Fourier Series for the Tangent Polynomials, Tangent–Bernoulli and Tangent–Genocchi Polynomials of Higher Order

Abstract

In this paper, the Fourier series expansion of Tangent polynomials of higher order is derived using the Cauchy residue theorem. Moreover, some variations of higher-order Tangent polynomials are defined by mixing the concept of Tangent polynomials with that of Bernoulli and Genocchi polynomials, Tangent–Bernoulli and Tangent–Genocchi polynomials. Furthermore, Fourier series expansions of these variations are also derived using the Cauchy residue theorem.

1. Introduction

In recent decades, several mathematicians have been attracted to work on some well-known special functions, numbers and polynomials (e.g., zeta functions, Bernoulli, Euler and Genocchi numbers and polynomials and derivative polynomials [1,2,3,4,5,6,7]) due to their wide-ranging applications, from number theory and combinatorics to other fields of applied mathematics. With these, different variations and generalizations of these functions, numbers and polynomials have appeared in the literature. Some variations were constructed by mixing the concepts of two or three special functions, numbers or polynomials. For instance, the poly-Bernoulli numbers and polynomials in [6,8,9,10,11] were constructed by mixing the concepts of polylogarithm and Bernoulli numbers and polynomials. Moreover, the Apostol–Genocchi polynomials, Frobenius–Euler polynomials, Frobenius–Genocchi polynomials and Apostol–Frobenius-type poly–Genocchi polynomials in [12,13,14,15] are constructed by mixing the concepts of Apostol, Frobenius, Genocchi and Euler polynomials. Another interesting mixture of special polynomials can be constructed by joining the concepts of Tangent polynomials with Bernoulli and Genocchi polynomials.

The Tangent polynomials $T_n\left(x\right)$ of degree n with complex argument x are defined as coefficients of the following generating function (see [16,17])

$$\sum _{n=0}^{\infty }{T}_n\left(x\right)\frac{z^n}{n!}=\left(\frac{2}{e^{2z}+1}\right)e^{xz},$$

(1)

where $T_n\left(0\right)=T_n$, the Tangent numbers defined as coefficient of the following series expansion of the tangent function $tanx$ (see [12]):

$$tanz=\sum _{n=0}^{\infty }{(-1)}^{n+1}{T}_{2n+1}\frac{z^{2n+1}}{(2n+1)!},$$

(2)

with $T_0=1$ and $T_{2n}=0,n\in \mathbb{N}$. The first few values of Tangent polynomials are given as follows:

$$T_0\left(x\right)=1,T_1\left(x\right)=x-1,T_2\left(x\right)=x^2-2x,T_3=x^3-3x^2+2,$$

$$T_4\left(x\right)=x^4-4x^3+8x,T_5\left(x\right)=x^5-5x^4+20x^2-16,$$

$$T_6\left(x\right)=x^6-6x^5+40x^3-96x,$$

where

$$T_0\left(0\right)=1,T_1\left(0\right)=-1,T_2=0,T_3\left(0\right)=-2,T_4\left(0\right)=0,T_5=-16,T_6\left(0\right)=0,$$

which are exactly the values of Tangent numbers.

For $r\in \mathbb{N}$, the higher-order tangent polynomials, $T_n^r\left(x\right)$$(n\ge 0)$, are defined by the following generating function (see [18])

$${\left(\frac{2}{e^{2t}+1}\right)}^r{e}^{xt}=\sum _{n=0}^{\infty }{T}_n^r\left(x\right)\frac{t^n}{n!},\left|2t\right|<\pi .$$

(3)

When $r=1$, the above equation gives the generating function for the tangent polynomials in (2) (see [16]).

The Tangent–Berrnoulli polynomials and Tangent–Genocchi polynomials of higher order, which are variations of the Tangent polynomials, are defined below.

Definition 1.

The Tangent–Bernoulli polynomials of higher order denoted by ${\left(TB\right)}_n^r\left(x\right),r\ge 2$ are defined by means of the following generating function

$${\left(\frac{t}{e^{2t}-1}\right)}^r{e}^{xt}=\sum _{n=0}^{\infty }{\left(TB\right)}_n^r\left(x\right)\frac{t^n}{n!},\left|t\right|<\pi .$$

(4)

Definition 2.

The Tangent–Genocchi polynomials of higher order denoted by ${\left(TG\right)}_n^r\left(x\right),r\ge 2$ are defined by means of the following generating function

$${\left(\frac{2t}{e^{2t}+1}\right)}^r{e}^{xt}=\sum _{n=0}^{\infty }{\left(TG\right)}_n^r\left(x\right)\frac{t^n}{n!},\left|t\right|<\pi .$$

(5)

In this paper, the researchers derived the Fourier series expansion of the Tangent polynomials, Tangent–Bernoulli and Tangent–Genocchi polynomials of order r, $r\in {\mathbb{Z}}^{+}$.

2. Methods

This study was facilitated by the use of Residue Theory; the Cauchy Residue Theorem [19] was applied to solve the contour integral representation of the polynomials.

3. Results and Discussions

In this section, we provide a detailed discussion of the process of deriving the Fourier series expansion of higher-order Tangent polynomials as well as the Fourier series expansions of higher-order Tangent–Bernoulli and Tangent–Genocchi polynomials using the method of Bayad [20].

3.1. Higher Order Tangent Polynomials

To derive the Fourier series expansion of Tangent polynomials of higher order, the Cauchy Integral Formula is applied to (3) to yield

$$\frac{T_n^r\left(x\right)}{n!}=\frac{1}{2\pi i}{\int }_C{\left(\frac{2}{e^{2t}+1}\right)}^r{e}^{xt}\frac{dt}{t^{n+1}},$$

(6)

where C is a circle about zero with radius $<\pi /2$. Consider the function

$$f\left(t\right)={\left(\frac{2}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

This function has a pole of order $n-r+1$ at $t=0$ and poles of order r at $t_k$ where $t_k$ are zeros of $e^{2t}+1$. The values of $t_k$ are obtained as follows:

$$\begin{array}{cc}\hfill e^{2t}+1& =0\hfill \\ \hfill log\left(e^{2t}\right)& =log\left(-1\right)\hfill \\ \hfill 2t& =log(-1)=ln|-1|+iArg(-1)+2k\pi i\hfill \\ \hfill & =\pi i+2k\pi i\hfill \\ \hfill t_k& :=t=(2k+1)\frac{\pi }{2}i,fork\in \mathbb{Z}\hfill \end{array}$$

(7)

Notice that the nearest $t_k$ are $t_0$ and $t_{-1}$. Hence, the contour in (6) must have a radius of less than $|t_0|=|t_{-1}|=\frac{\pi }{2}$. Now, let $C_N$ be the circle about 0 of radius $(2N+1)\frac{\pi }{2}-ϵ$, where $0<ϵ<1$. This range of $ϵ$ will ensure that the circle $C_N$ is between the circles $C_{N-1}$ and $C_N$. Then, using Cauchy Residue Theorem [19], we have

$$\frac{1}{2\pi i}{\int }_{C_N}f\left(t\right)dt=\sum _{k<N}Res\left(f\left(t\right),t=t_k\right)+Res\left(f\left(t\right),t=0\right).$$

(8)

Taking the limit as $N\to \infty$,

$$\underset{N\to \infty }{lim}\frac{1}{2\pi i}{\int }_{C_N}f\left(t\right)dt=\sum _{k\in \mathbb{Z}}Res\left(f\left(t\right),t=t_k\right)+Res\left(f\left(t\right),t=0\right).$$

(9)

We prove the following lemma:

Lemma 1.

