# TRIGONOMETRIC MELNIKOV PRODUCTS: EXTENSION, DERIVATION AND EVALUATION

Robert Reynolds

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Robert Reynolds. TRIGONOMETRIC MELNIKOV PRODUCTS: EXTENSION, DERIVATION AND EVALUATION: TRIGONOMETRIC MELNIKOV PRODUCTS: EXTENSION, DERIVATION AND EVALUATION. 2023. [https://hal.science/hal-04017422](https://hal.science/hal-04017422)

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# TRIGONOMETRIC MELNIKOV PRODUCTS: EXTENSION, DERIVATION AND EVALUATION

Robert Reynolds

**ABSTRACT.** Trigonometric Melnikov products are considered. These products are taken over both infinite and finite domains which involve trigonometric functions where the range of evaluation of parameters involved are general complex numbers. This property represents an extension of the infinite product of elementary functions previously published in current literature. These products are derived from the sums of the Hurwitz-Lerch Zeta and incomplete gamma functions which can be simplified to composite functions in special cases of integer values of the parameters involved. Symmetric plots are produced to showcase interesting features of special cases of infinite product forms.

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## 1. Introduction

In the work of Melnikov [1], an innovative technique was developed for obtaining infinite product representations for some elementary functions. The technique was based on the comparison of alternative expressions of Green's functions constructed by two different methods. Some standard boundary value problems were considered posed for two-dimensional Laplace equation on regions of a regular configuration. The study of trigonometric functions spans a wide range of scientific areas such Greens functions [2] and the Laplace equation [3] to name a few. Infinite products are studied in the work of Chamberland et al. [4], where convergent infinite products were used in applications involving multiplicative partitions, entries in Ramanujan's notebooks, the Chowla-Selberg formula, and the Thue-Morse sequence.

In this paper the contour integral method in [5] is applied to the Valean sum listed equation (3.59) in [6], to produce analogous sums involving the Hurwitz-Lerch Zeta function. This work's overarching theme is that one may express finite and infinite sums, as well as products involving special functions and their composite functions in terms of special functions. Note that the contrast of a series' infinite form with its partial sum, might have intriguing analysis properties. A definite integral of an infinite product is taken and written down in terms Catalan's constant K.

<footnote n="2020">2020 Mathematics Subject Classification. Primary 30E20, 33-01, 33-03, 33-04.

Key words and phrases. Melnikov product, trigonometric function, Cauchy integral, Hurwitz-Lerch Zeta function.</footnote>
Our preliminaries comprise three equations starting with a trigonometric sum given by

$$ (1.1) \qquad \sum_{j=0}^{\infty} \left( 2^{-j} \sec(2^{-j}x) - 2^{-j} \cot\left(\frac{1}{4}(2^{1-j}x + \pi)\right) \right) = \frac{1}{x} - 2\cot(2x) $$

where $x \in \mathbb{C}$. Equation (1.1) is derived by looking at the trigonometric difference equation $\cot(b/2) - \cot(b)$ then replacing $b \to b + \pi/2$. Next replace $x \to 2^{-j}x$ and taking the sum over $j \in [0, n]$ and simplify and then taking the limit as $n \to \infty$ and simplify. The second and third equations involve a contour integral method applied to equation (3.303) in [6] and (1.1). Let $a, k, m$ and $w$ be general complex numbers and $j \in [0, \infty)$, the contour integral form [5] of the Vălean-Cotangent-Secant sum is given by

$$ (1.2) \qquad \frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} \left( -2^{-j} a^w w^{-k-1} \cot \left( \frac{1}{4} (\pi - 2^{-j}(m+w)) \right) \right. \\ \qquad \left. + 2a^w w^{-k-1} \csc (2^{-j}(m+w)) + 2^{-j} a^w w^{-k-1} \sec (2^{-j-1}(m+w)) \right. \\ \qquad \left. - 2a^w w^{-k-1} \cosh (2^{-j}(m+w)) \right) dw \\ \qquad = \frac{1}{2\pi i} \int_C \left( 2a^w w^{-k-1} \coth(m+w) - \frac{2a^w w^{-k-1}}{m+w} \right) dw $$

and

$$ (1.3) \qquad \frac{1}{2\pi i} \int_C \sum_{j=0}^{n} 2^{-j} a^w w^{-k-1} \left( -\cot\left(\frac{1}{4}(2^{1-j}(m+w)+\pi)\right) \right. \\ \qquad \left. + \sec(2^{-j}(m+w)) + 2^j \left(\csc(2^{-j}(m+w)) - \cosh(2^{-j}(m+w))\right) \right) dw \\ \qquad = \frac{1}{2\pi i} \int_C \left( -a^w w^{-k-1}(\cot(m+w) + 2\cot(2(m+w))) \right. \\ \qquad \left. + 2^{-n} a^w w^{-k-1} (2^n \cot(2^{-n-1}(m+w)) + \cot(2^{-n}(m+w))) \right. \\ \qquad \left. + a^w w^{-k-1} (\coth(m+w) - \coth(2^{-n-1}(m+w))) \right) dw $$

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

$$ (1.4) \qquad \frac{y^k}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C \frac{e^{wy}}{w^{k+1}} dw, $$

where $y, w \in \mathbb{C}$ and $C$ is 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 (1.4) then multiplies both sides by a function, then takes the sum of both sides. This yields a sum in terms of a contour integral. Then multiply both sides of equation (1.4) by another function and take the infinite sum of both sides such that the contour integral of both equations are the same.

## 2. The Hurwitz-Lerch Zeta function

Use equation (1.11.3) in [7] where $\Phi(z, s, v)$ is the Hurwitz-Lerch Zeta function which is a generalization of the Hurwitz zeta $\zeta(s, v)$ and Polylogarithm functions $\mathrm{Li}_n(z)$. The Lerch function has a series representation given by
$$ (2.1) \qquad \Phi(z, s, v) = \sum_{n=0}^{\infty} (v+n)^{-s} z^n $$

where $|z|< 1, v \neq 0, -1, -2, -3, \dots$, and is continued analytically by its integral representation given by

$$ (2.2) \qquad \Phi(z, s, v) = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-vt}}{1 - ze^{-t}} dt = \frac{1}{\Gamma(s)} \int_{0}^{\infty} \frac{t^{s-1} e^{-(v-1)t}}{e^t - z} dt $$

where $Re(v) > 0$, and either $|z| \le 1, z \ne 1, Re(s) > 0$, or $z = 1, Re(s) > 1$. Almost all Huwitz-Lerch Zeta functions have an asymmetrical zero distribution [8].

### 3. Derivation Of The Infinite Sums Of The Contour Integral Representation

In this section use the Cauchy integral formula and the stated contour integral method to derive the infinite sum of the Hurwitz-Lerch Zeta functions in terms of its contour integral representation.

