Is Catalan’s Constant Rational?

Abstract

This paper employs a contour integral method to derive and evaluate the infinite sum of the Euler polynomial expressed in terms of the Hurwitz Zeta function. We provide formulae for several classes of infinite sums of the Euler polynomial in terms of the Riemann Zeta function and fundamental mathematical constants, including Catalan’s constant. This representation of Catalan’s constant suggests it could be rational.

1. Introduction

Studies involving infinite sums of higher transcendental functions such as the Harmonic numbers in terms of Catalan’s constant and other fundamental constants were conducted by Ramanujan [1] and further explored by Sitaramachandra Rao [2] and Flajolet et al. [3]. Euler polynomials appear in statistical physics as well as in semi-classical approximations to quantum probability distributions (Ballentine and McRae [4]).

Studies involving the infinite sum of Euler’s polynomial expressed in terms of the Hurwitz Zeta function are not available in the literature to the best of our knowledge. In this work, we look to expand upon previous work featuring the finite sum of special functions. We derived the infinite sum of the Euler polynomial over general indices expressed in terms of the difference of two Hurwitz Zeta functions. An immediate consequence of this representation is the ability to write down Catalan’s constant in terms of the infinite sum of the ratio of integers. We are also able to derive the infinite product of the exponential of the Euler polynomial in terms of the Beta function by using the derivative of the Hurwitz Zeta function.

We proceed by using the contour integral method [5] applied to Equation (24.2.6) in [6] to yield the contour integral representation given by

$$\begin{array}{c}\frac{1}{2\pi i}{\int }_C{\displaystyle \sum _{p=0}^{\infty }}\frac{{\pi }^p{a}^w{E}_p\left(x\right)w^{-k+p-1}}{p!}dw\\ =\frac{1}{2\pi i}{\int }_C{a}^w{w}^{-k-1}\sec \mathrm{h}\left(\frac{\pi w}{2}\right)\left(cosh\left(\frac{\pi w}{2}-\pi wx\right)-sinh\left(\frac{\pi w}{2}-\pi wx\right)\right)dw\end{array}$$

(1)

where $a,x,k\in \mathbb{C},Re\left(w\pi \right(1/2-x\left)\right)>0$. Using Equation (1), the main theorem to be derived and evaluated is obtained as follows

$$\sum _{p=0}^{\infty }\frac{a^{-p}{E}_p\left(x\right){(-k)}_p}{\mathsf{\Gamma }(p+1)}=2^{k+1}{a}^{-k}\left(\zeta \left(-k,\frac{a+x}{2}\right)-\zeta \left(-k,\frac{1}{2}(a+x+1)\right)\right)$$

(2)

where the variables $k,a,x$ are general complex numbers and the Pochhammer symbol ${(-k)}_p$ is given in Equation (5.2.5) in [6]. This new expression is then used to derive special cases in terms of the Riemann Zeta function, Catalan’s constant and Apery’s constant. The derivations follow the method used by us in [5]. This method involves using a form of the generalized Cauchy’s integral formula given by

$$\frac{y^k}{\mathsf{\Gamma }(k+1)}=\frac{1}{2\pi i}{\int }_C\frac{e^{wy}}{w^{k+1}}dw,$$

(3)

where $y,w\in \mathbb{C}$ and C represent in general an open contour in the complex plane where the bilinear concomitant [5] has the same value at the end points of the contour. This method involves using a form of Equation (3), then multiplying both sides by a function, after which the definite integral of both sides is used. This yields a definite integral in terms of a contour integral. Then, we multiply both sides of Equation (3) by another function and take the infinite sum of both sides such that the contour integrals of both equations are the same.