For $0<x\le r$, $n\ge 1$ and fixed $r\in {\mathbb{Z}}^{+}$, as $N\to \infty$,

$${\int }_{C_N}f\left(t\right)dt\to 0,$$

where $C_N$ is a circle about 0 with radius $R=(2N+1)\frac{\pi }{2}-ϵ$, $0<ϵ<1$.

Proof. 

Taking the modulus of the integral gives

$$\left|{\int }_{C_N}{\left(\frac{2}{e^{2t}+1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|\le 2^r{\int }_{C_N}\frac{|e^{xt}|}{|e^{2t}{+1|}^r}·\frac{\left|dt\right|}{|t^{n+1}|},witht\in C_N,\left|t\right|=(2N+1)\frac{\pi }{2}-ϵ.$$

Note that, with $t=a+ib$, $a=Re\left(t\right)$ and $b=Im\left(t\right)$, we have

$$\begin{array}{cc}\hfill |e^{2t}+1|& =|e^{2(a+ib)}+1|=|e^{2a}{e}^{2ib}+1|\hfill \\ \hfill & =|e^{2a}(cos2b+isin2b)+1|=|e^{2a}cos2b+1+e^{2a}isin2b|\hfill \\ \hfill & =\sqrt{{(e^{2a}cos2b+1)}^2+{(e^{2a}isin2b)}^2}\hfill \\ \hfill |e^{2t}{+1|}^r& =e^{2ar}{\left(\sqrt{1+\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}}\right)}^r.\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill \frac{|e^{xt}|}{|e^{2t}{+1|}^r}& =\frac{e^{xa}}{e^{2ar}{\left(1+\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}=\frac{1}{e^{a(2r-x)}{\left(1+\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}.\hfill \end{array}$$

Now, with $0<x\le r$, $a(2r-x)\ge a(2r-r)=ar$. Hence,

$$\frac{1}{e^{a(2r-x)}}\le \frac{1}{e^{ar}}.$$

Thus,

$$\frac{|e^{xt}|}{|e^{2t}{+1|}^r}\le \frac{1}{e^{ar}}\frac{1}{{\left(1+\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}=\frac{1}{{\left(e^{2a}+2cos2b+e^{-2a}\right)}^{\frac{r}{2}}}.$$

Let $B={\left(e^{2a}+2cos2b+e^{-2a}\right)}^{\frac{r}{2}}$. With $t=R{e}^{i\theta }=R(cos\theta +isin\theta )$, $a=[(2N+1)\frac{\pi }{2}-ϵ]cos\theta$, $0\le \theta \le \pi$, we consider three cases:

• Case 1. When $cos\theta <0$, $e^{2a}\to 0$ and $e^{-2a}\to +\infty$ as $N\to +\infty$, which means $B\to +\infty$ and so $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ is bounded;
• Case 2. When $cos\theta >0$, $e^{2a}\to +\infty$ and $e^{-2a}\to 0$ as $N\to +\infty$, which means $B\to +\infty$ and so $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ is bounded;
• Case 3. When $cos\theta =0$, $e^{2a}=e^{-2a}=1$, which means $B=2+2cos2b$. For $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ to be bounded, $2+2cos2b\ne 0$, where

  $$b=\left[(2N+1)\frac{\pi }{2}-ϵ\right]sin\theta =\left[(2N+1)\frac{\pi }{2}-ϵ\right](\pm 1).$$
  Note that

  $$\begin{array}{cc}\hfill 2+2cos2b& =2+2cos2(\pm 1)\left[(2N+1)\frac{\pi }{2}-ϵ\right]\hfill \\ \hfill & =2+2cos2\left[(2N+1)\frac{\pi }{2}-ϵ\right].\hfill \end{array}$$
  The last expression is zero if, and only if,

  $$cos2\left[(2N+1)\frac{\pi }{2}-ϵ\right]=-1$$

  $$\begin{array}{cc}\hfill & ⟺2\left[(2N+1)\frac{\pi }{2}-ϵ\right]=(2M+1)\pi ,M\in \mathbb{Z}\hfill \\ \hfill & ⟺\left[(2N+1)\frac{\pi }{2}-ϵ\right]=(2M+1)\frac{\pi }{2}\hfill \\ \hfill & ⟺2(N-M)\frac{\pi }{2}=ϵ\hfill \\ \hfill & ⟺(N-M)\pi =ϵ,\hfill \end{array}$$

  which is impossible because $0<ϵ<1$. Hence, $B\ne 0$. That is, $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ is bounded.

Thus, in any case,

$$\frac{|e^{xt}|}{|e^{2t}{+1|}^r}\le K,$$

for some constant K, $\forall t\in C_N$. Consequently,

$$\begin{array}{cc}\hfill \left|{\int }_{C_N}{\left(\frac{2}{e^{2t}+1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|& \le 2^r{\int }_{C_N}\frac{|e^{xt}|}{|e^{2t}{+1|}^r}·\frac{\left|dt\right|}{|t^{n+1}|}\hfill \\ \hfill & \le 2^r{\int }_{C_N}K\frac{\left|dt\right|}{|t^{n+1}|}\hfill \\ \hfill & =\frac{2^{r}K}{{\left(\left(2N+1\right)\frac{\pi }{2}-ϵ\right)}^{n+1}}·\left[\left(2N+1\right)\frac{\pi }{2}-ϵ\right]\left(2\pi \right)\hfill \\ \hfill & <\frac{2^{r+1}K\pi }{{\left[\left(2N+1\right)\frac{\pi }{2}-ϵ\right]}^n}\to 0asN\to \infty forn\ge 1.\hfill \end{array}$$

□

Using Lemma 1, Equation (9) becomes

$$\begin{array}{cc}\hfill 0& =\sum _{k\in \mathbb{Z}}Res\left(f\left(t\right),t=t_k\right)+Res\left(f\left(t\right),t=0\right)\hfill \\ \hfill 0& =Res\left(f\left(t\right),t=0\right)+\sum _{k\in \mathbb{Z}}Res\left(f\left(t\right),t=t_k\right)\hfill \\ \hfill 0& =\frac{T_n^r\left(x\right)}{n!}+\sum _{k\in \mathbb{Z}}Res\left(f\left(t\right),t=t_k\right).\hfill \end{array}$$

Thus, we have

$$T_n^r=-n!\sum _{k\in \mathbb{Z}}Res\left(f\left(t\right),t=t_k\right).$$

(10)

The Fourier series for the tangent polynomials of higher order is given in the following theorem.