**3.1. Left-hand side first contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) - i2^{-j-1}(y+1)$ then multiply both sides by $i2^{1-j}e^{-i2^{-j-1}(y+1)(m-\pi 2^j)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ \begin{align} \sum_{j=0}^{\infty} \frac{i2^{-j+k+1} (-i2^{-j-2})^k e^{2i(\frac{\pi}{4}-2^{-j-2}m)} \Phi\left(e^{2i(\frac{\pi}{4}-2^{-j-2}m)}, -k, i2^{j+1} \log(a)+1\right)}{\Gamma(k+1)} & \nonumber \\ &= \frac{1}{2\pi i} \sum_{y=0}^{\infty} \sum_{j=0}^{\infty} \int_C 2^{1-j} a^w w^{-k-1} e^{\frac{1}{2}i(\pi y - 2^{-j}(y+1)(m+w))} dw & \nonumber \\ &= \frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} \sum_{y=0}^{\infty} 2^{1-j} a^w w^{-k-1} e^{\frac{1}{2}i(\pi y - 2^{-j}(y+1)(m+w))} dw & \nonumber \\ &= -\frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} \left( 2^{-j} a^w w^{-k-1} \cot\left(\frac{\pi}{4} - 2^{-j-2}(m+w)\right) + i2^{-j} a^w w^{-k-1} \right) dw & \end{align} $$

from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177 in [10] as the summands are of bounded measure over the space $C \times [0, \infty) \times [0, \infty)$.

**3.2. Left-hand side first additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $-i2^{-j}$ and simplify to get

$$ (3.2) \qquad -\frac{i2^{-j} \log^k(a)}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C i2^{-j} a^w w^{-k-1} dw $$
**3.3. Left-hand side second contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)+i2^{-j}(2y+1)$ then multiply both sides by $-4ie^{i2^{-j}m(2y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (3.3) \quad -\sum_{j=0}^{\infty} \frac{i2^{k+2} (i2^{-j})^k e^{i2^{-j}m} \Phi(e^{i2^{1-j}m}, -k, \frac{1}{2}(1-i2^j \log(a)))}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \sum_{j=0}^{\infty} \int_C 4ia^w w^{-k-1} e^{i2^{-j}(2y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} \sum_{y=0}^{\infty} 4ia^w w^{-k-1} e^{i2^{-j}(2y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} 2a^w w^{-k-1} \csc(2^{-j}(m+w)) dw $$

**3.4. Left-hand side third contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 2^{-j}(2y+1)$ then multiply both sides by $4e^{2^{-j}m(2y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (3.4) \quad \sum_{j=0}^{\infty} \frac{2^{k+2} (2^{-j})^k e^{2^{-j}m} \Phi(e^{2^{1-j}m}, -k, \frac{1}{2}(2^j \log(a) + 1))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{y=0}^{\infty} \sum_{j=0}^{\infty} \int_C 4a^w w^{-k-1} e^{2^{-j}(2y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} \sum_{y=0}^{\infty} 4a^w w^{-k-1} e^{2^{-j}(2y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} 2a^w w^{-k-1} \csc(2^{-j}(m+w)) dw $$

<footnote n="28">28 from equation (1.232.3) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order</footnote>

<footnote n="29">29 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177</footnote>

<footnote n="30">30 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty) \times [0, \infty)$.</footnote>

<footnote n="31">31 from equation (1.232.3) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order</footnote>

<footnote n="32">32 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177</footnote>

<footnote n="33">33 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty) \times [0, \infty)$.</footnote>
**3.5. Left-hand side fourth contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + i2^{-j-1}(2y+1)$ then multiply both sides by $2^{1-j}(-1)^y e^{i2^{-j-1}m(2y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (3.5) \quad -\sum_{j=0}^{\infty} \frac{2i (i2^{-j})^{k+1} e^{i2^{-j-1}m} \Phi(-e^{i2^{-j}m}, -k, \frac{1}{2} - i2^j \log(a))}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \sum_{j=0}^{\infty} \int_C 2^{1-j} a^w w^{-k-1} \exp \left( \frac{1}{2}i (2^{-j}(2y+1)(m+w) + 2\pi y) \right) dw \\ = -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} \sum_{j=0}^{\infty} 2^{1-j} a^w w^{-k-1} \exp \left( \frac{1}{2}i (2^{-j}(2y+1)(m+w) + 2\pi y) \right) dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{\infty} 2^{-j} a^w w^{-k-1} \sec (2^{-j-1}(m+w)) dw $$

34 from equation (1.232.2) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

35 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177

36 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty) \times [0, \infty)$.

**3.6. Right-hand side first contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + x$ then multiply both sides by $e^{mx}$ and take the infinite integral over $x \in [0, \infty)$ and simplify in terms of the incomplete gamma function to get

$$ (3.6) \quad \frac{2a^{-m}(-m)^{-k-1}\Gamma(k+1, -m \log(a))}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C \frac{2a^w w^{-k-1}}{m+w} dw $$

37 from equation (3.382.4) in [9] where $|\arg \log(a)| < \pi$ and $Re(m+w) < 0$.

**3.7. Right-hand side second contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 2(y+1)$ then multiply both sides by $-4e^{2bm(y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch zeta function to get

$$ (3.7) \quad -\frac{2^{k+2} e^{2m} \Phi\left(e^{2m}, -k, \frac{\log(a)}{2} + 1\right)}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 4a^w w^{-k-1} e^{2(y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 4a^w w^{-k-1} e^{2(y+1)(m+w)} dw $$
$$= \frac{1}{2\pi i} \int_C (2a^w w^{-k-1} \coth(m+w) + 2a^w w^{-k-1}) dw$$

from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sum to converge. Apply Fubini's theorem for integrals and sums, see page 178 in [10] as the summand is of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

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**3.8. Right-hand side second additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by 2 and simplify to get

$$(3.8) \qquad \frac{2 \log^k(a)}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C 2a^w w^{-k-1} dw$$

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42

**4. Derivation Of The Finite Sums Of The Contour Integral**
**Representation**

In this section the Cauchy integral formula is used and the stated contour integral method to derive the finite sum of the Hurwitz-Lerch Zeta functions in terms of its contour integral representation.

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**4.1. Left-hand side first contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + i2^{-j}(y+1)$ then multiply both sides by $i2^{1-j}e^{i2^{-j}(y+1)(\pi 2^{j-1}+m)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, n]$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$(4.1) \qquad \sum_{j=0}^{n} \frac{i2^{-j+k+1} (i2^{-j-1})^k e^{2i(2^{-j-1}m+\frac{\pi}{4})} \Phi(e^{2i(2^{-j-1}m+\frac{\pi}{4})}, -k, 1-i2^j \log(a))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{j=0}^{n} \sum_{y=0}^{\infty} \int_C i2^{1-j} a^w w^{-k-1} \exp(i2^{-j-1}(y+1)(\pi 2^j + 2(m+w))) dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{n} \sum_{y=0}^{\infty} i2^{1-j} a^w w^{-k-1} \exp(i2^{-j-1}(y+1)(\pi 2^j + 2(m+w))) dw \\ = -\frac{1}{2\pi i} \int_C \sum_{j=0}^{n} \left(2^{-j} a^w w^{-k-1} \cot\left(2^{-j-1}(m+w) + \frac{\pi}{4}\right) + i2^{-j} a^w w^{-k-1}\right) dw$$

from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, n] \times [0, \infty)$.