2. Left-Hand Side Contour Integral

We use the method in [5]. The cut and contour are in the first quadrant of the complex w-plane with $Re\left(w\right)>0$. The cut approaches the origin from the interior of the first quadrant and reaches infinity vertically; the contour moves round the origin with zero radius and is on the opposite side of the cut. Using a generalization of Cauchy’s integral formula (3), we first replace $y\to log\left(a\right),k\to k-p$; then, we multiply both sides by $\frac{{\pi }^p}{p!}{E}_p\left(x\right)$ and take the infinite sum over $p\in [0,\infty )$ to obtain

$$\sum _{p=0}^{\infty }\frac{{\pi }^p{E}_p\left(x\right){log}^{k-p}\left(a\right)}{p!(k-p)!}=\frac{1}{2\pi i}{\int }_C\sum _{p=0}^{\infty }\frac{{\pi }^p{a}^w{E}_p\left(x\right)w^{-k+p-1}}{p!}dw$$

(4)

where $Re\left(w\right)>0$. We apply Fubini’s theorem for multiple integrals and sums (see p. 178 in [7]) as the summand is of bounded measure over the space $\mathbb{C}\times [0,\infty )$.

3. The Hurwitz Zeta Function

The Hurwitz zeta function (25.11) (i) in [6] is defined by the infinite sum

$$\zeta (s,a)=\sum _{n=0}^{\infty }\frac{1}{{(n+a)}^s},$$

(5)

where $\zeta (s,a)$ has a meromorphic continuation in the s-plane, its only singularity in $\mathbb{C}$ is a simple pole at $s=1$ with residue 1. As a function of a, with $s(\ne 1)$ fixed, $\zeta (s,a)$ is analytic in the half-plane $Re\left(a\right)>0$.

The Hurwitz zeta function is continued analytically with a definite integral representation (25.11.25) in [6] given by

$$\zeta (s,a)=\frac{1}{\mathsf{\Gamma }\left(s\right)}{\int }_0^{\infty }\frac{x^{s-1}{e}^{-ax}}{1-e^{-x}}dx,$$

(6)

where $Re\left(s\right)>1,Re\left(a\right)>0$.

4. Contour Integral Representations for the Hurwitz Zeta Function

4.1. Derivation of the First Contour Integral

We use the method in [5]. Using Equation (3), we first form a new equation by replacing y with $y+x$. Next, a second equation is formed by replacing x with $-x$ and the second equation and the first are added, followed by replacing x with $\pi (1/2-x)$ to obtain

$$\frac{{\left(\pi \left(\frac{1}{2}-x\right)+y\right)}^k+{\left(\pi \left(x-\frac{1}{2}\right)+y\right)}^k}{\mathsf{\Gamma }(k+1)}=\frac{1}{2\pi i}{\int }_{C}2w^{-k-1}{e}^{wy}cosh\left(\frac{1}{2}\pi w(1-2x)\right)dw$$

(7)

Next, replace y with $log\left(a\right)+\frac{1}{2}\pi (2y+1)$ and multiply both sides by ${(-1)}^y$; then, take the infinite sum over $y\in [0,\infty )$ and simplify this in terms of the Hurwitz–Lerch Zeta function to obtain

$$\begin{array}{c}\frac{{\left(2\pi \right)}^k}{\mathsf{\Gamma }(k+1)}\left(\zeta \left(-k,\frac{-\pi x+log\left(a\right)+\pi }{2\pi }\right)-\zeta \left(-k,\frac{1}{2}\left(-x+\frac{log\left(a\right)}{\pi }+2\right)\right)\right.\hfill \\ \left.+\zeta \left(-k,\frac{1}{2}\left(x+\frac{log\left(a\right)}{\pi }\right)\right)-\zeta \left(-k,\frac{1}{2}\left(x+\frac{log\left(a\right)}{\pi }+1\right)\right)\right)\hfill \\ ={\displaystyle \sum _{y=0}^{\infty }}\frac{1}{2\pi i}{\int }_C{(-1)}^y{a}^w{w}^{-k-1}\left(e^{2\pi wx}+e^{\pi w}\right)e^{\pi w(y-x)}dw\hfill \\ =\frac{1}{2\pi i}{\int }_C{\displaystyle \sum _{y=0}^{\infty }}{(-1)}^y{a}^w{w}^{-k-1}\left(e^{2\pi wx}+e^{\pi w}\right)e^{\pi w(y-x)}dw\hfill \\ =\frac{1}{2\pi i}{\int }_C{a}^w{w}^{-k-1}\sec \mathrm{h}\left(\frac{\pi w}{2}\right)cosh\left(\frac{1}{2}\pi w(1-2x)\right)dw\hfill \end{array}$$

