On Euler-Ramanujan formula, Dirichlet series and minimal surfaces

TL;DR

This paper explores connections between Euler-Ramanujan identities, Dirichlet series, and minimal surfaces, deriving functional equations and identities through complex analysis and geometric representations.

Contribution

It introduces new identities linking Dirichlet series with minimal surface representations and derives their functional equations, expanding the understanding of these mathematical structures.

Findings

Derived functional equations for Dirichlet series related to Euler-Ramanujan identities

Established identities involving Dirichlet series and minimal surfaces

Reformulated Euler-Ramanujan identities in terms of Dirichlet series

Abstract

In this paper, we rewrite two forms of an Euler-Ramanujan identity in terms of certain Dirichlet series and derive functional equation of the latter. We also use the Weierstrass-Enneper representation of minimal surfaces to obtain some identities involving these Dirichlet series and one complex parameter.

Figures

Equations

$$\frac{\cos (X+A)}{\cos A}=\prod _{k=1}^{\infty }\left(1-\frac{X}{(k-\frac{1}{2})\pi -A}\right)\left(1+\frac{X}{(k-\frac{1}{2})\pi +A}\right).$$

$$\log \left(\frac{\cos (X+A)}{\cos A}\right)=\sum _{k=1}^{\infty }\log \left\{\left(\frac{(k-\frac{1}{2})\pi -(X+A)}{(k-\frac{1}{2})\pi -A}\right)\left(\frac{(k-\frac{1}{2})\pi +(X+A)}{(k-\frac{1}{2})\pi +A}\right)\right\}.$$

$$\log \left(\frac{\cos y}{\cos x}\right)$$

$$\log \left(\frac{\cos y}{\cos x}\right)=\sum _{k=1}^{\infty }[\log (1+\frac{x^2}{c_k^2-x^2})+\log (1-\frac{y^2}{c_k^2})].$$

$$\log \left(\frac{\cos y}{\cos x}\right)$$

$$\mathrm{Re}\log \left(\frac{1+{\zeta }^2}{1-{\zeta }^2}\right)$$

$$X_{\theta }(u,v)=(u+v\cos \theta ,v\sin \theta ,\log \frac{\cos v}{\cos u}),$$

$$\frac{\cos (\frac{y}{\sin \theta })}{\cos (x-y\cot \theta )}=\prod _{k=1}^{\infty }\left(\frac{c_k^2-\frac{y^2}{{\sin }^2\theta }}{c_k^2-(x-y\cot \theta {)}^2}\right),$$

$$\log \left(\frac{\cos (\frac{y}{\sin \theta })}{\cos (x-y\cot \theta )}\right)=\sum _{k=1}^{\infty }\log \left(\frac{c_k^2-\frac{y^2}{{\sin }^2\theta }}{c_k^2-(x-y\cot \theta {)}^2}\right).$$

$$\log \left(\frac{\cos (\frac{y}{\sin \theta })}{\cos (x-y\cot \theta )}\right)=\sum _{k=1}^{\infty }[L_k(1,\frac{(x-y\cot \theta {)}^2}{c_k^2-(x-y\cot \theta {)}^2})-M_k(1,(\frac{y}{c_k\sin \theta }{)}^2)]$$

$${\tan }^{-1}\omega =\frac{i}{2}(\log (1-i\omega )-\log (1+i\omega ))$$

$${\tan }^{-1}\frac{y}{x}=\frac{i}{2}\left\{\log \left(1-i\frac{y}{x}\right)-\log \left(1+i\frac{y}{x}\right)\right\}$$

$${\tan }^{-1}\frac{y}{x}$$

$$-\frac{\pi }{2}+\mathrm{Im}(\log \zeta )=\frac{-1}{2}\left\{L\left(1,\frac{\mathrm{Re}\left(\zeta -\frac{1}{\zeta }\right)}{-\mathrm{Im}\left(\zeta +\frac{1}{\zeta }\right)}\right)-M\left(1,-\frac{\mathrm{Re}\left(\zeta -\frac{1}{\zeta }\right)}{-\mathrm{Im}\left(\zeta +\frac{1}{\zeta }\right)}\right)\right\}.$$