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47

48

**4.2. Left-hand side first additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $-i2^{-j}$ and simplify to get

$$(4.2) \qquad -\frac{i2^{-j} \log^k(a)}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C i2^{-j} a^w w^{-k-1} dw$$
**4.3. Left-hand side second contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + i2^{-j}(y+1)$ then multiply both sides by $i2^{1-j}e^{i2^{-j}(y+1)(\pi 2^{j-1}+m)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, n]$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.3) \quad -\sum_{j=0}^{n} \frac{i2^{k+1} (i2^{-j})^k e^{i2^{-j}m} \Phi(e^{i2^{1-j}m}, -k, \frac{1}{2}(1-i2^j \log(a)))}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \sum_{j=0}^{n} \sum_{y=0}^{\infty} \int_C 2ia^w w^{-k-1} e^{i2^{-j}(2y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{j=0}^{n} \sum_{y=0}^{\infty} 2ia^w w^{-k-1} e^{i2^{-j}(2y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C a^w w^{-k-1} \csc(2^{-j}(m+w)) dw $$

49 from equation (1.232.3) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

50 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177

51 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, n] \times [0, \infty)$.

**4.4. Left-hand side third contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 2^{-j}(2y+1)$ then multiply both sides by $2e^{2^{-j}m(2y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and $j \in [0, n]$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.4) \quad \sum_{j=0}^{n} \frac{2^{k+1} (2^{-j})^k e^{2^{-j}m} \Phi(e^{2^{1-j}m}, -k, \frac{1}{2}(2^j \log(a) + 1))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{j=0}^{n} \sum_{y=0}^{\infty} \int_C 2a^w w^{-k-1} e^{2^{-j}(2y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{n} \sum_{y=0}^{\infty} 2a^w w^{-k-1} e^{2^{-j}(2y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C a^w w^{-k-1} \cosh(2^{-j}(m+w)) dw $$

52 from equation (1.232.3) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

53 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177

54 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, n] \times [0, \infty)$.

**4.5. Left-hand side fourth contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + i2^{-j}(2y+1)$ then multiply both sides by $2^{1-j}(-1)^y e^{i2^{-j}m(2y+1)}$ and take the infinite sums over $y \in [0, \infty)$ and
$j \in [0, n]$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.5) \quad \sum_{j=0}^{n} \frac{2^{-j+k+1} (i2^{-j})^k e^{i2^{-j}m} \Phi(-e^{i2^{1-j}m}, -k, \frac{1}{2}(1-i2^j \log(a)))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{j=0}^{n} \sum_{y=0}^{\infty} \int_C 2^{1-j} a^w w^{-k-1} e^{i(2^{-j}(2y+1)(m+w)+\pi y)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{j=0}^{n} \sum_{y=0}^{\infty} 2^{1-j} a^w w^{-k-1} e^{i(2^{-j}(2y+1)(m+w)+\pi y)} dw \\ = \frac{1}{2\pi i} \int_C 2^{-j} a^w w^{-k-1} \sec(2^{-j}(m+w)) dw $$

55 from equation (1.232.2) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

56 for the sums to converge. Apply Tonelli's theorem for multiple sums, see page 177

57 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, n] \times [0, \infty)$.

**4.6. Right-hand side first contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 2i(y+1)$ then multiply both sides by $2ie^{2im(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.6) \quad \sum_{j=0}^{n} \frac{(2i)^{k+1} e^{2im} \Phi(e^{2im}, -k, 1 - \frac{1}{2}i \log(a))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 2ia^w w^{-k-1} e^{2i(y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2ia^w w^{-k-1} e^{2i(y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C (a^w w^{-k-1} \cot(m+w) + ia^w w^{-k-1}) dw $$

58 from equation (1.232.2) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

59 for the sums to converge. Apply Fubini's theorem for sums and integrals, see page

60 178 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

**4.7. Right-hand side first additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $-i$ and simplify to get

$$ (4.7) \quad -\frac{i \log^k(a)}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C ia^w w^{-k-1} dw $$
**4.8. Right-hand side second contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 4iy + 4i$ then multiply both sides by $4ie^{4im(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.8) \quad \sum_{j=0}^{n} \frac{i4^{k+1} e^{\frac{1}{2}i(\pi k+8m)} \Phi(e^{4im}, -k, 1 - \frac{1}{4}i \log(a))}{\Gamma(k+1)} \\ = \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 4ia^w w^{-k-1} e^{4i(y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 4ia^w w^{-k-1} e^{4i(y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C (2a^w w^{-k-1} \cot(2(m+w)) + 2ia^w w^{-k-1}) dw $$

**4.9. Right-hand side second additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $-2i$ and simplify to get

$$ (4.9) \quad -\frac{2i \log^k(a)}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C 2ia^w w^{-k-1} dw $$

**4.10. Right-hand side third contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)+i2^{-n}y+i2^{-n}$ then multiply both sides by $-2ie^{im2^{-n}(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ (4.10) \quad -\sum_{j=0}^{n} \frac{2i (i2^{-n})^k e^{im2^{-n}} \Phi(e^{i2^{-n}m}, -k, 1 - i2^n \log(a))}{\Gamma(k+1)} \\ = -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 2ia^w w^{-k-1} e^{i2^{-n}(y+1)(m+w)} dw \\ = -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2ia^w w^{-k-1} e^{i2^{-n}(y+1)(m+w)} dw \\ = \frac{1}{2\pi i} \int_C (a^w w^{-k-1} \cot(2^{-n-1}(m+w)) + ia^w w^{-k-1}) dw $$

<footnote n="61">61 from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order</footnote>

<footnote n="62">62 for the sums to converge. Apply Fubini's theorem for sums and integrals, see page</footnote>

<footnote n="63">63 178 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.</footnote>
**4.11. Right-hand side third additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $i$ and simplify to get

$$ (4.11) \qquad \frac{i \log^k(a)}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C ia^w w^{-k-1} dw $$