Theorem 1.

For $0<x\le r$,

$$\begin{array}{cc}\hfill T_n^r\left(x\right)=2·n!{\left(\frac{2}{\pi }\right)}^{r+n}& \sum _{k=0}^{\infty }\sum _{j=0}^{r-1}{(-1)}^j\left(\genfrac{}{}{0pt}{}{r+n-j-1}{r-j-1}\right)\frac{{\pi }^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & \times \frac{cos\left[(2k+1)\pi x/2-(r+n-j)\pi /2\right]}{{(2k+1)}^{r+n-j}},\hfill \end{array}$$

(11)

where $B_j^r\left(\frac{x}{2}\right)$ denotes the Bernoulli polynomials of order r defined by

$${\left(\frac{w}{e^w-1}\right)}^r{e}^{xw}=\sum _{j=0}^{\infty }{B}_j^r\left(x\right)\frac{w^n}{n!}.$$

Proof. 

For $r\ge 2$,

$$Res\left(f\left(t\right),t=t_k\right)=\frac{1}{(r-1)!}\underset{t\to t_k}{lim}\frac{d^{r-1}}{d{t}^{r-1}}{(t-t_k)}^r{\left(\frac{2}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

Since $e^{-2t_k}=e^{-2(2k+1)\frac{\pi }{2}i}=e^{-(2k+1)\pi i}=-1$, we can write ${\left(e^{2t}+1\right)}^r$ as

$$\begin{array}{cc}\hfill {\left(e^{2t}+1\right)}^r& ={(-1)}^r{\left(e^{2t}(-1)-1\right)}^r\hfill \\ \hfill & ={(-1)}^r{\left(e^{2t}·e^{-2t_k}-1\right)}^r\hfill \\ \hfill & ={(-1)}^r{\left(e^{2(t-t_k)}-1\right)}^r.\hfill \end{array}$$

Then, we have

$$\begin{array}{cc}\hfill {(t-t_k)}^r{\left(\frac{2}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}& =\frac{2^r{(t-t_k)}^r}{{(-1)}^r{\left(e^{2(t-t_k)}-1\right)}^r}·\frac{e^{xt}}{t^{n+1}}\hfill \\ \hfill & ={(-1)}^r\frac{{\left(2(t-t_k)\right)}^r}{{(e^{2(t-t_k)}-1)}^r}·\frac{e^{xt}}{t^{n+1}}\hfill \\ \hfill & ={(-1)}^r\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n+1)},\hfill \end{array}$$

where $B_n^r$ denotes the Bernoulli numbers of order r defined in the theorem.

$${\left(\frac{w}{e^w-1}\right)}^r=\sum _{n=0}^{\infty }{B}_n^r\frac{w^n}{n!}.$$

Applying the Leibniz Rule on differentiation,

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left\{{(t-t_k)}^r{\left(\frac{2}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}\right\}\hfill \\ \hfill & =\frac{d^{r-1}}{d{t}^{r-1}}\left\{{(-1)}^r\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n+1)}\right\}\hfill \\ \hfill & ={(-1)}^r\frac{d^{r-1}}{d{t}^{r-1}}\left\{\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)t^{-(n+1)}\right\}\hfill \\ \hfill & ={(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n+1)}·\frac{d^j}{d{t}^j}\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^j}{d{t}^j}& \left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{2^n{(t-t_k)}^n)}{n!}\right)\hfill \\ \hfill & =\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{e}^{xt}\sum _{n=l}^{\infty }{B}_n^r\frac{2^n}{n!}{\left(n\right)}_l{(t-t_k)}^{n-l}\hfill \\ \hfill & =e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!},\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left({(t-t_k)}^r{\left(\frac{2}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}\right)\hfill \\ \hfill & ={(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n+1)}\hfill \\ \hfill & \times e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}.\hfill \end{array}$$

Thus,

$$\begin{array}{cc}\hfill Res\left(f\left(t\right),t=t_k\right)=\frac{1}{(r-1)!}& \underset{t\to t_k}{lim}{(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n+1)}\hfill \\ \hfill & \times \underset{t\to t_k}{lim}{e}^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}.\hfill \end{array}$$

Note that $B_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}\to 0$ as $t\to t_k$ except when $n=l$. Using the notation ${\left(n\right)}_l$ defined by

$${\left(n\right)}_l=\left\{\begin{array}{cc}n(n-1)\dots (n-l+1),\hfill & n\ge l\hfill \\ 1,\hfill & l=0\hfill \\ 0,\hfill & n<l,\hfill \end{array}\right.$$

this gives

$$\begin{array}{cc}\hfill Res\left(f\left(t\right),t=t_k\right)& =\frac{1}{(r-1)!}{(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right){(-1)}^{r-1-j}{(n+r-1-j)}_{r-1-j}{t}_k^{-(n+r-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\frac{{(-1)}^r}{(r-1)!}\sum _{j=0}^{r-1}\frac{(r-1)!}{j!(r-1-j)!}{(-1)}^{r-1-j}{(n+r-1-j)}_{r-1-j}{t}_k^{-(n+r-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}\left(\genfrac{}{}{0pt}{}{n+r-1-j}{r-1-j}\right)\frac{t_k^{j-n-r}}{j!}{e}^{x{t}_k}{2}^j\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)\frac{x^{j-l}}{2^{j-l}}{B}_l^r.\hfill \end{array}$$

Recall from [21] that $B_j^r\left(\frac{x}{2}\right)={\sum }_{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)B_l^r{\left(\frac{x}{2}\right)}^{j-l}$. Thus,

$$\begin{array}{cc}\hfill Res\left(f\left(t\right),t=t_k\right)& =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n+r-1-j}{r-1-j}\right)\frac{t_k^{j-n-r}}{j!}{e}^{x{t}_k}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n+r-1-j}{r-1-j}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{n+r-j}}\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{r+n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{r+n-j}}.\hfill \end{array}$$