$${\tan }^{-1}(\tanh X\cot A)={\tan }^{-1}\left(\frac{X}{A}\right)+\sum _{k=1}^{\infty }\left({\tan }^{-1}\left(\frac{X}{k\pi +A}\right)-{\tan }^{-1}\left(\frac{X}{k\pi +A}\right)\right).$$

$${\tan }^{-1}\left(\frac{2X}{\pi }\right)=\sum _{k=1}^{\infty }\left({\tan }^{-1}\left(\frac{X}{c_k}\right)-{\tan }^{-1}\left(\frac{X}{d_k}\right)\right),$$

$${\tan }^{-1}\left(\frac{y}{x}\right)=\sum _{k=1}^{\infty }\left({\tan }^{-1}\left(\frac{e_{k}y}{x}\right)-{\tan }^{-1}\left(\frac{f_{k}y}{x}\right)\right),$$

$$L(1,\frac{y}{x})-M(1,-\frac{y}{x})=\sum _{k=1}^{\infty }(L_k(1,\frac{e_{k}y}{x})-M_k(1,-\frac{e_{k}y}{x}))-(L_k(1,\frac{f_{k}y}{x})-M_k(1,-\frac{f_{k}y}{x}))$$

$$L_k(s,a)=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{a^n}{n^s},wherea=\frac{x^2}{c_k^2-x^2},0<a<1,s\in \mathbb{C}$$

$$M_k(s,a)=\sum _{n=1}^{\infty }(-1{)}^n\frac{a^n}{n^s},wherea=\frac{y^2}{c_k^2},0<a<1,s\in \mathbb{C}.$$

$$L_k(s,a)=\sum _{n=1}^{\infty }(-1{)}^{n-1}\frac{a^n}{n^s},0<a<1,s=\sigma +ib,\sigma ,b\in \mathbb{R}.$$

$$\sum _{n=1}^{\infty }\frac{|a{|}^n}{n^{\sigma }}=\sum _{n=1}^{\infty }\frac{1}{|c{|}^n{n}^{\sigma }},a=1/c,c>1.$$

$$\underset{n\to \infty }{\lim }|\frac{a_{n+1}}{a_n}|=\underset{n\to \infty }{\lim }\frac{|c{|}^n{n}^{\sigma }}{|c{|}^{n+1}(n+1{)}^{\sigma }}=\underset{n\to \infty }{\lim }\frac{1}{|c|}\frac{1}{(1+1/n{)}^{\sigma }}=\frac{1}{c}<1.$$

$$\gamma (s)$$

$$=\sum _{n=1}^{\infty }[F(n)-F(n-1)]\frac{a^n}{n^s}$$

$$=\sum _{n=1}^{\infty }F(n)\frac{a^n}{n^s}-\sum _{n=1}^{\infty }F(n)\frac{a^{n+1}}{(n+1{)}^s}-F(0).a$$

$$\displaystyle=\sum_{n=1}^{\infty}F(n)\int_{n}^{n+1}\Big{(}\frac{a^{x}\ln a}{x^{s}}-\frac{sa^{x}}{x^{s+1}}\Big{)}dx$$

$$=\ln a\sum _{n=1}^{\infty }{\int }_n^{n+1}\frac{a^{x}F(x)}{x^s}-s\sum _{n=1}^{\infty }{\int }_n^{n+1}\frac{a^{x}F(x)}{x^{s+1}}dx$$

$$=\ln a{\int }_1^{\infty }\frac{a^{x}F(x)}{x^s}-s{\int }_1^{\infty }\frac{a^{x}F(x)}{x^{s+1}}dx.$$

$$Letf(x)$$

$$=\ln a{\int }_1^{\infty }\frac{a^x(F(x)-1/2)}{x^s}dx+\frac{\ln a}{2}{\int }_1^{\infty }\frac{a^x}{x^s}dx$$

$$=I_1(s)+I_2(s)(say),and$$

$$g(x)$$

$$=s{\int }_1^{\infty }\frac{a^x(F(x)-1/2)}{x^{s+1}}dx+\frac{s}{2}{\int }_1^{\infty }\frac{a^x}{x^{s+1}}dx$$