**4.12. Right-hand side fourth contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + i2^{1-n}y + i2^{1-n}$ then multiply both sides by $-i2^{1-n}e^{im2^{1-n}(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ \begin{align} (4.12) \quad & -\sum_{j=0}^{n} \frac{(i2^{1-n})^{k+1} e^{im2^{1-n}} \Phi(e^{i2^{1-n}m}, -k, 1 - i2^{1-n} \log(a))}{\Gamma(k+1)} \nonumber \\ &= -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C i2^{1-n} a^w w^{-k-1} e^{i2^{1-n}(y+1)(m+w)} dw \nonumber \\ &= -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C i2^{1-n} a^w w^{-k-1} e^{i2^{1-n}(y+1)(m+w)} dw \nonumber \\ &= \frac{1}{2\pi i} \int_C (2^{-n} a^w w^{-k-1} \cot(2^{-n}(m+w)) + i2^{-n} a^w w^{-k-1}) dw \end{align} $$

**4.13. Right-hand side fourth additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ then multiply both sides by $i2^{-n}$ and simplify to get

$$ (4.13) \qquad \frac{i2^{-n} \log^k(a)}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C i2^{-n} a^w w^{-k-1} dw $$

**4.14. Right-hand side fifth contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a) + 2y + 2$ then multiply both sides by $-2e^{2m(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ \begin{align} (4.14) \quad & -\sum_{j=0}^{n} \frac{2^{k+1} e^{2m} \Phi\left(e^{2m}, -k, \frac{\log(a)}{2} + 1\right)}{\Gamma(k+1)} \nonumber \\ &= -\frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 2a^w w^{-k-1} e^{2(y+1)(m+w)} dw \end{align} $$

<footnote n="64">64 from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. Apply Fubini's theorem for sums and integrals, see page 65
66 178 in [10] as the summands are of bounded measure over the space $C \times [0, \infty)$.</footnote>

<footnote n="67">67 from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order for the sums to converge. Apply Fubini's theorem for sums and integrals, see page 68
69 178 in [10] as the summands are of bounded measure over the space $C \times [0, \infty)$.</footnote>
$$ 
\begin{aligned}
&= -\frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2a^w w^{-k-1} e^{2(y+1)(m+w)} dw \\
&= \frac{1}{2\pi i} \int_C (a^w w^{-k-1} \coth(m+w) + a^w w^{-k-1}) dw
\end{aligned}
$$

70 from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

71 for the sums to converge. Apply Fubini's theorem for sums and integrals, see page

72 178 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

**4.15. Right-hand side fifth additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ and simplify to get

$$ (4.15) \qquad \frac{\log^k(a)}{\Gamma(k+1)} = \frac{1}{2\pi i} \int_C a^w w^{-k-1} dw $$

**4.16. Right-hand side sixth contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)+2^{-n}y+2^{-n}$ then multiply both sides by $2e^{m2^{-n}(y+1)}$ and take the infinite sum over $y \in [0, \infty)$ and simplify in terms of the Hurwitz-Lerch Zeta function to get

$$ 
\begin{aligned}
& (4.16) \qquad \sum_{j=0}^{n} \frac{2(2^{-n})^k e^{m2^{-n}} \Phi(e^{2^{-n}m}, -k, 2^n \log(a) + 1)}{\Gamma(k+1)} \\
&= \frac{1}{2\pi i} \sum_{y=0}^{\infty} \int_C 2a^w w^{-k-1} e^{2^{-n}(y+1)(m+w)} dw \\
&= \frac{1}{2\pi i} \int_C \sum_{y=0}^{\infty} 2a^w w^{-k-1} e^{2^{-n}(y+1)(m+w)} dw \\
&= -\frac{1}{2\pi i} \int_C (a^w w^{-k-1} \coth(2^{-n-1}(m+w)) - a^w w^{-k-1}) dw
\end{aligned}
$$

73 from equation (1.232.1) in [9] where $Re(m+w) > 0$ and $Im(m+w) > 0$ in order

74 for the sums to converge. Apply Fubini's theorem for sums and integrals, see page

75 178 in [10] as the summands are of bounded measure over the space $\mathbb{C} \times [0, \infty)$.

**4.17. Right-hand side sixth additional contour integral.** Using a generalization of Cauchy's integral formula (1.4), first replace $y \to \log(a)$ multiply both sides by $-1$ and simplify to get

$$ (4.17) \qquad -\frac{\log^k(a)}{\Gamma(k+1)} = -\frac{1}{2\pi i} \int_C a^w w^{-k-1} dw $$
5. Infinite and Finite Sum of Hurwitz-Lerch Zeta Functions in Terms
of Incomplete Gamma and Hurwitz-Lerch Zeta Functions

THEOREM 5.1. For all $k, a, m \in \mathbb{C}$ then,

$$
\begin{align*}
& (5.1) \quad \sum_{j=0}^{\infty} \left( \begin{aligned}[t]
& -2^{-j-k+1} (-i2^{-j})^k e^{-i2^{-j-1}m} \Phi \left( ie^{-i2^{-j-1}m}, -k, i2^{j+1} \log(a) + 1 \right) \\
& -i2^{k+2} (i2^{-j})^k e^{i2^{-j}m} \Phi \left( e^{i2^{1-j}m}, -k, \frac{1}{2} (1 - i2^j \log(a)) \right) \\
& +2^{k+2} (2^{-j})^k e^{2^{-j}m} \Phi \left( e^{2^{1-j}m}, -k, \frac{1}{2} (2^j \log(a) + 1) \right) \\
& +2^{1-j} (i2^{-j})^k e^{i2^{-j-1}m} \Phi \left( -e^{i2^{-j}m}, -k, \frac{1}{2} - i2^j \log(a) \right) + i2^{-j} \log^k(a)
\end{aligned} \right) \\
& = 2a^{-m} (-m)^{-k-1} \Gamma(k+1, -m \log(a)) - 2^{k+2} e^{2m} \Phi \left( e^{2m}, -k, \frac{\log(a)}{2} + 1 \right) - 2 \log^k(a)
\end{align*}
$$

PROOF. Since the addition of the right-hand sides of equations (3.1), (3.2), (3.3), (3.4) and (3.5) is equivalent to the addition of the right-hand sides of equations (3.6), (3.7) and (3.8), then equate the left-hand sides and apply equations (3.303) in [6] and (1.1) and simplify the Gamma function to yield the stated result. $\square$