(8)

from Equation (1.232.2) in [8], where $Im\left(w\right)>0$ in order for the sum to converge. We apply Fubini’s theorem for integrals and sums (see p. 178 in [7]) as the summand is of bounded measure over the space $\mathbb{C}\times [0,\infty )$.

4.2. Derivation of the Second Contour Integral

We use the method in [5]. Using Equation (3), we first form a new equation by replacing y with $y+x$. Next, we form a second equation by replacing x with $-x$ and subtracting the second equation from the first, followed by replacing x with $\pi (1/2-x)$ to obtain

$$\frac{{\left(\pi \left(x-\frac{1}{2}\right)+y\right)}^k-{\left(\pi \left(\frac{1}{2}-x\right)+y\right)}^k}{\mathsf{\Gamma }(k+1)}=-\frac{1}{2\pi i}{\int }_{C}2w^{-k-1}{e}^{wy}sinh\left(\frac{1}{2}\pi w(1-2x)\right)dw$$

(9)

Next, we replace y with $log\left(a\right)+\frac{1}{2}\pi (2y+1)$ and multiply both sides by ${(-1)}^y$; we then take the infinite sum over $y\in [0,\infty )$ and simplify it in terms of the Hurwitz–Lerch Zeta function to obtain

$$\begin{array}{c}\frac{{\left(2\pi \right)}^k}{\mathsf{\Gamma }(k+1)}\left(-\zeta \left(-k,\frac{-\pi x+log\left(a\right)+\pi }{2\pi }\right)+\zeta \left(-k,\frac{1}{2}\left(-x+\frac{log\left(a\right)}{\pi }+2\right)\right)\right.\\ \left.+\zeta \left(-k,\frac{1}{2}\left(x+\frac{log\left(a\right)}{\pi }\right)\right)-\zeta \left(-k,\frac{1}{2}\left(x+\frac{log\left(a\right)}{\pi }+1\right)\right)\right)\\ ={\displaystyle \sum _{y=0}^{\infty }}\frac{1}{2\pi i}{\int }_C{(-1)}^{y+1}{w}^{-k-1}\left(e^{\pi w(1-2x)}-1\right)e^{w(log(a)+\pi (x+y\left)\right)}dw\\ =\frac{1}{2\pi i}{\int }_C{\displaystyle \sum _{y=0}^{\infty }}{(-1)}^{y+1}{w}^{-k-1}\left(e^{\pi w(1-2x)}-1\right)e^{w(log(a)+\pi (x+y\left)\right)}dw\\ =-\frac{1}{2\pi i}{\int }_C{a}^w{w}^{-k-1}\mathrm{sech}\left(\frac{\pi w}{2}\right)sinh\left(\frac{1}{2}\pi w(1-2x)\right)dw\end{array}$$

(10)

from Equation (1.232.2) in [8], where $Im\left(w\right)>0$ in order for the sum to converge. We apply Fubini’s theorem for integrals and sums, (see p. 178 in [7]) as the summand is of bounded measure over the space $\mathbb{C}\times [0,\infty )$.