Full text

1. Introduction

In this article, we explore two different forms of the Euler-Ramanujan identities and their relationship to certain Dirichlet series and minimal surfaces. The use of Weierstrass-Enneper representation of minimal surfaces in E-R identities first appeared in  [2]. When we rewrite the identities in terms of sums of Dirichlet series we get relationships between the latter and sometimes using the W-E representation of minimal surfaces the relationship is in terms of one complex parameter. The minimal surfaces which are involved are a family of Scherk’s surfaces and helicoids which are minimal surfaces of translation. In section $2$ of this paper we devote ourselves to relationships between sums of Dirichlet series. In section $3$ we study the properties of one of the Dirichlet series and derive a functional equation of the latter.

2. Rewriting a form of the Euler-Ramanujan identity in terms of Dirichlet series

We have Euler-Ramanujan’s identity,  [6], Example $(1)$ page $38$, where $X$, $A$ are complex variables, $A$ is not an odd multiple of $\pi/2$:

[TABLE]

We take complex logarithm on both sides, to get:

[TABLE]

Let $c_{k}=(k-\frac{1}{2})\pi$. Replace $X+A$ by $y$ and $A$ by $x$ in the identity where $x$ (not an odd multiple of $\frac{\pi}{2}$) and $y$ are real parameters, we have,

[TABLE]

or,

[TABLE]

Let us take the special case when $|\frac{x^{2}}{c_{k}^{2}-x^{2}}|<1$ and $|\frac{y^{2}}{c_{k}^{2}}|<1$.

Then,

[TABLE]

where $L_{k}(s,a)=\sum_{n}((-1)^{n+1}(\frac{x^{2}}{c_{k}^{2}-x^{2}})^{n}/n^{s})$ is a Dirichlet series with a real paramater $a=\frac{x^{2}}{c_{k}^{2}-x^{2}}$ and an integer $k$, $M_{k}(s,b)=\sum_{n}(\frac{y^{2}}{c_{k}^{2}})^{n}/n^{s}$ is a Dirichlet series with a real parameter $b=\frac{y^{2}}{c_{k}^{2}}$ and an integer $k$ and $L_{k}(1,\frac{x^{2}}{c_{k}^{2}-x^{2}})$ and $M_{k}(1,\frac{y^{2}}{c_{k}^{2}})$ are these Dirichlet series evaluated at $s=1$. We note that $|\frac{x^{2}}{c_{k}^{2}-x^{2}}|<1$ and $|\frac{y^{2}}{c_{k}^{2}}|<1$ for all $k$ is equivalent to $-\frac{\pi}{2\sqrt{2}}<x<\frac{\pi}{2\sqrt{2}}$ and $-\sqrt{\pi/2}<y<\sqrt{\pi/2}$.

Thus we have the following proposition:

Proposition 2.1**.**

$\log\left(\frac{\cos{y}}{\cos{x}}\right)=\sum_{k=1}^{\infty}[L_{k}(1,\frac{x^{2}}{c_{k}^{2}-x^{2}})-M_{k}(1,\frac{y^{2}}{c_{k}^{2}})]$* for $-\frac{\pi}{2\sqrt{2}}<x<\frac{\pi}{2\sqrt{2}}$ and $-\sqrt{\pi/2}<y<\sqrt{\pi/2}$.*

In  [2], using the Weierstrass-Enneper representation of minimal surfaces we derived the following way of writing the equation $z=\log\left(\frac{\cos{y}}{\cos{x}}\right)$ (Scherk’s minimal surface) in parametric form (in terms of a complex parameter $\zeta$), $x(\zeta,\bar{\zeta})=2{\rm Re}\tan^{-1}(\zeta),\;y(\zeta,\bar{\zeta})=-{\rm Im}\log\left(\frac{1+\zeta}{1-\zeta}\right)$ and $z(\zeta,\bar{\zeta})={\rm Re}\log\left(\frac{1+\zeta^{2}}{1-\zeta^{2}}\right)$. This parametrization fails precisely at $\zeta=\pm 1,\pm i$.