With $t_k=\frac{1}{2}(2k+1)\pi i$, we get

$$\begin{array}{cc}\hfill Res\left(f\left(t\right),t=t_k\right)& =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{r+n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{\frac{1}{2}(2k+1)\pi ix}}{{\left(\frac{1}{2}(2k+1)\pi i\right)}^{r+n-j}}\hfill \end{array}$$

$$=\frac{1}{{\left(\frac{1}{2}\pi i\right)}^{r+n}}\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{r+n-j-1}{r-j-1}\right)\frac{{\left(\frac{1}{2}\pi i\right)}^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\frac{e^{\frac{1}{2}(2k+1)\pi ix}}{{(2k+1)}^{r+n-j}}.$$

This gives

$$T_n^r\left(x\right)=n!{\left(\frac{2}{\pi i}\right)}^{r+n}\sum _{k\in \mathbb{Z}}\left(\sum _{j=0}^{r-1}{(-1)}^j\left(\genfrac{}{}{0pt}{}{r+n-j-1}{r-j-1}\right)\frac{{\left(\pi i\right)}^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\right)\frac{e^{\frac{1}{2}(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{r+n-j}}.$$

(12)

Now, from (12), we look at

$$i^{-(r+n-j)}\sum _{k\in \mathbb{Z}}\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{r+n-j}}.$$

(13)

Noting that $i^{-(r+n-j)}=e^{-(r+n-j)\pi i/2}$ and ${(-1)}^{r+n-j}=e^{(r+n-j)\pi i}$, we see that

$$i^{-(r+n-j)}\sum _{k\in \mathbb{Z}}\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{r+n-j}}$$

$$\begin{array}{cc}\hfill & =i^{-(r+n-j)}\left\{\sum _{k=0}^{\infty }\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{r+n-j}}+{(-1)}^{r+n-j}\sum _{k=0}^{\infty }\frac{e^{-(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{r+n-j}}\right\}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{e^{\left[(2k+1)x/2-(r+n-j)/2\right]\pi i}+e^{-\left[(2k+1)x/2-(r+n-j)/2\right]\pi i}}{{(2k+1)}^{r+n-j}}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{2cos\left[(2k+1)\pi x/2-(r+n-j)\pi /2\right]}{{(2k+1)}^{r+n-j}}\hfill \\ \hfill & =2\sum _{k=0}^{\infty }\frac{cos\left[(2k+1)\pi x/2-(r+n-j)\pi /2\right]}{{(2k+1)}^{r+n-j}}.\hfill \end{array}$$

(14)

Replacing (13) with (14) and substituting to (12), we get the desired formula (11). □

3.2. Higher-Order Tangent–Bernoulli Polynomials

The Fourier series of higher-order Tangent–Bernoulli polynomials is derived similarly to that in Section 3.1. Applying the Cauchy Integral Formula [19] to (4) yields

$$\frac{{\left(TB\right)}_n^r\left(x\right)}{n!}=\frac{1}{2\pi i}{\int }_C{\left(\frac{t}{e^{2t}-1}\right)}^r{e}^{xt}\frac{dt}{t^{n+1}},$$

(15)

where C is a circle about zero with radius $<\pi$. Consider the function

$$g\left(t\right)={\left(\frac{t}{e^{2t}-1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

This function has a pole of order $n+1$ at $t=0$ and poles of order r at $t_k$, where $t_k$ are zeros of $e^{2t}+1$. The values of $t_k$ are obtained as follows:

$$\begin{array}{cc}\hfill e^{2t}-1& =0\hfill \\ \hfill log\left(e^{2t}\right)& =log\left(1\right)\hfill \\ \hfill 2t& =log\left(1\right)=ln\left|1\right|+iArg\left(1\right)+2k\pi i\hfill \\ \hfill & =2k\pi i\hfill \\ \hfill t_k& :=t=k\pi i,fork\in \mathbb{Z},k\ne 0.\hfill \end{array}$$

(16)

Note that the poles nearest to zero are $\pm \pi$. Hence, the circle C must have a radius $<\pi$.

Now, let $C{\ast }_N$ be the circle about 0 of radius $N\pi -ϵ$, where $0<ϵ<1$. This range of $ϵ$ will ensure that the circle $C{\ast }_N$ is between the circles $C{\ast }_{N-1}$ and $C{\ast }_N$. Then, using Cauchy Residue Theorem [19], we have

$$\frac{1}{2\pi i}{\int }_{C{\ast }_N}g\left(t\right)dt=\sum _{k<N}Res\left(g\left(t\right),t=t_k\right)+Res\left(g\left(t\right),t=0\right).$$

(17)

Taking the limit as $N\to \infty$,

$$\underset{N\to \infty }{lim}\frac{1}{2\pi i}{\int }_{C{\ast }_N}g\left(t\right)dt=\sum _{k\in \mathbb{Z}}Res\left(g\left(t\right),t=t_k\right).$$

(18)

We prove the following lemma.

Lemma 2.

For $0<x\le r$ and $n\ge r$, as $N\to \infty$,

$${\int }_{C{\ast }_N}g\left(t\right)dt\to 0,$$

where $C_N$ is a circle about 0 with radius $R=N\pi -ϵ$, $0<ϵ<1$.

Proof. 

Taking the modulus of the integral gives

$$\left|{\int }_{C{\ast }_N}{\left(\frac{t}{e^{2t}-1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|\le {\int }_{C{\ast }_N}\frac{|e^{xt}|}{|e^{2t}{-1|}^r}·\frac{\left|dt\right|}{|t^{n-r+1}|},witht\in C_N,\left|t\right|=N\pi -ϵ.$$

Note that, with $t=a+ib$, $a=Re\left(t\right)$ and $b=Im\left(t\right)$, we have

$$\begin{array}{cc}\hfill |e^{2t}-1|& =|e^{2(a+ib)}+1|=|e^{2a}{e}^{2ib}-1|\hfill \\ \hfill & =|e^{2a}(cos2b+isin2b)-1|=|e^{2a}cos2b-1+e^{2a}isin2b|\hfill \\ \hfill & =\sqrt{{(e^{2a}cos2b-1)}^2+{(e^{2a}isin2b)}^2}\hfill \\ \hfill |e^{2t}{-1|}^r& =e^{2ar}{\left(\sqrt{1-\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}}\right)}^r.\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill \frac{|e^{xt}|}{|e^{2t}{-1|}^r}& =\frac{e^{xa}}{e^{2ar}{\left(1+\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}=\frac{1}{e^{a(2r-x)}{\left(1-\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}.\hfill \end{array}$$

