{k=1}^{n}\frac{k-\theta}{\theta+k} \Biggr) \Biggl(1+\sum_{k=1}^{n} \frac{\theta}{\theta+k} \Biggr).$$
(22)

### Theorem 14

For $$r,l,n \in\mathbb{N}$$, $$0 \leq m \leq n$$, $$\beta\neq0$$.

When $$\theta\notin\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \Biggl(- \biggl(-\frac{\beta}{\theta} \biggr)^{r}+\sum _{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac {\beta^{r}}{\theta(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta )\cdots(s_{l}-\theta)} \Biggr). \end{aligned}
(23)

When $$\theta\in\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta } \Biggl( \biggl(- \frac{\beta}{\theta} \biggr)^{r}+\sum_{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\beta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}
(24)

### Proof

By means of Theorem 1 and the residues calculus, we obtain the following.

• When $$\theta\notin\{0,1, 2,\ldots, n\}$$, taking $$f(z)= (\frac{\beta}{z-\theta} )^{r}$$ in (8), and $$\mathcal{C}$$ is a positively oriented, encloses the poles $$0, 1, 2,\ldots,n$$ and no others.

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta } \biggr)^{r}\,dz\\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac{(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-n)}\frac{\beta^{r}}{(z-\theta)^{r}} \\ &\quad=\frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(z-\theta)^{r-1}\bigr]\frac {(z+1+m)(z+2+m)\cdots(z+n)\beta^{r}}{z(z-1)\cdots(z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r-1}\bigr]\frac{(z+1+m+\theta )(z+2+m+\theta)\cdots(z+n+\theta)\beta^{r}}{(z+\theta)(z+\theta -1)\cdots(z+\theta-n)} \\ &\quad= (-1)^{n+1}\frac{n!(\theta-n-1){!}(m+1+\theta)_{n-m}}{(n-m)!\theta {!}}\bigl[z^{r-1}\bigr] \frac{(1+\frac{z}{m+1+\theta})\cdots(1+\frac {z}{n+\theta})\beta^{r}}{(1+\frac{z}{\theta})(1+\frac{z}{\theta -1})\cdots(1+\frac{z}{\theta-n})} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \bigl[z^{r-l-1}\bigr] \Biggl(1+\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{l}-\theta)} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \bigl[z^{r-l-1}\bigr] \Biggl(1+\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times \Biggl((-1)^{l}\frac{\beta^{r}}{\theta^{l+1}}+\sum _{j=1}^{l} \ \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \Biggr) \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{m+\theta } \Biggl(- \biggl(-\frac{\beta}{\theta} \biggr)^{r}+\sum _{j=1}^{r-1}\ \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\beta ^{r}}{\theta(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac {\beta^{r}}{\theta(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta )\cdots(s_{l}-\theta)} \Biggr). \end{aligned}

• When $$\theta\in\{0,1, 2,\ldots, n\}$$, taking $$f(z)= (\frac{\beta}{z-\theta} )^{r}$$ in (8), and $$\mathcal {C}$$ is a positively oriented, encloses the poles $$0, 1, \ldots,\theta -1,\theta+1,\ldots,n$$ and no others.

\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{m+k} \biggl(\frac{\beta}{k-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n}n!}{2\pi{i}(n-m)!}\int_{\mathcal{C}}\frac {(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta } \biggr)^{r}\,dz \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac{(m+z+1)_{n-m} }{z(z-1)\cdots(z-n)} \biggl( \frac{\beta}{z-\theta} \biggr)^{r} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\mathop{\operatorname {Res}}\limits_{z=\theta}\frac {(m+z+1)_{n-m}}{z(z-1)\cdots(z-\theta+1)(z-\theta-1)\cdots (z-n)}\frac{\beta^{r}}{(z-\theta)^{r+1}} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[(z-\theta)^{r}\bigr]\frac {(z+1+m)(z+2+m)\cdots(z+n)\beta^{r}}{z(z-1)\cdots(z-\theta +1)(z-\theta-1)\cdots(z-n)} \\ &\quad= \frac{(-1)^{n+1}n!}{(n-m)!}\bigl[z^{r}\bigr]\frac{(z+1+m+\theta )(z+2+m+\theta)\cdots(z+n+\theta)\beta^{r}}{(z+\theta)(z+\theta -1)\cdots(z+1)(z-1)\cdots(z+\theta-n)} \\ &\quad= (-1)^{\theta+1}\frac{n!(m+1+\theta)_{n-m}}{(n-m)!\theta {!}(n-\theta){!}}\bigl[z^{r}\bigr] \frac{(1+\frac{z}{m+1+\theta})\cdots (1+\frac{z}{n+\theta})\beta^{r}}{(1+\frac{z}{\theta})(1+\frac {z}{\theta-1})\cdots(1+z)(1-z)\cdots(1-\frac{z}{n-\theta})} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta }\bigl[z^{r-l} \bigr] \Biggl(1+\sum_{i=1}^{n-m}\sum _{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times\mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n}\frac{\beta ^{r}}{(s_{1}-\theta)\cdots(s_{l}-\theta)} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta }\bigl[z^{r-l} \bigr] \Biggl(1+\sum_{i=1}^{n-m}\sum _{m+1\leq k_{1}<\cdots< k_{i}\leq n}\frac{z^{i}}{(k_{1}+\theta)\cdots(k_{i}+\theta)} \Biggr) \\ &\qquad{} \times \Biggl(\frac{\beta^{r}}{(-\theta)^{l}}+\sum_{j=1}^{l} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \Biggr) \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{m+\theta } \Biggl( \biggl(- \frac{\beta}{\theta} \biggr)^{r}+\sum_{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq\cdots \leq s_{j}\leq n}\frac{\beta^{r}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n-m} \sum_{m+1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n}}_{s_{1},\ldots,s_{l}\neq\theta} \frac{\beta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}

This proof is complete. □

### Corollary 15

For $$r,n \in\mathbb{N}$$.