Using the log version of the E-R identity , for $\zeta\neq\pm 1,\pm i$ and belonging to a small domain in ${\mathbb{C}}$, the expression in terms of Dirichlet series look as follows:

[TABLE]

The condition that $|\frac{x^{2}}{c_{k}^{2}-x^{2}}|<1$ and $|\frac{y^{2}}{c_{k}^{2}}|<1$ for all $k$ is satisfied if $|\zeta|<\frac{1}{2}$. This can be seen as follows. For $\zeta\in\mathbb{C}$ such that $|\zeta|<\frac{1}{2}$, we have $-2<2\frac{|\zeta|\cos{\theta}}{1-|\zeta|^{2}}<2,$ which implies $-\tan{(\frac{\pi}{2\sqrt{2}})}<2\frac{|\zeta|\cos{\theta}}{1-|\zeta|^{2}}<\tan{(\frac{\pi}{2\sqrt{2}})}.$ In other words, $-\tan{(\frac{\pi}{2\sqrt{2}})}<\frac{\zeta+\bar{\zeta}}{1-|\zeta|^{2}}<\tan{(\frac{\pi}{2\sqrt{2}})}$ and hence $-\frac{\pi}{2\sqrt{2}}<\tan^{-1}(\zeta)+\tan^{-1}(\bar{\zeta})<\frac{\pi}{2\sqrt{2}},$ or $-\frac{\pi}{2\sqrt{2}}<2\rm{Re}\tan^{-1}(\zeta)<\frac{\pi}{2\sqrt{2}}.$ This implies $|\frac{x^{2}}{c_{k}^{2}-x^{2}}|<1$ where $x=2\rm{Re}\tan^{-1}(\zeta)$. From this it follows that $|\frac{x^{2}}{c_{k}^{2}-x^{2}}|<1$ for all $k$, $y^{2}/c_{k}^{2}<1$ does not give any additional constraint on $\zeta$.

Thus, we have the following proposition:

Proposition 2.2**.**

For $\zeta\in{\mathbb{C}}$ such that $|\zeta|<\frac{1}{2}$,

[TABLE]

2.1. More examples of minimal surfaces and their relation to Dirichlet series

In this section, we will revisit the Scherk’s surface, [7], and the helicoid, [4, 5]. In fact, we consider a one parameter family of Scherk’s type surface. We show that all these surfaces will give reise to a Dirichlet series decomposition. We consider the case of helicoid separately.

We consider a one parameter family of minimal translation surfaces (Scherk’s type surfaces),

[TABLE]

where $\theta\in\mathbb{R}$. Indeed, for a fixed $\theta$, $X_{\theta}(u,v)=\alpha(u)+\beta_{\theta}(v)$, where $\alpha(u)=(u,0,-\log{\cos{u}})$ and $\beta_{\theta}(v)=(v\cos{\theta},v\sin{\theta},\log{\cos{v}})$ and hence it is a minimal surface of translation (see remark at the end of this section).

Remark 2.3*.*

Observe that $\theta=0$ corresponds to a plane in the above family and $\theta=\frac{\pi}{2}$ corresponds to the classical Scherk’s surface, i.e., $X_{\frac{\pi}{2}}(u,v)=(u,v,\log{\frac{\cos v}{\cos u}})$ (discussed in the previous section).

The $\theta$-family of minimal translation surfaces gives us a family of Euler-Ramanujan’s identities. For $\theta\neq\pm\frac{(2n+1)\pi}{2}$, the corresponding identity we call it a twisted Euler-Ramanujan’s identity. Indeed, put $u+v\cos{\theta}=x$ and $v\sin{\theta}=y$, then using the Euler-Ramanujan’s identity we obtain

[TABLE]

where $x-y\cot{\theta}$ is not an odd multiple of $\frac{\pi}{2}$. Taking $\log$ on both the sides, we get

[TABLE]

When we take $\theta$ as an odd multiple of $\frac{\pi}{2}$ in the above identity, we get back the identity (1). Following the same idea (as in Scherk’s surface case), we have a Dirichlet series decomposition in this case as well, which is as follows:

Proposition 2.4**.**

[TABLE]

where $L_{k}(1,\frac{(x-y\cot\theta)^{2}}{c_{k}^{2}-(x-y\cot\theta)^{2}})$ is a Dirichlet series (evaluated at $s=1$) with a real paramater $a=\frac{(x-y\cot\theta)^{2}}{c_{k}^{2}-(x-y\cot\theta)^{2}}$ for a fixed $\theta$ and an integer $k$, $M_{k}(s,-(\frac{y}{c_{k}\sin\theta})^{2})$, ($\theta$-fixed), is a Dirichlet series (evaluated at $s=1$ ) with a real parameter $b=-(\frac{y}{c_{k}\sin\theta})^{2}$ and an integer $k$ as before.

Now, let us consider the non parametric representation of helicoid which is given by $z=\tan^{-1}\frac{y}{x}$ and also recall the identity

[TABLE]

where $\omega$ is a complex number. Now if we put $\omega=\frac{y}{x}$ in the above identity, we get

[TABLE]

The above expression helps us to write helicoid as a sum of two Dirichlet series evaluated at a specific value. When $|y|<|x|$ we can express the identity (2) as

[TABLE]

where $L(s,\frac{y}{x})=\sum_{k=1}^{\infty}\frac{i^{k+1}}{k^{s}}(\frac{y}{x})^{k}$ and $M(s,-\frac{y}{x})=\sum_{k=1}^{\infty}\frac{i^{k+1}}{k^{s}}(-\frac{y}{x})^{k}$ are Dirichlet series which are in turn evaluated at $s=1$. Thus we have the following proposition:

Proposition 2.5**.**

When $|y|<|x|$, $\tan^{-1}\frac{y}{x}=\frac{1}{2}[L(1,\frac{y}{x})-M(1,-\frac{y}{x})]$.

The Weierstrass-Enneper representation of the helicoid in terms of the complex parameter is given by (see for instance [2]), $x(\zeta,\bar{\zeta})=-\frac{1}{2}{\rm Im}\left(\zeta+\frac{1}{\zeta}\right)$, $y(\zeta,\bar{\zeta})=\frac{1}{2}{\rm Re}\left(\zeta-\frac{1}{\zeta}\right),$ and $z(\zeta,\bar{\zeta})=-\frac{\pi}{2}+{\rm Im}(\log{\zeta})$. The condition $|y|<|x|$ translates to $|\rm{Re}(\zeta-\frac{1}{\zeta})|<|\rm{Im}(\zeta+\frac{1}{\zeta})|$ which is satisfied if $|\zeta|<1$. Now it is easy to see that we have

Proposition 2.6**.**

For $\zeta\in{\mathbb{C}}$, such that $|\zeta|<1$

[TABLE]

2.2. Another Euler-Ramanujan identity

Next we look at another identity. For $X$ and $A$ real, we have (see entry $11$ in [6])

[TABLE]

When $A=\frac{\pi}{2}$ the identity (4) reduces to the following identity

[TABLE]

where $c_{k}=(k-\frac{1}{2})\pi$ and $d_{k}=(k+\frac{1}{2})\pi$. Next we put $\frac{2X}{\pi}=\frac{y}{x}$ in (5) to obtain

[TABLE]

where $e_{k}=\frac{\pi}{2c_{k}}$ and $f_{k}=\frac{\pi}{2d_{k}}$. Thus we have

Proposition 2.7**.**

[TABLE]

where $L_{k}(1,\frac{e_{k}y}{x})=L(1,\frac{e_{k}y}{x})$ and $M_{k}(1,-\frac{e_{k}y}{x})=M(1,-\frac{f_{k}y}{x})$.

Remark 2.8*.*

Our objects of interest here happen to be minimal surfaces of translation. In the past, there has been a good amount of interest in knowing what are all the minimal translation surfaces in $\mathbb{R}^{3}$. See for instance Lopez and Hasanis,  [3]. It would be interesting to see if number-theoretical identities like the E-R identities are available for all minimal surfaces of translation.