THEOREM 5.2. For all $k, a, m \in \mathbb{C}$ then,

$$
\begin{align*}
& (5.2) \quad \sum_{j=0}^{n} \left( \begin{aligned}[t]
& -2^{-j+k+1} (i2^{-j-1})^k e^{i2^{-j}m} \Phi \left( ie^{i2^{-j}m}, -k, 1 - i2^j \log(a) \right) \\
& + 2^{-j+k+1} (i2^{-j})^k e^{i2^{-j}m} \Phi \left( -e^{i2^{1-j}m}, -k, \frac{1}{2} (1 - i2^j \log(a)) \right) \\
& - i2^{k+1} (i2^{-j})^k e^{i2^{-j}m} \Phi \left( e^{i2^{1-j}m}, -k, \frac{1}{2} (1 - i2^j \log(a)) \right) \\
& + 2^{k+1} (2^{-j})^k e^{2^{-j}m} \Phi \left( e^{2^{1-j}m}, -k, \frac{1}{2} (2^j \log(a) + 1) \right) + i2^{-j} \log^k(a)
\end{aligned} \right) \\
& = 2^{1-n} (i2^{-n})^{k-1} e^{im2^{-n}} \Phi \left( e^{i2^{-n}m}, -k, 1 - i2^n \log(a) \right) \\
& \quad - i2^{1-n} (i2^{1-n})^k e^{im2^{1-n}} \Phi \left( e^{i2^{1-n}m}, -k, 1 - i2^{n-1} \log(a) \right) \\
& \quad + 2 (2^{-n})^k e^{m2^{-n}} \Phi \left( e^{2^{-n}m}, -k, 2^n \log(a) + 1 \right) \\
& \quad + i2^{k+1} e^{\frac{1}{2}i(\pi k + 4m)} \Phi \left( e^{2im}, -k, 1 - \frac{1}{2}i \log(a) \right) \\
& \quad + i2^{2k+2} e^{\frac{1}{2}i(\pi k + 8m)} \Phi \left( e^{4im}, -k, 1 - \frac{1}{4}i \log(a) \right) \\
& \quad - 2^{k+1} e^{2m} \Phi \left( e^{2m}, -k, \frac{\log(a)}{2} + 1 \right) - i2^{-n} \log^k(a) + 2i \log^k(a)
\end{align*}
$$
PROOF. Since the addition of the right-hand sides of equations (4.1), (4.2) (4.3), (4.4) and (4.5) is equivalent to the addition of the right-hand sides of equations (4.6), (4.7), (4.8), (4.9), (4.10), (4.11), (4.12), (4.13), (4.14), (4.15), (4.16) and (4.17), then equate the left-hand sides and apply equation (3.303) in [6] and (1.1) and simplify the Gamma function to yield the stated result. □

## Main Results and Special Cases

### 6. The Degenerate Cases

In this brief section the degenerate cases for both the infinite and finite sums are derived.

**EXAMPLE 6.1.** The degenerate case for the infinite sum.

$$ (6.1) \quad \sum_{j=0}^{\infty} -2^{-j-1} \tan(2^{-j-1}m) + \csc(2^{-j}m) - \csch(2^{-j}m) = \coth(m) - \frac{1}{m} $$

PROOF. Use equation (5.1) and set $k = 0$ and simplify using entry (2) in Table of Section (64:12) in [12]. □

**EXAMPLE 6.2.** The degenerate case for the finite sum.

$$ (6.2) \quad \begin{aligned} & \sum_{j=0}^{n} 2^{-j} \tan(2^{-j}m) + \csc(2^{-j}m) - \csch(2^{-j}m) \\ & = \cot(m2^{-n-1}) + 2^{-n} \cot(m2^{-n}) - \coth(m2^{-n-1}) + \tan(m) - 2\cot(m) + \coth(m) \end{aligned} $$

PROOF. Use equation (5.2) and set $k = 0$ and simplify using entry (2) in Table of Section (64:12) in [12]. □

### 7. Extended Melnikov Infinite Product

Melnikov [1], published work involving a technique developed for obtaining infinite product representations for some elementary functions. In his work many infinite products were derived and evaluated. In this section an alternate form for equation (23) in [1] is given where the parameters are general complex numbers, thereby expanding upon the formula derived by Melnikov. In this section the symmetric plots of the right-hand side of equation (7.1) when $r \to \alpha, m \to 1/\alpha$ for real and complex ranges are produced.

**THEOREM 7.1.** For all $m, r \in \mathbb{C}$ then,

$$ (7.1) \quad \prod_{j=0}^{\infty} \frac{\cosh(2^{-1-j}r)}{\cosh(2^{-1-j}m)} \left( \frac{\tan(2^{-1-j}m) \tanh(2^{-1-j}r)}{\tan(2^{-1-j}r) \tanh(2^{-1-j}m)} \right)^{2j} = \frac{m \sin(r)}{r \sin(m)} $$

PROOF. Use equation (5.1) and form a second equation by replacing $m \to r$ take the difference of both these equations then set $a = 1$ and simplify in terms of the Polylogarithm function using equation (64:12:2) in [12]. Next apply L'Hopital's rule to the right-hand side as $k \to -1$ and simplify in terms of the logarithmic function using entry (1) in Table of Section (64:12) in [12]. Next take the exponential function of both sides and simplify in terms of the cosine, tangent and hyperbolic cosine and tangent functions. Replace $m \to im, r \to ir$ and simplify. To get the form quoted in [1] simply replace $r \to At, m \to Bt$. □
<table><caption>TABLE 1. Table of Infinite Products in terms of Constants</caption><tr><td>∞ &prod;<sub>j=0</sub> cosh(π2<sup>-j-2</sup>) sech(1/3π2<sup>-j-1</sup>) (tan(1/3π2<sup>-j-1</sup>) cot(π2<sup>-j-2</sup>) tanh(π2<sup>-j-2</sup>) coth(1/3π2<sup>-j-1</sup>))<sup>2j</sup></td><td>4/3√3</td></tr><tr><td>∞ &prod;<sub>j=0</sub> cosh(π2<sup>-j-3</sup>) sech(1/5π2<sup>-j-1</sup>) (tan(1/5π2<sup>-j-1</sup>) cot(π2<sup>-j-3</sup>) tanh(π2<sup>-j-3</sup>) coth(1/5π2<sup>-j-1</sup>))<sup>2j</sup></td><td>4/5√(1 + 1/√5)</td></tr><tr><td>∞ &prod;<sub>j=0</sub> cosh(1/3π2<sup>-j-1</sup>) sech(1/3π2<sup>-j-2</sup>) (tan(1/3π2<sup>-j-2</sup>) cot(1/3π2<sup>-j-1</sup>) tanh(1/3π2<sup>-j-1</sup>) coth(1/3π2<sup>-j-2</sup>))<sup>2j</sup></td><td>√3/2</td></tr><tr><td>∞ &prod;<sub>j=0</sub> cosh(π2<sup>-j-2</sup>) sech(π2<sup>-j-4</sup>) (tan(π2<sup>-j-4</sup>) cot(π2<sup>-j-2</sup>) tanh(π2<sup>-j-2</sup>) coth(π2<sup>-j-4</sup>))<sup>2j</sup></td><td>1/(2√2-√2)</td></tr><tr><td>∞ &prod;<sub>j=0</sub> cosh(π2<sup>-j-3</sup>) sech(3/5π2<sup>-j-1</sup>) (tan(3/5π2<sup>-j-1</sup>) cot(π2<sup>-j-3</sup>) tanh(π2<sup>-j-3</sup>) coth(3/5π2<sup>-j-1</sup>))<sup>2j</sup></td><td>24/5√(5+√5)</td></tr></table>