5. The Infinite Sum of the Euler Polynomial in Terms of Hurwitz Zeta Function

In this section, we will derive the finite sum of the Euler polynomial in terms of the Hurwitz Zeta function. Some of the functions used in this section are the Polygamma function, obtained with $\zeta (n+1,a)=\frac{{(-1)}^{n+1}{\psi }^{\left(n\right)}\left(a\right)}{n!}$, which is Equation (25.11.12) in [6]. Another equation we will use is ${\zeta }^{\prime }(0,a)=log\left(\mathsf{\Gamma }\left(a\right)\right)-\frac{1}{2}log\left(2\pi \right)$, which is Equation (25.11.18) in [6].

Theorem 1.

For all $k,a,x\in \mathbb{C},Re\left(a\right)>2\pi$ then,

$$\sum _{p=0}^{\infty }\frac{a^{-p}{E}_p\left(x\right){(-k)}_p}{\mathsf{\Gamma }(p+1)}=2^{k+1}{a}^{-k}\left(\zeta \left(-k,\frac{a+x}{2}\right)-\zeta \left(-k,\frac{1}{2}(a+x+1)\right)\right)$$

(11)

Proof. 

When applying Equation (1), the right-hand side of relation (4) is equal to the addition of the right-hand sides of Equations (8) and (10); hence, the left-hand sides of the equations are identical too. Simplification with the Gamma function and Pochhammer symbol yields the desired conclusion. □

Example 1.

$$\sum _{p=0}^{\infty }{(-1)}^p(p+1)a^{-p}{E}_p\left(x\right)=\frac{1}{2}{a}^2\left({\psi }^{\left(1\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(1\right)}\left(\frac{1}{2}(a+x+1)\right)\right)$$

(12)

Proof. 

Use Equation (11) and set $k=-2$ and simplify using Equation (25.11.12) in [6]. □

5.1. Catalan’s Constant in Terms of the Euler’s Polynomial

In this section, we will use Equation (12) to derive a closed form formula in terms of Catalan’s constant. We will also use Wolfram Mathematica to assist in the numerical evaluation, approximation and actual value of Catalan’s constant.

Example 2.

Catalan’s constant in terms of the infinite sum of the Euler Polynomial for all $a\in {\mathbb{Z}}^{+}$.

$$\sum _{p=0}^{\infty }2{(-1)}^p(p+1){\left(\frac{1}{a}\right)}^{p+2}{E}_p\left(-\frac{1}{2}\right)={\psi }^{\left(1\right)}\left(\frac{a}{2}-\frac{1}{4}\right)-{\psi }^{\left(1\right)}\left(\frac{a}{2}+\frac{1}{4}\right)$$

(13)

Proof. 

Use Equation (12) and set $k=-2$, $x=-1/2$ and simplify the Hurwitz Zeta function. □

Example 3.

Catalan’s constant C.

$$\sum _{p=0}^{\infty }{\left(-\frac{1}{14}\right)}^p(p+1)E_p\left(-\frac{1}{2}\right)\approx \frac{\mathrm{82,124,372,321,267,331,872}}{\mathrm{57,134,420,638,205,625}}-1568C$$

(14)

Proof. 

Use Equation (12), set $k=-2,a=14,x=-1/2$ and simplify the Hurwitz zeta function using Equation (2.2.1.2.7) in [9]. □

Example 4.

The Digamma Function ${\psi }^{\left(0\right)}\left(z\right)$.

$$\sum _{p=0}^{\infty }{\left(-\frac{1}{a}\right)}^{p+1}{E}_p\left(x\right)={\psi }^{\left(0\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(0\right)}\left(\frac{1}{2}(a+x+1)\right)$$

(15)

Proof. 

Use Equation (11) and apply l’Hopitals’ rule as $k\to -1$ and simplify the Hurwitz Zeta function using Equation (25.4.1) in [6]. □

Example 5.

The Riemann Zeta Function $\zeta \left(z\right)$.

$$\sum _{p=0}^{\infty }\frac{{(-1)}^p{(2-x)}^{-p}{E}_p\left(x\right){(-k)}_p}{\mathsf{\Gamma }(p+1)}={(2-x)}^{-k}\left(\left(2^{k+2}-2\right)\zeta (-k)+2\right)$$

(16)

Proof. 