3. Properties of the Dirichlet series

We consider the general form of the Dirichlet series defined in the previous section

[TABLE]

3.1. Convergence of the series

We show that the series is convergent for all $s\in\mathbb{C}$ by showing its absolute convergence.

[TABLE]

Now, $|n^{s}|=|n^{\sigma}||n^{ib}|=n^{\sigma}|e^{ib\ln n}|=n^{\sigma}$.

The absolute series then is

[TABLE]

Then the ratio test gives

[TABLE]

Thus $L_{k}(s,a)$ is convergent on $0<a<1$, for each $k$. Similar is the case for $M_{k}(s,a)$.

3.2. Functional equation

In this section, we follow a technique of Hardy, expounded in $[1]$.

Our general Dirichlet series is $\gamma(s)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^{n}}{n^{s}}$, where, $0<a<1,~{}s\in\mathbb{C}$.

Then, defining $F(x)=\sum_{n\leq x}(-1)^{n-1}=\dfrac{1-{(-1)}^{m}}{2},~{}m<x<m+1$, we have,

[TABLE]

[TABLE]

The integrals can all be seen to be convergent for all $s$.

$F(x)-\frac{1}{2}$ is bounded, say by $M$. Then, taking $a=\frac{1}{c},c>1$ we can see that $\int_{1}^{\infty}|\frac{a^{x}(F(x)-\frac{1}{2})}{x^{s}}|dx=\int_{1}^{\infty}|\frac{(F(x)-\frac{1}{2})}{|c|^{x}x^{\sigma}}|dx\leq M\int_{1}^{\infty}\frac{1}{c^{x}x^{\sigma}}dx$ (for $\sigma>0$), which is convergent. $I_{1}$ and similarly $I_{2}$ are thus absolutely convergent since $c^{x}$ is exponential growth and $x^{\sigma}$ has only polynomial growth. $I_{3}$ and $I_{4}$ can similarly be concluded to be convergent.

We try to evaluate $I_{4}$ first. Let $x\ln a=u\Rightarrow du=\ln a~{}dx$. Then $u$ ranges from $\ln a$ to $-\infty$ as $x$ ranges from $1$ to $\infty$.

Then,

[TABLE]

Similar to $I_{4}(s)$, $I_{2}(s)$ will then be $-\frac{(-\ln(a))^{s}}{2}~{}\Gamma(-(s-1),-\ln a)$, where $\Gamma$ is the incomplete gamma function.

We now calculate $I_{3}$. Since the function $F(x)$ is piecewise continuous of period 2, we calculate its Fourier expansion. Then, $F(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}(a_{n}cos\frac{n\pi x}{L}+b_{n}sin\frac{{n\pi x}}{L})$, where,

$a_{0}=\frac{1}{L}\int_{0}^{2L}F(x)cos\frac{0\pi x}{L}dx=\int_{0}^{2}F(x)dx=\int_{1}^{2}dx=1$

$a_{m}==\frac{1}{L}\int_{0}^{2L}F(x)cos\frac{m\pi x}{L}dx=0$

$b_{n}=\frac{1}{L}\int_{0}^{2L}F(x)sin\frac{n\pi x}{L}dx=\frac{-2}{n\pi}$ for $n$ odd.

Therefore, $F(x)=\frac{1}{2}+\sum_{k=0}^{\infty}\dfrac{-2sin(2k\pi x+\pi x)}{(2k+1)\pi}$.

Substituting $F(x)-\frac{1}{2}$ by its Fourier expansion, we have,

[TABLE]

We now try to evaluate $I_{5k}(s)=\int_{0}^{\infty}\frac{a^{\frac{t}{\pi(2k+1)}}\sin(t)}{t^{s+1}}~{}dt$.

Let $\rm{Re}(s)<0$.

$I_{5k}$ can be evaluated in terms of a complex gamma function as follows:

[TABLE]

where $C_{\pm}=\frac{ln(a)}{\pi(2k+1)}\pm i$.