<figure><img src="image_2.png" /><figcaption>FIGURE 1. Plot of $\frac{\sinh(\alpha)}{\alpha^2 \sinh(\frac{1}{\alpha})}$, $\alpha \in \mathbb{R}$.</figcaption></figure>
<figure><img src="image_3.png" /><figcaption>FIGURE 2. Plot of $Re \left( \frac{\sinh(\alpha)}{\alpha^2 \sinh\left(\frac{1}{\alpha}\right)} \right)$, $\alpha \in \mathbb{C}$.</figcaption></figure>

<figure><img src="image_4.png" /><figcaption>FIGURE 3. Plot of $Im \left( \frac{\sinh(\alpha)}{\alpha^2 \sinh\left(\frac{1}{\alpha}\right)} \right)$, $\alpha \in \mathbb{C}$.</figcaption></figure>
<figure><img src="image_5.png" /><figcaption>FIGURE 4. Plot of $\left| \left( \frac{\sinh(\alpha)}{\alpha^2 \sinh\left(\frac{1}{\alpha}\right)} \right) \right|$, $\alpha \in \mathbb{C}$.</figcaption></figure>

**7.1. Special Cases of the Extended Melnikov Infinite Product.** In this section the evaluation of infinite product forms for the Sine and Cosine function which are alternate forms featuring parameters with complex numbers to equations (17), (20) and (30) in [1]. These derived formulae are derived by using equation (7.1) replacing $m \to \pi/2$, $m \to \pi/2$, $r \to r - \pi/2$ and $m \to r + \pi/2$ respectively.

$$ (7.2) \qquad \frac{2r}{\pi} \prod_{j=0}^{\infty} \frac{\cosh(2^{-1-j}r)}{\cosh(2^{-2-j}\pi)} \left( \frac{\tan(2^{-2-j}\pi) \tanh(2^{-1-j}r)}{\tan(2^{-1-j}r) \tanh(2^{-2-j}\pi)} \right)^{2j} = \sin(r) $$

$$ (7.3) \qquad \frac{\pi - 2r}{\pi} \prod_{j=0}^{\infty} \frac{\cosh(2^{-2-j}(\pi - 2r))}{\cosh(2^{-2-j}\pi)} \left( \frac{\tan(2^{-2-j}\pi) \tanh(2^{-2-j}(\pi - 2r))}{\tan(2^{-2-j}(\pi - 2r)) \tanh(2^{-2-j}\pi)} \right)^{2j} = \cos(r) $$

$$ (7.4) \qquad \frac{2r}{(\pi + 2r)} \prod_{j=0}^{\infty} \frac{\cosh(2^{-1-j}r)}{\cosh(2^{-2-j}(\pi + 2r))} \left( \frac{\tan(2^{-2-j}(\pi + 2r)) \tanh(2^{-1-j}r)}{\tan(2^{-1-j}r) \tanh(2^{-2-j}(\pi + 2r))} \right)^{2j} = \tan(r) $$

$$ (7.5) \qquad \frac{\alpha}{\beta} \prod_{j=0}^{\infty} \frac{\cosh(2^{-1-j}z\alpha)}{\cosh(2^{-1-j}z\beta)} \left( \frac{\tan(2^{-1-j}z\beta) \tanh(2^{-1-j}z\alpha)}{\tan(2^{-1-j}z\alpha) \tanh(2^{-1-j}z\beta)} \right)^{2j} = \frac{\sin(z\alpha)}{\sin(z\beta)} $$

**7.2. Catalan's Constant, K.** In this section the derivation of two methods of arriving at Catalan's constant are evaluated using the integration over an infinite product with factors containing the Cosine, Tangent and Hyperbolic Cosine and Tangent functions with Powers of 2. Using equation (7.1), set $r = \pi/2$ and take the definite integral of both sides to get
$$ (7.6) \quad \frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \prod_{j=0}^{\infty} \frac{\cosh(2^{-2-j}\pi)}{\cosh(2^{-1-j}m)} \left( \frac{\tan(2^{-1-j}m) \tanh(2^{-2-j}\pi)}{\tan(2^{-2-j}\pi) \tanh(2^{-1-j}m)} \right)^{2j} dm \\ = \frac{\pi}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cosh(\frac{\pi}{2})}{\cosh(\frac{m}{2})} \left( \frac{\tan(\frac{m}{2}) \tanh(\frac{\pi}{2})}{\tan(\frac{\pi}{2}) \tanh(\frac{m}{2})} \right)^{2^0} \cdot \frac{\cosh(\frac{\pi}{2^3})}{\cosh(\frac{m}{2^2})} \left( \frac{\tan(\frac{m}{2^2}) \tanh(\frac{\pi}{2^3})}{\tan(\frac{\pi}{2^3}) \tanh(\frac{m}{2^2})} \right)^{2^1} \dots dm \\ = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{m}{\sin(m)} dm = 4K $$

where the integral representation for Catalan's constant, K is from Glaisher [13].
Next applying the limit as $r \to 0$ to equation (7.1) we write

$$ (7.7) \quad \lim_{r \to 0} \left( \prod_{j=0}^{\infty} \frac{\cos(2^{-1-j}r)}{\cos(2^{-1-j}m)} \left( \frac{\tan(2^{-1-j}r) \tanh(2^{-1-j}m)}{\tan(2^{-1-j}m) \tanh(2^{-1-j}r)} \right)^{2j} \right) = \frac{m \sinh(r)}{r \sinh(m)} $$

then simplify and taking the definite integral over $m \in [0, \pi i/2]$ and simplify to get
(7.8)

$$ \int_0^{\pi i/2} \prod_{j=0}^{\infty} \frac{1}{\cos(2^{-1-j}m)} \left( \frac{\tanh(2^{-1-j}m)}{\tan(2^{-1-j}m)} \right)^{2j} dm = \int_0^{\pi i/2} \frac{m}{\sinh(m)} dm = 2iK $$

## 8. Special Case Involving the Exponential of Trigonometric Functions

In this section equation (5.1) is evaluated to yield the exponential of trigonometric functions in terms of the exponential of trigonometric functions. A special case is also evaluated in terms of the exponential of the reciprocal of $\pi$.