Use Equation (12) and set $a=2-x$ and simplify the Hurwitz Zeta function using Equation (25.11.2) in [6]. □

Example 6.

An example from Knuth’s series in terms of $\zeta (1/2)$.

$$\sum _{p=0}^{\infty }\frac{{(-1)}^p{\left(\frac{1}{2}\right)}_p{(2-x)}^{-p}{E}_p\left(x\right)}{\mathsf{\Gamma }(p+1)}=\sqrt{2-x}\left(2\left(\sqrt{2}-1\right)\zeta \left(\frac{1}{2}\right)+2\right)$$

(17)

Proof. 

Use Equation (16) and set $k=-1/2$ and simplify the Riemann Zeta function. See [10] for another form for $\zeta (1/2)$. □

Example 7.

An example in terms of $\zeta (3/2)$.

$$\sum _{p=0}^{\infty }\frac{{(-1)}^p{(2-x)}^{-p}\mathsf{\Gamma }\left(p-\frac{1}{2}\right)E_p\left(x\right)}{\mathsf{\Gamma }(p+1)}=\frac{-\zeta \left(\frac{3}{2}\right)+2\sqrt{2}\zeta \left(\frac{3}{2}\right)-4\pi }{\sqrt{\pi }\sqrt{2-x}}$$

(18)

Proof. 

Use Equation (16), set $k=-3/2$ and simplify the Riemann Zeta function. □

Example 8.

An example in terms of $log\left(2\right)$.

$$\sum _{p=0}^{\infty }{\left(-\frac{1}{12}\right)}^p{E}_p(-10)=-24(log\left(2\right)-1)$$

(19)

Proof. 

Use Equation (16) and apply l’Hopitals rule as $k\to -1$ and simplify the Riemann Zeta function using Equation (25.4.1) in [6]. □

Example 9.

An example in terms of Apéry’s $\zeta \left(3\right)$.

$$\sum _{p=0}^{\infty }{\left(-\frac{1}{20}\right)}^p(p+1)(p+2)E_p\left(5\right)\approx \mathrm{24,000}\zeta \left(3\right)-\frac{\mathrm{138,375,992,572,893,310,789,654,802,183}}{\mathrm{4,796,680,795,562,969,871,654,096}}$$

(20)

Proof. 

Use Equation (11) and set $k=-3,a=20,x=5$ and simplify the Hurwitz Zeta function using Equation (25.11.1) in [6]. □

Example 10.

A Double Sum in Terms of the Polygamma Function.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{q=0}^{\infty }}q{(-1)}^{p+q}{a}^{-p-q}{E}_p\left(x\right)E_{q-1}(-x)\\ =\frac{1}{2}{a}^2\left({\psi }^{\left(0\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(0\right)}\left(\frac{1}{2}(a+x+1)\right)\right)\left({\psi }^{\left(1\right)}\left(\frac{a-x}{2}\right)-{\psi }^{\left(1\right)}\left(\frac{1}{2}(a-x+1)\right)\right)\end{array}$$

(21)

Proof. 

Use Equation (11) and apply l’Hoptial’s rule as $k\to -1$ and simplify this using Equation (25.11.12) in [6] to form a second equation. Use the second equation and take the first partial derivative with respect to x and replace x with $-x$ and p with q; then, multiply Equation (11) again with $q-1$ and $E_{q-1/2}$ and simplify. □

Example 11.

A double sum in terms of Catalan’s constant C.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}{2}^{1-q}q{E}_{q-1}{E}_p\left(-\frac{1}{2}\right){\left(-\frac{1}{11}\right)}^{p+q}\\ \approx \frac{121}{2}\left(\frac{\mathrm{44,257,352}}{\mathrm{14,549,535}}-\pi \right)\left(\frac{\mathrm{3,105,874,656,398,608}}{\mathrm{211,688,968,716,225}}-16C\right)\end{array}$$

(22)

Proof. 