Let $u=C_{+}t$ and $w=C_{-}t$. Then making change to these complex variables and multiplying by $(-1)^{-s-1}$ when needed, we get

[TABLE]

where $\gamma_{1}$ is from [math] to $C_{+}\infty$ along the line $y=\frac{(2k+1)\pi}{ln(a)}x$ and $\gamma_{2}$ is from [math] to $C_{-}\infty$ along the line $y=\frac{-(2k+1)\pi}{ln(a)}x$. Take contours shown in the diagrams below (with $\gamma_{1}=e$ in the first diagram and $\gamma_{2}=e$ in the second diagram) and let the radius of the circular arc grow bigger. It can be shown that the first integral evaluates to $\int_{\infty}^{0}u^{-s-1}e^{-u}du=-\Gamma(-s)$ and so does the second one, if ${\rm Re}(s)<0$. Here $\Gamma(-s)\equiv\int_{0}^{\infty}u^{-s-1}e^{-u}du$ and has poles along non-positive integers for $\rm{Re}(s)<0$. It is easy to check that $I_{5k}(s)=-\Gamma(-s)/((\frac{\ln(a)}{(2k+1)\pi})^{2}+1)$.

In this case, $I_{3}(s)=\frac{-2s}{\pi}\sum_{k=0}^{\infty}\frac{\pi^{s}}{(2k+1)^{1-s}}I_{5k}(s)=\frac{2s}{\pi}\sum_{k=0}^{\infty}\frac{\pi^{s}}{(2k+1)^{1-s}((\frac{\ln(a)}{(2k+1)\pi})^{2}+1)}\Gamma(-s)$.

$I_{1}(s)=\frac{\ln(a)}{s-1}I_{3}(s-1)=\frac{2\ln(a)}{\pi}\sum_{k=0}^{\infty}\frac{\pi^{s-1}}{(2k+1)^{2-s}((\frac{\ln(a)}{(2k+1)\pi})^{2}+1)}\Gamma(-s+1)$

Thus

[TABLE]

Let $N_{k}=(2k+1)$ and $A=\frac{-ln(a)}{\pi}$. Let us restrict ourselves to $e^{-\pi}<a<1$ such that 0¡ $\frac{A}{N_{k}}<1$ for all $k$. Let $\tilde{\zeta}(s)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}$.

Notice that

[TABLE]

and

[TABLE]

Thus, the following proposition gives us a functional relationship between the Dirichlet series and $\tilde{\zeta}$ function.

Proposition 3.1**.**

Let $A=\frac{-ln(a)}{\pi}$ and $\tilde{\zeta}(s)=\sum_{k=0}^{\infty}\frac{1}{(2k+1)^{s}}$. For $\rm{Re}(s)<0$ and $e^{-\pi}<a<1$, we have the following functional equation.

[TABLE]

The infinite sums converge since $0<A<1$.

3.3. Essential singularity at $\infty$

We consider the limit of our function as $s\rightarrow\infty$ along two directions. $L_{k}(s,a)=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^{n}}{n^{s}},~{}where~{}a=\frac{x^{2}}{c_{k}^{2}-x^{2}},0<a<1,~{}s=\sigma+it\in\mathbb{C}$.

When $\rm{Im}(s)$ is fixed and $\rm{Re}(s)\rightarrow\infty$ :

[TABLE]

Therefore, $\sum_{n=1}^{\infty}(-1)^{n-1}\frac{a^{n}}{n^{s}}\rightarrow a$ as $\sigma\rightarrow\infty$.

When $\rm{Re}(s)$ is fixed and $\rm{Im}(s)\rightarrow\infty$ :

If possible, let $L_{k}(\frac{1}{2}+it,a)\rightarrow a$ as $t\rightarrow\infty$. Then given $\epsilon>0,\exists t_{0}\in{\mathbb{R}}$ such that

[TABLE]

which is not true for sufficiently small $\epsilon$.

Thus we arrive at a contradiction. The limit of $L_{k}(s,a)$ as $s\rightarrow\infty$ does not exist and the series has an essential singularity at infinity.

4. Acknowledgement:

The first author would like to thank Professor Sanoli Gun of IMSC, Chennai, India for interesting discussions.