### EXAMPLE 8.1.

$$ (8.1) \quad \prod_{j=0}^{\infty} \exp \left( 2^{-j} \left( \tan(2^{-j-1}x) - \tan\left(\frac{2^{-j-1}x}{\beta}\right) \right) \right) \\ \qquad \exp \left( -2 \left( -\csc\left(\frac{2^{-j}x}{\beta}\right) + \csc\left(\frac{2^{-j}x}{\beta}\right) + \csc(2^{-j}x) \right) + 2\csc(2^{-j}x) \right) \\ \qquad 2 \log \left( e^{\frac{i(\beta-1)2^{-j-1}x}{\beta}} \sec(2^{-j-1}x) \cos\left(\frac{2^{-j-1}x}{\beta}\right) \left( \cot(2^{-j-1}x) \tanh(2^{-j-1}x) \tan\left(\frac{2^{-j-1}x}{\beta}\right) \coth\left(\frac{2^{-j-1}x}{\beta}\right) \right)^{2j} \right) - \frac{i(\beta-1)2^{-j}x}{\beta} \\ = \left( \cosh\left( \frac{2\left(\beta - x \coth\left(\frac{x}{\beta}\right) + x \coth(x) - 1\right)}{x} \right) + \sinh\left( \frac{-2\beta + 2x \coth\left(\frac{x}{\beta}\right) - 2x \coth(x) + 2}{x} \right) \right) \\ a^{\log(\beta^2 \csc^2(x) \sinh^2(\frac{x}{\beta}))} $$

PROOF. Use equation (5.1) and set $k = 1, m = x$ and apply the method in section (8) in [14]. □

### EXAMPLE 8.2.
$$
\begin{align*}
& (8.2) \quad \prod_{j=0}^{\infty} \exp \left( 2i \csc \left( 2^{-2-j}\pi \right) - 2i \csc \left( 2^{-1-j}\pi \right) - 2i \csch \left( 2^{-2-j}\pi \right) \right) \\
& \phantom{(8.2) \quad \prod_{j=0}^{\infty} \exp \left( 2i \csc \left( 2^{-2-j}\pi \right) - 2i \csc \left( 2^{-1-j}\pi \right) - 2i \csch \left( 2^{-2-j}\pi \right) \right)} \\
& \quad \exp \left( 2i \csch \left( 2^{-1-j}\pi \right) - i 2^{-j} \tanh \left( 2^{-3-j}\pi \right) + i 2^{-j} \tanh \left( 2^{-2-j}\pi \right) \right) \\
& \quad \left( \frac{\cosh \left( 2^{-3-j}\pi \right)}{\cosh \left( 2^{-2-j}\pi \right)} \right)^2 \left( \frac{\tan \left( 2^{-2-j}\pi \right) \tanh \left( 2^{-3-j}\pi \right)}{\tan \left( 2^{-3-j}\pi \right) \tanh \left( 2^{-2-j}\pi \right)} \right)^{2^{1+j}} \\
&= 2e^{-\frac{2i(-2+\pi)}{\pi}}
\end{align*}
$$

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PROOF. Use equation (8.1) and set $a = e, \beta = 2, x = \frac{\pi i}{2}$ and simplify. $\square$

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## 9. Extended Melnikov Finite Product

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In this section the derivation of the finite product formula forms for the extended Melnikov form. This formula could be used to as a part of the product in the stduy of the infinite form given by equation (7.1).

EXAMPLE 9.1.

$$
\begin{align*}
(9.1) \quad & \prod_{j=0}^{n} \frac{\cos (2^{-j}m)}{\cos (2^{-j}r)} \left( \frac{\tan (2^{-1-j}r) \tanh (2^{-1-j}m)}{\tan (2^{-1-j}m) \tanh (2^{-1-j}r)} \right)^{2^j} \\
& = \frac{\cos(m) \sin (2^{-n}r) \sin^2(m) \sinh(r)}{\cos(r) \sin (2^{-n}m) \sin^2(r) \sinh(m)} \left( \frac{\sin (2^{-1-n}r) \sinh (2^{-1-n}m)}{\sin (2^{-1-n}m) \sinh (2^{-1-n}r)} \right)^{2^{1+n}}
\end{align*}
$$

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PROOF. Use equation (5.2) and form a second equation by replacing $m \to r$ to take the difference of both these equations then set $a = 1$ and simplify in terms of the Polylogarithm function using equation (64:12:2) in [12]. Next apply L'Hopital's rule to the right-hand side as $k \to -1$ and simplify in terms of the logarithmic function using entry (1) in Table of Section (64:12) in [12]. Next take the exponential function of both sides and simplify in terms of the cosine, tangent and hyperbolic cosine and tangent functions. The final form is derived by forming a second equation by replacing $m \to -m$ and $r \to -r$ and multiplying both and simplifying the trigonometric functions. $\square$

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## 10. The Infinite Product of Quotient Cosine Functions

In this section the derivation of an infinite product formula involving the ratio of the product of cosine functions. An equation used in this derivation is

$$
(10.1) \qquad \lim_{n \to \infty} \frac{\sin(2^{-1-n}r)}{\sin(2^{-1-n}m)} = \frac{r}{m}
$$

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where an application of l'Hopital's rule is used to derive the limit.

EXAMPLE 10.1.

$$
(10.2) \quad \prod_{j=0}^{\infty} \frac{\cos (2^{-1-j}r) \cos (2^{-j}r)}{\cos (2^{-1-j}m) \cos (2^{-j}m)} = \left(\frac{m}{r}\right)^2 \frac{\sin(2r) \sin(r)}{\sin(2m) \sin(m)}
$$

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PROOF. Use the hyperbolic form of equation (7.1) and divide it by equation (9.1) and take the limit as $n \to \infty$ and simplify. $\square$
## 11. Infinite Product Of The Quotient Of Hyperbolic Cosine, Sine And Tangent Functions Raised To A Complex Power

In this example we will look at the infinite product of the quotient of trigonometric functions raised to a complex power expressed in terms of the quotient of trigonometric functions raised to a complex power. In section (5.8) in [11] is a list of very interesting infinite products of involving elementary functions. In particular equation (5.8.5) in [11] features a generalized infinite product of quotient polynomials expressed in terms of quotient Gamma functions. In this section we will look at a generalized infinite product raised to a complex power whereby expanding upon previous infinite product literature. Note $Re(\beta) > Re(x)$ in order for the product to converge.

### EXAMPLE 11.1.

$$ (11.1) \quad \prod_{j=0}^{\infty} \frac{\cos(2^{-1-j}x)}{\cos\left(\frac{2^{-1-j}x}{\beta}\right)} \left( \frac{\cos\left(\frac{2^{-1-j}x}{\beta^2}\right)}{\cos\left(\frac{2^{-1-j}x}{\beta}\right)} \right)^{\beta} \left( \frac{\tan(2^{-1-j}x) \tanh\left(\frac{2^{-1-j}x}{\beta}\right)}{\tan\left(\frac{2^{-1-j}x}{\beta}\right) \tanh(2^{-1-j}x)} \right)^{2j} \\ \left( \frac{\tan\left(\frac{2^{-1-j}x}{\beta^2}\right) \tanh\left(\frac{2^{-1-j}x}{\beta}\right)}{\tan\left(\frac{2^{-1-j}x}{\beta}\right) \tanh\left(\frac{2^{-1-j}x}{\beta^2}\right)} \right)^{2j\beta} $$

$$ = \beta^{\beta-1} \frac{\sinh(x)}{\sinh\left(\frac{x}{\beta}\right)} \left( \frac{\cosh\left(\frac{x}{\beta^2}\right) \tanh\left(\frac{x}{\beta}\right)}{\cosh\left(\frac{x}{\beta}\right) \tanh\left(\frac{x}{\beta}\right)} \right)^{\beta} $$