Use Equation (21) and set $k=-3,a=11,x=-1/2$ and simplify. □

5.2. An Open Problem

$$\sum _{p=0}^{\infty }{\left(-\frac{1}{11}\right)}^p(1+p)(E_p(-2)-E_p\left(2\right))\left(-1+H_{1+p}\right)=\frac{121}{\mathrm{16,200}}log\left(\frac{2^{774}{5}^{324}}{3^{575}{11}^{149}}\right)$$

(23)

5.3. The Derivative with Respect to k

Proposition 1.

For all $k,a,x\in \mathbb{C},Re\left(k\right)\le 0$ then,

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right){(-k)}_p({\psi }^{\left(0\right)}(-k)-{\psi }^{\left(0\right)}(p-k))}{\mathsf{\Gamma }(p+1)}\\ =2^{k+1}{\left(\frac{1}{a}\right)}^k\left(-{\zeta }^{\prime }\left(-k,\frac{a+x}{2}\right)+{\zeta }^{\prime }\left(-k,\frac{1}{2}(a+x+1)\right)\right.\\ \left.+log\left(\frac{2}{a}\right)\left(\zeta \left(-k,\frac{a+x}{2}\right)-\zeta \left(-k,\frac{1}{2}(a+x+1)\right)\right)\right)\end{array}$$

(24)

Proof. 

In Equation (11), use the first partial derivative with respect to k and simplify. □

Example 12.

The Log-gamma Function $log\left(\mathsf{\Gamma }\right(z\left)\right)$.

$$\sum _{p=1}^{\infty }{\left(-\frac{1}{a}\right)}^p\frac{E_p\left(x\right)}{p}=2log\left(\frac{\sqrt{\frac{a}{2}}\mathsf{\Gamma }\left(\frac{a+x}{2}\right)}{\mathsf{\Gamma }\left(\frac{1}{2}(1+a+x)\right)}\right)$$

(25)

Proof. 

Use Equation (24) to apply l’Hopital’s rule as $k\to 0$ and simplify using Equation (25.11.18) in [6]. □

Example 13.

Product Representation in terms of Quotient Gamma Functions.

$$\prod _{p=1}^{\infty }{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\frac{\sqrt{a}\mathsf{\Gamma }\left(\frac{a+x}{2}\right)}{\sqrt{2}\mathsf{\Gamma }\left(\frac{1}{2}(a+x+1)\right)}$$

(26)

Proof. 

In Equation (25), take the exponential of both sides and simplify. □

Example 14.

Product Representation in terms of A Definite Integral.

$$\prod _{p=1}^{\infty }{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\sqrt{\frac{2a}{\pi }}{\int }_0^{\pi /2}{cos}^{a+x-1}\left(\theta \right)d\theta$$

(27)

Proof. 

Use Equation (25) and simplify using Equation (5.12.2) in [6]. Note that other definite integrals can be derived using Equations (3.667.1), (3.688.14), (4.532.5), (4.561) and (6.575.2) in [8]. Similar types of product representations for definite integrals include Equation (6.441) in [8] and Equation (2.5.46.1) in [11]. □

Example 15.

A Product Representation Using The Duplication Formula.

$$\prod _{p=1}^{\infty }{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\frac{\sqrt{\pi }\sqrt{a}{2}^{-a-x+\frac{1}{2}}\mathsf{\Gamma }(a+x)}{\mathsf{\Gamma }{\left(\frac{a+x}{2}+\frac{1}{2}\right)}^2}$$

(28)

Proof. 

Use Equation (25) and simplify using Equation (5.5.5) in [6]. □

Example 16.