Now, with $0<x\le r$, $a(2r-x)\ge a(2r-r)=ar$. Hence,

$$\frac{1}{e^{a(2r-x)}}\le \frac{1}{e^{ar}}.$$

Thus,

$$\frac{|e^{xt}|}{|e^{2t}{-1|}^r}\le \frac{1}{e^{ar}}\frac{1}{{\left(1-\frac{2cos2b}{e^{2a}}+\frac{1}{e^{4a}}\right)}^{\frac{r}{2}}}=\frac{1}{{\left(e^{2a}-2cos2b+e^{-2a}\right)}^{\frac{r}{2}}}.$$

Let $B={\left(e^{2a}-2cos2b+e^{-2a}\right)}^{\frac{r}{2}}$. With $t=R{e}^{i\theta }=R(cos\theta +isin\theta )$, $a=[N\pi -ϵ]cos\theta$, $0\le \theta \le \pi$, we consider three cases:

• Case 1. When $cos\theta <0$, $e^{2a}\to 0$ and $e^{-2a}\to +\infty$ as $N\to +\infty$, which means $B\to +\infty$ and so $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ is bounded;
• Case 2. When $cos\theta >0$, $e^{2a}\to +\infty$ and $e^{-2a}\to 0$ as $N\to +\infty$, which means $B\to +\infty$ and so $\frac{|e^{xt}|}{|e^{2t}{+1|}^r}$ is bounded;
• Case 3. When $cos\theta =0$, $e^{2a}=e^{-2a}=1$, which means $B=2-2cos2b$. For $\frac{|e^{xt}|}{|e^{2t}{-1|}^r}$ to be bounded, $2-2cos2b\ne 0$, where

  $$b=[N\pi -ϵ]sin\theta =[N\pi -ϵ](\pm 1).$$
  Note that

  $$\begin{array}{cc}\hfill 2-2cos2b& =2-2cos2(\pm 1)[N\pi -ϵ]\hfill \\ \hfill & =2-2cos2[N\pi -ϵ].\hfill \end{array}$$
  The last expression is zero if and only if

  $$cos2[N\pi -ϵ]=1$$

  $$\begin{array}{cc}\hfill & ⟺2[N\pi -ϵ]=2M\pi ,M\in \mathbb{Z}\hfill \\ \hfill & ⟺N\pi -ϵ=M\pi \hfill \\ \hfill & ⟺(N-M)\pi =ϵ\hfill \end{array}$$

  which is impossible because $0<ϵ<1$. Hence, $B\ne 0$. That is, $\frac{|e^{xt}|}{|e^{2t}{-1|}^r}$ is bounded.

Thus, in any case,

$$\frac{|e^{xt}|}{|e^{2t}{-1|}^r}\le K_1,$$

for some constant $K_1$, $\forall t\in C_N$. Consequently,

$$\begin{array}{cc}\hfill \left|{\int }_{C{\ast }_N}{\left(\frac{t}{e^{2t}-1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|& \le K_1{\int }_{C{\ast }_N}\frac{\left|dt\right|}{|t^{n-r+1}|}\hfill \\ \hfill & =K_1\frac{(N\pi -ϵ)2\pi }{{(N\pi -ϵ)}^{n-r+1}}\hfill \\ \hfill & =\frac{2K_1\pi }{{(N\pi -ϵ)}^{n-r}}\to 0asN\to \infty forn\ge r.\hfill \end{array}$$

□

Applying Lemma 2, Equation (18) becomes

$$\begin{array}{cc}\hfill 0& =\sum _{k\in \mathbb{Z},k\ne 0}Res\left(g\left(t\right),t=t_k\right)+Res\left(g\left(t\right),t=0\right)\hfill \\ \hfill 0& =Res\left(g\left(t\right),t=0\right)+\sum _{k\in \mathbb{Z},k\ne 0}Res\left(g\left(t\right),t=t_k\right)\hfill \\ \hfill 0& =\frac{{\left(TB\right)}_n^r\left(x\right)}{n!}+\sum _{k\in \mathbb{Z},k\ne 0}Res\left(g\left(t\right),t=t_k\right).\hfill \end{array}$$

Thus, we have

$${\left(TB\right)}_n^r=-n!\sum _{k\in \mathbb{Z},k\ne 0}Res\left(g\left(t\right),t=t_k\right).$$

(19)

The Fourier series for the Tangent–Bernoulli polynomials ${\left(TB\right)}_n^r\left(x\right)$ of higher order is given in the following theorem.

Theorem 2.

For $0<x\le r$,

$$\begin{array}{cc}\hfill {\left(TB\right)}_n^r\left(x\right)& =\frac{n!}{{\pi }^n}\sum _{k=0}^{\infty }\sum _{j=0}^{r-1}{(-1)}^{r-j}\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{{\pi }^j}{2^{r-j-1}j!}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & \times \frac{cos\left[kx-(n-j)/2\right]\pi }{k^{n-j}},\hfill \end{array}$$

(20)

where $B_j^r\left(\frac{x}{2}\right)$ denotes the Bernoulli polynomials of order r defined by

$${\left(\frac{w}{e^w-1}\right)}^r{e}^{xw}=\sum _{n=0}^{\infty }{B}_n^r\left(x\right)\frac{w^n}{n!}.$$

Proof. 

For $r\ge 2$,

$$Res\left(g\left(t\right),t=t_k\right)=\frac{1}{(r-1)!}\underset{t\to t_k}{lim}\frac{d^{r-1}}{d{t}^{r-1}}{(t-t_k)}^r{\left(\frac{t}{e^{2t}-1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

Since $e^{-2t_k}=e^{-2k\pi i}=1$, we can write ${\left(e^{2t}-1\right)}^r$ as

$$\begin{array}{cc}\hfill {\left(e^{2t}-1\right)}^r& ={\left(e^{2t}\left(1\right)-1\right)}^r\hfill \\ \hfill & ={\left(e^{2t}·e^{-2t_k}-1\right)}^r\hfill \\ \hfill & ={\left(e^{2(t-t_k)}-1\right)}^r.\hfill \end{array}$$