When $$\theta\notin\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta }{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n}\binom{\theta-1}{n}^{-1} \Biggl(1-\sum _{j=1}^{r-1} \sum_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n} \frac{\theta ^{j}}{(s_{1}-\theta)\cdots(s_{j}-\theta)} \Biggr). \end{aligned}
(25)

When $$\theta\in\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta} \Biggl((-1)^{r}+\sum _{j=1}^{r} \mathop{\sum_{s_{1},\ldots,s_{j}\neq\theta}}_{ 1\leq s_{1}\leq \cdots\leq s_{j}\leq n} \frac{\theta^{j}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \Biggr). \end{aligned}
(26)

### Proof

Take $$m=n$$, $$\beta= \theta$$ in Theorem 14 and define that $$\prod_{k=n+1}^{n}=1$$, $$\sum_{i=1}^{0}=0$$, then Corollary 15 follows. □

### Corollary 16

For $$r,l,n \in\mathbb{N}$$.

When $$\theta\notin\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0}^{n}(-1)^{k} \binom{n}{k}\binom{n+k}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{n+1}\binom{\theta-1}{n}^{-1}\binom{n+\theta}{\theta} \Biggl( (-1 )^{r+1}+\sum_{j=1}^{r-1} \sum _{1\leq s_{1}\leq \cdots\leq s_{j}\leq n}\frac{\theta^{j}}{(s_{1}-\theta)\cdots (s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r-1}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots< k_{i}\leq n} \sum_{0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta ^{r-1}}{(k_{1}+\theta)\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots (s_{l}-\theta)} \Biggr). \end{aligned}
(27)

When $$\theta\in\{0,1, 2,\ldots, n\}$$, we have

\begin{aligned} &\sum_{k=0,k \neq\theta}^{n}(-1)^{k} \binom{n}{k} \binom{n+k}{k} \biggl(\frac{\theta}{k-\theta} \biggr)^{r} \\ &\quad= (-1)^{\theta+1}\binom{n}{\theta}\binom{n+\theta}{\theta} \\ &\qquad{}\times\Biggl( (-1 )^{r}+\sum_{j=1}^{r} \mathop{\sum _{s_{1},\ldots ,s_{j}\neq\theta}}_{1\leq s_{1}\leq\cdots\leq s_{j}\leq n}\frac {\theta^{j}}{(s_{1}-\theta)\cdots(s_{j}-\theta)} \\ &\qquad{} +\sum_{i+l=r}\sum_{i=1}^{n} \sum_{1\leq k_{1}<\cdots< k_{i}\leq n} \mathop{\sum_{s_{1},\ldots,s_{l}\neq\theta}}_{ 0\leq s_{1}\leq\cdots\leq s_{l}\leq n} \frac{\theta^{r}}{(k_{1}+\theta )\cdots(k_{i}+\theta)(s_{1}-\theta)\cdots(s_{l}-\theta)} \Biggr). \end{aligned}
(28)

### Proof

Take $$m=0$$, $$\beta= \theta$$ in Theorem 14 and define that $$\prod_{k=n+1}^{n}=1$$, $$\sum_{i=1}^{0}=0$$, then Corollary 16 follows. □

### Remark 17

By applying Bell polynomials (3) and (4), we can deduce all the results of [18] and [19]. Our results are another representation of the corresponding results in [18] and [19]. For example, compare formula (24) with $$m=0$$, $$\theta=M$$, $$\beta=1$$ and the following formula [19]: For $$0\leq M\leq n$$,

\begin{aligned} &\sum_{0\leq k \leq n,k \neq M}(-1)^{k-1} \binom{n}{k} \binom{n+k}{k}\frac{1}{(k-M)^{r}} \\ &\quad=(-1)^{M}\binom{n}{M}\binom{M+n}{M}\sum _{m_{1}+2m_{2}+3m_{3}+\cdots=r}\frac{1}{m_{1}!m_{2}!m_{3}! \cdots} \biggl(\frac {x_{1}}{1} \biggr)^{m_{1}} \biggl(\frac{x_{2}}{2} \biggr)^{m_{2}} \biggl( \frac {x_{3}}{3} \biggr)^{m_{3}}\cdots, \end{aligned}
(29)

where

\begin{aligned} x_{k}=(-1)^{k-1}H_{M+n}^{(k)}-2H_{M}^{(k)}+H_{n-M}^{(k)},\quad H_{n}^{(r)}=\sum_{k=1}^{n} \frac{1}{k^{r}},\ n,r \in\mathbb{N}. \end{aligned}

Obviously, formula (24) includes the above formula (29) and the main result of [18], i.e., formula (24) is an extension of them.

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## Acknowledgements

The authors express their sincere gratitude to referee for many valuable comments and suggestions. The present investigation was supported by Natural Science Foundation Project of Chongqing, China under Grant CSTC2011JJA00024, Research Project of Science and Technology of Chongqing Education Commission, China under Grant KJ120625, Fund of Chongqing Normal University, China under Grant 10XLR017 and 2011XLZ07.

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Correspondence to Qiu-Ming Luo.

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Liu, TT., Luo, QM. An extension for Rice’s integral and applications. J Inequal Appl 2015, 29 (2015). https://doi.org/10.1186/s13660-015-0557-z