The n-dimensional volume of a Euclidean ball of radius R in n-dimensional Euclidean space is given by Equation (5.19.4) in [6]. The volume can also be expressed in terms of an $(n-1)$-ball using the following one-dimension recurrence relation:

$$\prod _{p=1}^{\infty }{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\frac{\sqrt{a}\mathsf{\Gamma }\left(\frac{a+x}{2}\right)}{\sqrt{2}\mathsf{\Gamma }\left(\frac{1}{2}(a+x+1)\right)}=\sqrt{\frac{a}{2R}}\frac{V_{a+x-1}\left(R\right)}{V_{a+x-2}\left(R\right)}$$

(29)

Proof. 

Use Equation (25) and simplify using Equation (5.19.4) in [6]. □

Example 17.

The Harmonic Number Function $H_n$.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}{\left(-\frac{1}{a}\right)}^p{H}_p{E}_p\left(x\right)=a\left(-{\gamma }_1\left(\frac{a+x}{2}\right)+{\gamma }_1\left(\frac{1}{2}(a+x+1)\right)\right.\\ \left.+log\left(\frac{2}{a}\right)\left({\psi }^{\left(0\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(0\right)}\left(\frac{1}{2}(a+x+1)\right)\right)\right)\end{array}$$

(30)

Proof. 

Use Equation (24) to apply l’Hopital’s rule as $k\to -1$ and simplify using equations (25.2.5) and (25.11.21) in [6]. □

Example 18.

An Euler Polynomial Identity.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}\frac{\left(1+{(-1)}^p\right){\left(\frac{1}{a}\right)}^p{(-k)}_p{E}_p\left(x\right)}{\mathsf{\Gamma }(1+p)}=2^{1+k}{\left(\frac{1}{a}\right)}^k\left(\zeta \left(-k,\frac{1}{2}(1+a-x)\right)-\zeta \left(-k,\frac{1}{2}(2+a-x)\right)\right.\\ \left.+\zeta \left(-k,\frac{a+x}{2}\right)-\zeta \left(-k,\frac{1}{2}(1+a+x)\right)\right)\end{array}$$

(31)

Proof. 

Use Equation (11) and apply Equation (24.4.4) in [6] and simplify. □

Example 19.

Appell Sequence and the multiplication theorem.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{j=0}^{2m}}\frac{{(-1)}^{j+p}{e}^{-a\pi p}{(1+2m)}^p{E}_p\left(\frac{j+x}{1+2m}\right){(-k)}_p}{\mathsf{\Gamma }(1+p)}\\ =2^{1+k}{e}^{-a\pi k}\left(\zeta \left(-k,\frac{1}{2}\left(e^{a\pi }+x\right)\right)-\zeta \left(-k,\frac{1}{2}\left(1+e^{a\pi }+x\right)\right)\right)\end{array}$$

(32)

Proof. 

Use Equation (11) and simplify the left-hand side using Equation (4.2.11) in [12]. □

Example 20.

Appell Sequence and the multiplication theorem and the Bernoulli polynomial.

$$\begin{array}{c}{\displaystyle \sum _{p=0}^{\infty }}{\displaystyle \sum _{j=0}^{2m-1}}\frac{{(-1)}^{j+p}{e}^{-a\pi p}{\left(2m\right)}^p{B}_{1+p}\left(\frac{j+x}{2m}\right){(-k)}_p}{\mathsf{\Gamma }(2+p)}\\ =2^k{e}^{-a\pi k}\left(-\zeta \left(-k,\frac{1}{2}\left(e^{a\pi }+x\right)\right)+\zeta \left(-k,\frac{1}{2}\left(1+e^{a\pi }+x\right)\right)\right)\end{array}$$

(33)

Proof. 