Then, we have

$$\begin{array}{cc}\hfill {(t-t_k)}^r{\left(\frac{1}{e^{2t}-1}\right)}^r\frac{e^{xt}}{t^{n-r+1}}& =\frac{{(t-t_k)}^r}{{\left(e^{2(t-t_k)}-1\right)}^r}·\frac{e^{xt}}{t^{n-r+1}}\hfill \\ \hfill & =\frac{1}{2^r}\frac{{\left(\left(2(t-t_k)\right)\right)}^r}{{(e^{2(t-t_k)}-1)}^r}·\frac{e^{xt}}{t^{n-r+1}}\hfill \\ \hfill & =\frac{1}{2^r}\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n-r+1)},\hfill \end{array}$$

where $B_n^r$ denotes the Bernoulli numbers of order r defined by the generating function

$${\left(\frac{w}{e^w-1}\right)}^r=\sum _{n=0}^{\infty }{B}_n^r\frac{w^n}{n!}.$$

Applying the Leibniz Rule,

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left\{{(t-t_k)}^r{\left(\frac{1}{e^{2t}-1}\right)}^r\frac{e^{xt}}{t^{n-r+1}}\right\}\hfill \\ \hfill & =\frac{d^{r-1}}{d{t}^{r-1}}\left\{\frac{1}{2^r}\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n-r+1)}\right\}\hfill \\ \hfill & =\frac{1}{2^r}\frac{d^{r-1}}{d{t}^{r-1}}\left\{\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)t^{-(n-r+1)}\right\}\hfill \\ \hfill & =\frac{1}{2^r}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}·\frac{d^j}{d{t}^j}\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^j}{d{t}^j}& \left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{2^n{(t-t_k)}^n)}{n!}\right)\hfill \\ \hfill & =\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{e}^{xt}\sum _{n=l}^{\infty }{B}_n^r\frac{2^n}{n!}{\left(n\right)}_l{(t-t_k)}^{n-l}\hfill \\ \hfill & =e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!},\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left({(t-t_k)}^r{\left(\frac{1}{e^{2t}-1}\right)}^r\frac{e^{xt}}{t^{n-r+1}}\right)\hfill \\ \hfill & =\frac{1}{2^r}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}\hfill \\ \hfill & \times e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!},\hfill \end{array}$$

$$\begin{array}{cc}\hfill Res\left(g\left(t\right),t=t_k\right)=\frac{1}{(r-1)!}& \underset{t\to t_k}{lim}\frac{1}{2^r}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}\hfill \\ \hfill & \times \underset{t\to t_k}{lim}{e}^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}.\hfill \end{array}$$

Note that $B_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}\to 0$ as $t\to t_k$ except when $n=l$.

$$\begin{array}{cc}\hfill Res\left(g\left(t\right),t=t_k\right)& =\frac{1}{(r-1)!}\frac{1}{2^r}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right){(-1)}^{r-1-j}{(n-r+r-1-j)}_{r-1-j}{t}_k^{-(n-r+r-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\frac{1}{2^r(r-1)!}\sum _{j=0}^{r-1}\frac{(r-1)!}{j!(r-1-j)!}{(-1)}^{r-1-j}{(n-1-j)}_{r-1-j}{t}_k^{-(n-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\frac{1}{2^r}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{t_k^{j-n}}{j!}{e}^{x{t}_k}{2}^j\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)\frac{x^{j-l}}{2^{j-l}}{B}_l^r.\hfill \end{array}$$

We recall that $B_j^r\left(\frac{x}{2}\right)={\sum }_{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)B_l^r{\left(\frac{x}{2}\right)}^{j-l}$. Hence,

$$\begin{array}{cc}\hfill Res\left(g\left(t\right),t=t_k\right)& =\frac{1}{2^r}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}{2}^j\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{t_k^{j-n}}{j!}{e}^{x{t}_k}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & =\frac{1}{2^r}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}{2}^j\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{n-j}}\hfill \\ \hfill & =\frac{1}{2^r}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{n-j}}.\hfill \end{array}$$

Taking $t_k=k\pi i$, we get

$$\begin{array}{cc}\hfill Res\left(g\left(t\right),t=t_k\right)& =\frac{1}{2^r}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{k\pi ix}}{{\left(k\pi i\right)}^{n-j}}.\hfill \end{array}$$

$$=\frac{1}{2^r{\left(\pi i\right)}^n}\sum _{j=0}^{r-1}{(-1)}^{r-j+1}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{{\left(\pi i\right)}^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\frac{e^{k\pi ix}}{k^{n-j}}.$$

Consequently,

$${\left(TB\right)}_n^r\left(x\right)=n!\frac{1}{2^r{\left(\pi i\right)}^n}\sum _{k\in \mathbb{Z},k\ne 0}\sum _{j=0}^{r-1}{(-1)}^{r-j}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{{\left(\pi i\right)}^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\frac{e^{k\pi ix}}{k^{n-j}}.$$

(21)

Now, we look at

$$i^{-(n-j)}\sum _{k\in \mathbb{Z},k\ne 0}\frac{e^{k\pi ix}}{k^{n-j}}.$$

(22)

Noting that $i^{-(n-j)}=e^{-(n-j)\pi i/2}$ and ${(-1)}^{n-j}=e^{(n-j)\pi i}$, we see that

$$\begin{array}{cc}\hfill i^{-(n-j)}\sum _{k\in \mathbb{Z},k\ne 0}\frac{e^{k\pi ix}}{k^{n-j}}& =i^{-(n-j)}\left\{\sum _{k=1}^{\infty }\frac{e^{k\pi ix}}{k^{n-j}}+{(-1)}^{n-j}\sum _{k=1}^{\infty }\frac{e^{k\pi ix}}{k^{n-j}}\right\}\hfill \\ \hfill & =\sum _{k=1}^{\infty }\frac{e^{\left[kx-(n-j)/2\right]\pi i}+e^{-\left[kx-(n-j)/2\right]\pi i}}{k^{n-j}}\hfill \\ \hfill & =2\sum _{k=1}^{\infty }\frac{cos\left[kx-(n-j)/2\right]\pi }{k^{n-j}}.\hfill \end{array}$$

(23)

Replacing (22) with (23) in (21), we get the desired formula (20). □

3.3. Higher-Order Tangent–Genocchi Polynomials

Here, Fourier series of the higher-order Tangent–Genocchi polynomials will similarly be derived.

Applying the Cauchy Integral Formula [19] to (5) yields

$$\frac{{\left(TG\right)}_n^r\left(x\right)}{n!}=\frac{1}{2\pi i}{\int }_C{\left(\frac{2t}{e^{2t}+1}\right)}^r{e}^{xt}\frac{dt}{t^{n+1}},$$

(24)

where C is a circle about zero with the radius $<\pi /2$. Consider the function

$$h\left(t\right)={\left(\frac{2t}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

This function has a pole of order $n+1$ at $t=0$ and poles of order r at $t_k$ where $t_k$ are zeros of $e^{2t}+1$. More precisely,

$$t_k:=t=(2k+1)\frac{\pi }{2}i,fork\in \mathbb{Z}.$$

(25)

Now, let $C_N$ be the circle about 0 of radius $(2N+1)\frac{\pi }{2}-ϵ$, where $0<ϵ<1$. This is the same $C_N$ used in Section 3.1. By the Cauchy Residue Theorem [19],

$$\frac{1}{2\pi i}{\int }_{C_N}h\left(t\right)dt=\sum _{k<N}Res\left(h\left(t\right),t=t_k\right)+Res\left(h\left(t\right),t=0\right).$$

(26)

Taking the limit as $N\to \infty$,

$$\underset{N\to \infty }{lim}\frac{1}{2\pi i}{\int }_{C_N}h\left(t\right)dt=\sum _{k\in \mathbb{Z}}Res\left(h\left(t\right),t=t_k\right).$$

(27)

We prove the following lemma:

Lemma 3.