PROOF. Use equation (5.1) and set $k = 3, a = 1, m = x$ and apply the method in section (8) in [14]. □

### 11.1. Table Of Infinite Products In Terms Of Constants.
In this section we evaluate equation (11.1) for various values of the parameters involved with many other possible since $x, \beta \in \mathbb{C}$. These derivations are left as an exercise for readers.
154 PROOF. Use equation (5.2) and set $n = 0$ set $n = 0, m = -\frac{\alpha}{2}, k = -s, a = e^a$ and simplify. $\square$

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EXAMPLE 12.2.

$$ (12.2) \quad \Phi \left( e^{2\pi i}, 2, 1 - \frac{i}{4} \right) = 8ie^{-3\pi i/2} \Phi \left( ie^{\pi i/2}, 2, 1 - i \right) + 8e^{-3\pi i/2} \Phi \left( e^{\pi i/2}, 2, 1 - i \right) \\ - 2ie^{-3\pi i/2} \Phi \left( -e^{\pi i}, 2, \frac{1}{2} - \frac{i}{2} \right) - 2e^{-3\pi i/2} \Phi \left( e^{\pi i}, 2, \frac{1}{2} - \frac{i}{2} \right) $$

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PROOF. Use equation (12.1) and set $\alpha = \pi$ and simplify in terms of the Hurwitz-Zeta function using entry (4) in Section (64:12) of [12]. Next set $s = 2, v = 3/2$ and simplify. $\square$

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EXAMPLE 12.3. The Difference of Derivative of Hurwitz-Lerch Zeta Functions in terms of Euler's constant $\gamma$ and Stieltjes constant $\gamma_1$ and generalized Stieltjes constant $\gamma_1(\beta)$ where $\beta \in \mathbb{R}_+$.

$$ (12.3) \quad \pi \left( 2 \left( -2\mathrm{Li}'_0(e^{-2\pi i}) + \mathrm{Li}'_0(ie^{-\pi i/2}) + \mathrm{Li}'_0(e^{-\pi i/2}) + \log \left( \frac{\Gamma\left(-\frac{3}{4}\right)}{\Gamma\left(-\frac{1}{4}\right)} \right) \right) \right. \\ \left. - \gamma + \log\left(\frac{9}{128\pi}\right) + \log(16) \coth(\pi) + \log(2) \left( \cosh\left(\frac{\pi}{2}\right) + i \sech\left(\frac{\pi}{2}\right) \right) \right. \\ \left. - \gamma_1\left(\frac{1}{4}\right) + \gamma_1\left(\frac{3}{4}\right) \right) \\ = 2e^{-\pi/2} \pi \left( \Phi' \left( e^{-\pi i}, 0, \frac{1}{2} \right) + i\Phi' \left( -e^{-\pi i}, 0, \frac{1}{2} \right) \right) $$

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PROOF. Use equation (5.2) and set $n = 0, a = 1, m = \pi i/2$ and simplify in terms of the polylogarithm function using entry (4) in Section (64:12) of [12]. Next take the first partial derivative with respect to $k$ and use l'Hopital's rule as $k \to 0$ and simplify. $\square$

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13. Conclusion

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In this paper, the derivation of new finite and infinite sums and products involving trigonometric functions using contour integration was conducted. The method applied in the derivation of the main theorem may be used to derive other sums and products in future work. Infinite forms were derived involving trigonometric functions and their partial sum counterpart. These types of evaluations allows the analysis of the infinite sum of functions and the partial sums which they are built upon. The results for special cases were summarized in the form of Table (2) and 2D and 3D plots involving a form of an infinite product. The results presented were numerically verified for both real and imaginary and complex values of the parameters in the integrals using Mathematica by Wolfram.

References

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[1] Yu.A. Melnikov, A new approach to the representation of trigonometric and hyperbolic functions by infinite products, Journal of Mathematical Analysis and Applications, Volume 344, Issue 1, 2008, Pages 521-534, https://doi.org/10.1016/j.jmaa.2008.03.010.

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[2] Y. A. Melnikov *New infinite product representations of some elementary functions*, Appl. Math. Sci., Vol. 2, 2008, no. 1-4, 81-97.

[3] Melnikov, Y.A. (2011). *Green's Functions for the Laplace Equation. In: Green's Functions and Infinite Products*. Birkhäuser Boston. https://doi.org/10.1007/978-0-8176-8280-4'3

[4] Marc Chamberland, Armin Straub, *On gamma quotients and infinite products*, Advances in Applied Mathematics, Volume 51, Issue 5, 2013, Pages 546-562, https://doi.org/10.1016/j.aam.2013.07.003.

[5] Reynolds, R.; Stauffer, A. *A Method for Evaluating Definite Integrals in Terms of Special Functions with Examples*. *Int. Math. Forum* **2020**, *15*, 235-244, doi:10.12988/imf.2020.91272

[6] Vălean, Cornel Ioan. (*Almost*) impossible integrals, sums, and series. Springer Nature Switzerland AG 2019.

[7] Erdélyi, A.; Magnus, W.; Oberhettinger, F.; Tricomi, F.G. *Higher Transcendental Functions*; McGraw-Hill Book Company, Inc.: New York, NY, USA; Toronto, ON, Canada; London, UK, 1953; Volume I.

[8] Laurincikas, A.; Garunkstis, R. *The Lerch Zeta-Function*; Springer Science & Business Media: New York, NY, USA, 2013; 189p.

[9] Gradshteyn, I.S.; Ryzhik, I.M. *Tables of Integrals, Series and Products*, 6th ed.; Academic Press: Cambridge, MA, USA, 2000.

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[11] Olver, F.W.J.; Lozier, D.W.; Boisvert, R.F.; Clark, C.W. (Eds.) *NIST Digital Library of Mathematical Functions*; U.S. Department of Commerce, National Institute of Standards and Technology: Washington, DC, USA; Cambridge University Press: Cambridge, UK, 2010; With 1 CD-ROM (Windows, Macintosh and UNIX). MR 2723248 (2012a:33001).

[12] Oldham, K.B.; Myland, J.C.; Spanier, J. *An Atlas of Functions: With Equator, the Atlas Function Calculator*, 2nd ed.; Springer: New York, NY, USA, 2009.

[13] Glaisher, J. W. L. "On a Numerical Continued Product.", Messenger Math. 6, 71-76, 1877.

[14] Reynolds, R., & Stauffer, A. (2022). *A Note on the Infinite Sum of the Lerch function*. European Journal of Pure and Applied Mathematics, 15(1), 158-168. https://doi.org/10.29020/nybg.ejpam.v15i1.4137

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(Robert Reynolds) DEPARTMENT OF MATHEMATICS AND STATISTICS, YORK UNIVERSITY, TORONTO, ON, CANADA, M3J1P3

*Email address*, Corresponding author: milver@my.yorku.ca