Use Equation (11) and simplify the left-hand side using Equation (4.2.11) in [12]. □

6. Discussion

In this paper, we have presented a method for deriving the infinite sum of the Euler polynomial in terms of special functions and mathematical constants using contour integration. In Section 5, we derived a closed form solution for the infinite sum of Euler’s polynomial in terms of the Hurwitz Zeta function. We also derived Catalan’s constant in terms of the infinite sum of the ratio of integers. We also expressed the Riemann Zeta function in terms of the infinite sum of Euler’s polynomial. The infinite sum of the ratio of integers was also used to derive Apéry’s constant $\zeta \left(3\right)$. Some interesting derivations in terms of the infinite product of the exponential function and Beta function were also obtained, which are new to the literature to the best of our knowledge. The ratio of the Gamma function in terms of the infinite product of the exponential function was also produced. A summary Table 1 was also produced for ease of reading. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram.

Table 1. Summary Table of Results.

${\displaystyle \sum _{p=0}^{\infty }}\frac{a^{-p}{E}_p\left(x\right){(-k)}_p}{\mathsf{\Gamma }(p+1)}=2^{k+1}{a}^{-k}\left(\zeta \left(-k,\frac{a+x}{2}\right)-\zeta \left(-k,\frac{1}{2}(a+x+1)\right)\right)$

${\displaystyle \sum _{p=0}^{\infty }}{(-1)}^p(p+1)a^{-p}{E}_p\left(x\right)=\frac{1}{2}{a}^2\left({\psi }^{\left(1\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(1\right)}\left(\frac{1}{2}(a+x+1)\right)\right)$

${\displaystyle \sum _{p=0}^{\infty }}{\left(-\frac{1}{a}\right)}^{p+1}{E}_p\left(x\right)={\psi }^{\left(0\right)}\left(\frac{a+x}{2}\right)-{\psi }^{\left(0\right)}\left(\frac{1}{2}(a+x+1)\right)$

${\displaystyle \sum _{p=0}^{\infty }}\frac{{(-1)}^p{(2-x)}^{-p}{E}_p\left(x\right){(-k)}_p}{\mathsf{\Gamma }(p+1)}={(2-x)}^{-k}\left(\left(2^{k+2}-2\right)\zeta (-k)+2\right)$

${\displaystyle \sum _{p=0}^{\infty }}{\left(-\frac{1}{12}\right)}^p{E}_p(-10)=-24(log\left(2\right)-1)$

${\displaystyle \sum _{p=1}^{\infty }}{\left(-\frac{1}{a}\right)}^p\frac{E_p\left(x\right)}{p}=2log\left(\frac{\sqrt{\frac{a}{2}}\mathsf{\Gamma }\left(\frac{a+x}{2}\right)}{\mathsf{\Gamma }\left(\frac{1}{2}(1+a+x)\right)}\right)$

${\displaystyle \prod _{p=1}^{\infty }}{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\frac{\sqrt{a}\mathsf{\Gamma }\left(\frac{a+x}{2}\right)}{\sqrt{2}\mathsf{\Gamma }\left(\frac{1}{2}(a+x+1)\right)}$

${\displaystyle \prod _{p=1}^{\infty }}{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\sqrt{\frac{2a}{\pi }}{\int }_0^{\pi /2}{cos}^{a+x-1}\left(\theta \right)d\theta$

${\displaystyle \prod _{p=1}^{\infty }}{e}^{\frac{{\left(-\frac{1}{a}\right)}^p{E}_p\left(x\right)}{2p}}=\frac{\sqrt{\pi }\sqrt{a}{2}^{-a-x+\frac{1}{2}}\mathsf{\Gamma }(a+x)}{\mathsf{\Gamma }{\left(\frac{a+x}{2}+\frac{1}{2}\right)}^2}$

7. Conclusions

One of our goals in this paper was to determine a new mathematical property of Catalan’s constant. Based on our work, we can write Catalan’s constant in terms of the ratio of an infinite number of positive integers. We were also able to achieve the same form for Apéry’s constant. We will be applying this method to derive other formulae in future work.