For $0<x\le r$, $n>r$ and $r>1$, as $N\to \infty$,

$${\int }_{C_N}h\left(t\right)dt\to 0.$$

where $C_N$ is a circle about 0 with radius $R=(2N+1)\frac{\pi }{2}-ϵ$, $0<ϵ<1$.

Proof. 

Taking the modulus of the integral gives

$$\left|{\int }_{C_N}{\left(\frac{2t}{e^{2t}+1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|\le 2^r{\int }_{C_N}\frac{|e^{xt}|}{|e^{2t}{+1|}^r}·\frac{\left|dt\right|}{|t^{n-r+1}|},witht\in C_N,\left|t\right|=(2N+1)\frac{\pi }{2}-ϵ.$$

As shown in the proof of Lemma 1,

$$\frac{|e^{xt}|}{|e^{2t}{+1|}^r}\le K$$

for some constant K, $\forall t\in C_N$. Consequently,

$$\begin{array}{cc}\hfill \left|{\int }_{C_N}{\left(\frac{2}{e^{2t}+1}\right)}^r·\frac{e^{xt}}{t^{n+1}}dt\right|& \le 2^r{\int }_{C_N}\frac{|e^{xt}|}{|e^{2t}{+1|}^r}·\frac{\left|dt\right|}{|t^{n-r+1}|}\hfill \\ \hfill & \le 2^r{\int }_{C_N}K\frac{\left|dt\right|}{|t^{n-r+1}|}\hfill \\ \hfill & =\frac{2^{r}K}{{\left(\left(2N+1\right)\frac{\pi }{2}-ϵ\right)}^{n-r+1}}·\left[\left(2N+1\right)\frac{\pi }{2}-ϵ\right]\left(2\pi \right)\hfill \\ \hfill & <\frac{2^{r+1}K\pi }{{\left[\left(2N+1\right)\frac{\pi }{2}-ϵ\right]}^{n-r}}\to 0asN\to \infty forn\ge r.\hfill \end{array}$$

□

Applying Lemma 3, Equation (27) becomes

$$\begin{array}{cc}\hfill 0& =\sum _{k\in \mathbb{Z}}Res\left(h\left(t\right),t=t_k\right)+Res\left(h\left(t\right),t=0\right)\hfill \\ \hfill 0& =Res\left(h\left(t\right),t=0\right)+\sum _{k\in \mathbb{Z}}Res\left(h\left(t\right),t=t_k\right)\hfill \\ \hfill 0& =\frac{{\left(TG\right)}_n^r\left(x\right)}{n!}+\sum _{k\in \mathbb{Z}}Res\left(h\left(t\right),t=t_k\right).\hfill \end{array}$$

Thus, we have

$${\left(TG\right)}_n^r=-n!\sum _{k\in \mathbb{Z}}Res\left(h\left(t\right),t=t_k\right).$$

(28)

The desired Fourier series for the tangent–Genocchi polynomials of higher order is given in the following theorem.

Theorem 3.

For $0<x\le r$, $n>r$ and $r>1$,

$$\begin{array}{cc}\hfill {\left(TG\right)}_n^r\left(x\right)& =n!2^{n-1}\sum _{k=0}^{\infty }\sum _{j=0}^{r-1}{(-1)}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{1}{{\pi }^{n-j}j!}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & \times \frac{cos\left[(2k+1)\pi x/2-(n-j)\pi /2\right]}{{(2k+1)}^{n-j}},\hfill \end{array}$$

(29)

where $B_j^r\left(\frac{x}{2}\right)$ denotes the Bernoulli polynomials of order r defined by

$${\left(\frac{w}{e^w-1}\right)}^r{e}^{xw}=\sum _{n=0}^{\infty }{B}_n^r\left(x\right)\frac{w^n}{n!}.$$

Proof. 

For $r\ge 2$,

$$Res\left(h\left(t\right),t=t_k\right)=\frac{1}{(r-1)!}\underset{t\to t_k}{lim}\frac{d^{r-1}}{d{t}^{r-1}}{(t-t_k)}^r{\left(\frac{2t}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}.$$

Since $e^{-2t_k}=e^{-2(2k+1)\frac{\pi }{2}i}=1$, we can write ${\left(e^{2t}+1\right)}^r$ as

$$\begin{array}{cc}\hfill {\left(e^{2t}+1\right)}^r& ={(-1)}^r{\left(e^{2t}(-1)-1\right)}^r\hfill \\ \hfill & ={(-1)}^r{\left(e^{2t}·e^{-2t_k}-1\right)}^r\hfill \\ \hfill & ={(-1)}^r{\left(e^{2(t-t_k)}-1\right)}^r.\hfill \end{array}$$

Then, we have

$$\begin{array}{cc}\hfill {(t-t_k)}^r{\left(\frac{2t}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}& =\frac{{\left(2(t-t_k)\right)}^r}{{(-1)}^r{\left(e^{2(t-t_k)}-1\right)}^r}·\frac{e^{xt}}{t^{n-r+1}}\hfill \\ \hfill & ={(-1)}^r\frac{{\left(\left(2(t-t_k)\right)\right)}^r}{{(e^{2(t-t_k)}-1)}^r}·\frac{e^{xt}}{t^{n-r+1}}\hfill \\ \hfill & ={(-1)}^r\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n-r+1)},\hfill \end{array}$$

where $B_n^r$ denotes the Bernoulli numbers of order r defined by the generating function

$${\left(\frac{w}{e^w-1}\right)}^r=\sum _{n=0}^{\infty }{B}_n^r\frac{w^n}{n!}.$$

Applying the Leibniz Rule for differentiation,

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left\{{(t-t_k)}^r{\left(\frac{2t}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}\right\}\hfill \\ \hfill & =\frac{d^{r-1}}{d{t}^{r-1}}\left\{{(-1)}^r\left(\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)e^{xt}{t}^{-(n-r+1)}\right\}\hfill \\ \hfill & ={(-1)}^r\frac{d^{r-1}}{d{t}^{r-1}}\left\{\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right)t^{-(n-r+1)}\right\}\hfill \\ \hfill & ={(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}·\frac{d^j}{d{t}^j}\left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{{\left(2(t-t_k)\right)}^n}{n!}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^j}{d{t}^j}& \left(e^{xt}\sum _{n=0}^{\infty }{B}_n^r\frac{2^n{(t-t_k)}^n)}{n!}\right)\hfill \\ \hfill & =\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{e}^{xt}\sum _{n=l}^{\infty }{B}_n^r\frac{2^n}{n!}{\left(n\right)}_l{(t-t_k)}^{n-l}\hfill \\ \hfill & =e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!},\hfill \end{array}$$

$$\begin{array}{cc}\hfill \frac{d^{r-1}}{d{t}^{r-1}}& \left({(t-t_k)}^r{\left(\frac{2t}{e^{2t}+1}\right)}^r\frac{e^{xt}}{t^{n+1}}\right)\hfill \\ \hfill & ={(-1)}^r\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}\hfill \\ \hfill & \times e^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!},\hfill \end{array}$$

$$\begin{array}{cc}\hfill Res\left(h\left(t\right),t=t_k\right)=\frac{{(-1)}^r}{(r-1)!}& \underset{t\to t_k}{lim}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right)\frac{d^{r-1-j}}{d{t}^{r-1-j}}{t}^{-(n-r+1)}\hfill \\ \hfill & \times \underset{t\to t_k}{lim}{e}^{xt}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}\sum _{n=l}^{\infty }{2}^n{B}_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}.\hfill \end{array}$$

Note that $B_n^r\frac{{(t-t_k)}^{n-l}}{(n-l)!}\to 0$ as $t\to t_k$ except when $n=l$. Thus,

$$\begin{array}{cc}\hfill Res\left(g\left(t\right),t=t_k\right)& =\frac{{(-1)}^r}{(r-1)!}\sum _{j=0}^{r-1}\left(\genfrac{}{}{0pt}{}{r-1}{j}\right){(-1)}^{r-1-j}{(n-r+r-1-j)}_{r-1-j}{t}_k^{-(n-r+r-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\frac{{(-1)}^r}{(r-1)!}\sum _{j=0}^{r-1}\frac{(r-1)!}{j!(r-1-j)!}{(-1)}^{r-1-j}{(n-1-j)}_{r-1-j}{t}_k^{-(n-j)}\hfill \\ \hfill & \times e^{x{t}_k}\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)x^{j-l}{2}^l{B}_l^r\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{t_k^{j-n}}{j!}{e}^{x{t}_k}{2}^j\sum _{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)\frac{x^{j-l}}{2^{j-l}}{B}_l^r.\hfill \end{array}$$

We recall that $B_j^r\left(\frac{x}{2}\right)={\sum }_{l=0}^j\left(\genfrac{}{}{0pt}{}{j}{l}\right)B_l^r{\left(\frac{x}{2}\right)}^{j-l}$. Hence,

$$\begin{array}{cc}\hfill Res\left(h\left(t\right),t=t_k\right)& =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{t_k^{j-n}}{j!}{e}^{x{t}_k}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n-1-j}{r-1-j}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{n-j}}\hfill \\ \hfill & =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{x{t}_k}}{t_k^{n-j}}.\hfill \end{array}$$

Taking $t_k=(2k+1)\frac{\pi }{2}i$, we get

$$\begin{array}{cc}\hfill Res\left(h\left(t\right),t=t_k\right)& =\sum _{j=0}^{r-1}{(-1)}^{j-1}{2}^j\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{B_j^r\left(\frac{x}{2}\right)}{j!}\frac{e^{(2k+1)\frac{\pi }{2}ix}}{{\left((2k+1)\frac{\pi }{2}i\right)}^{n-j}}\hfill \end{array}$$

$$={\left(\frac{2}{\pi i}\right)}^n\sum _{j=0}^{r-1}{(-1)}^{j-1}\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{{\left(\pi i\right)}^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\frac{e^{(2k+1)\frac{\pi }{2}ix}}{{\left(2k+1\right)}^{n-j}}.$$

Hence,

$$\begin{array}{cc}\hfill \frac{{\left(TG\right)}_n^r\left(x\right)}{n!}& =-{\left(\frac{2}{\pi }\right)}^n\sum _{j=0}^{r-1}{(-1)}^{j-1}\left(\genfrac{}{}{0pt}{}{n-j-1}{r-j-1}\right)\frac{{\pi }^j}{j!}{B}_j^r\left(\frac{x}{2}\right)\hfill \\ \hfill & i^{-(n-j)}\sum _{k\in \mathbb{Z}}\frac{e^{(2k+1)\frac{\pi }{2}.ix}}{{\left(2k+1\right)}^{n-j}}.\hfill \end{array}$$

(30)

Now, we look at

$$i^{-(n-j)}\sum _{k\in \mathbb{Z}}\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{n-j}}.$$

(31)

Noting that $i^{-(n-j)}=e^{-(n-j)\pi i/2}$ and ${(-1)}^{n-j}=e^{(n-j)\pi i}$, we have

$$\begin{array}{cc}\hfill & i^{-(n-j)}\sum _{k\in \mathbb{Z}}\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{n-j}}=i^{-(n-j)}\left\{\sum _{k=0}^{\infty }\frac{e^{(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{n-j}}\right.\hfill \\ \hfill & \left.+{(-1)}^{n-j}\sum _{k=0}^{\infty }\frac{e^{-(2k+1)\frac{\pi i}{2}x}}{{(2k+1)}^{n-j}}\right\}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{e^{\left[(2k+1)x/2-(n-j)/2\right]\pi i}+e^{-\left[(2k+1)x/2-(n-j)/2\right]\pi i}}{{(2k+1)}^{n-j}}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\frac{2cos\left[(2k+1)\pi x/2-(n-j)\pi /2\right]}{{(2k+1)}^{n-j}}\hfill \\ \hfill & =2\sum _{k=0}^{\infty }\frac{cos\left[(2k+1)\pi x/2-(n-j)\pi /2\right]}{{(2k+1)}^{n-j}}.\hfill \end{array}$$

(32)

Replacing (31) with (32) in (30), we get the desired formula (29). □

4. Conclusions and Recommendation

The Fourier series expansion of the Tangent polynomials of higher order was derived using the Cauchy Residue Theorem. Moreover, two variations of higher-order tangent polynomials were defined, namely, Tangent–Bernoulli and Tangent–Genocchi polynomials of higher order, and their Fourier series expansions were derived using Cauchy Residue theorem. For future research work, the authors recommend to derive the Fourier series expansion of Apostol–Frobenius Tangent, Apostol–Frobenius Tangent–Bernoulli and Apostol–Frobenius Tangent–Genocchi polynomials of higher order using the Cauchy Residue Theorem.