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Let $f:\ x\mapsto \sum_{n=0}^{\infty}{1 \over \cosh (nx)}$. We can see that $f$ is defined in $\Bbb{R}/\{0\}$ and $f$ is an even function.

It is easy to prove $f\to+\infty$ at $0^+$, and by using mathematics software I found that $$f(x)=_{x\to0^+}{\pi\over2x}+{1\over2}+o(1).$$ I can show that $\lim_{x\to0^+}xf(x)$ exists by proving $g:\ x\mapsto xf(x)$ is an increasing function in $(0,1)$, but I don't know how to give its exact value (which is supposed to be ${\pi\over2}$), furthermore I have no idea how to deal with the $1\over2$ in the formula above.

Any hint will be appreciated.

Aforest
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3 Answers3

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Here is an elementary derivation. By the Taylor's theorem, if $|\epsilon| \leq x$ then

$$ \frac{1}{\cosh(nx+\epsilon)} = \frac{1}{\cosh(nx)} - \frac{\sinh(nx)}{\cosh^2(nx)} \epsilon + \mathcal{O}\left(\frac{x^2}{\cosh(nx)}\right). $$

Here, the implicit bound of the error term is independent of both $x$ and $n$, owing to the exponential decay of the second derivative of $\operatorname{sech}$. Then rearranging and averaging both sides w.r.t. $\epsilon$ over the interval $[0,x]$, we obtain

$$ \frac{1}{\cosh(nx)} = \frac{1}{x} \int_{nx}^{(n+1)x} \frac{du}{\cosh u} + \frac{\sinh(nx)}{\cosh^2(nx)} \cdot \frac{x}{2} + \mathcal{O}\left(\frac{x^2}{\cosh(nx)}\right). $$

Applying similar trick to the second term in the RHS, we also obtain

$$ \frac{1}{\cosh(nx)} = \frac{1}{x} \int_{nx}^{(n+1)x} \frac{du}{\cosh u} + \frac{1}{2}\int_{nx}^{(n+1)x} \frac{\sinh u}{\cosh^2 u} \, du + \mathcal{O}\left(\frac{x^2}{\cosh(nx)}\right). $$

Summing over $n = 0, 1, \cdots$, we have

\begin{align*} f(x) &= \frac{1}{x}\int_{0}^{\infty} \frac{du}{\cosh u} + \frac{1}{2}\int_{0}^{\infty} \frac{\sinh u}{\cosh^2 u} \, du + \mathcal{O}\left(\sum_{n=0}^{\infty} \frac{x^2}{\cosh(nx)}\right) \\ &= \frac{\pi}{2x} + \frac{1}{2} + \mathcal{O}\left(\sum_{n=0}^{\infty} \frac{x^2}{\cosh(nx)}\right). \end{align*}

But it is easy to prove that $\sum_{n=0}^{\infty} \frac{x}{\cosh(nx)} \leq x + \int_{0}^{\infty} \frac{du}{\cosh u}$, so the above error term is $\mathcal{O}(x)$. Therefore we obtain

$$ f(x) = \frac{\pi}{2x} + \frac{1}{2} + \mathcal{O}(x). $$

Sangchul Lee
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    This is really cool +1 – Paramanand Singh Dec 07 '17 at 14:04
  • I think the identity in question is equivalent to the transformation formula for theta functions and your solution can then be used to prove the transformation formula for theta functions independent of the Poisson summation formula. – Paramanand Singh Dec 07 '17 at 14:15
  • @ParamanandSingh ??? The transformation for theta functions is equivalent to the Poisson summation formula ie. to the Fourier series. How do you plan to avoid that ? – reuns Dec 07 '17 at 14:28
  • @reuns : Ramanujan developed his theory of theta functions without making use of Poisson formula. His approach was based on algebraic identities between theta functions and transformation of hypergeometric functions. However I admit that Poisson formula gives the shortest proof of theta transformation formula. – Paramanand Singh Dec 07 '17 at 15:04
  • @ParamanandSingh, Thank you for the comment! Perhaps my computation is best understood in the context of Euler-Maclaurin formula. – Sangchul Lee Dec 07 '17 at 17:07
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This result belongs more properly to the theory of theta functions and elliptic integrals and the following approach is based on the same.

Let $q=e^{-x} $ and then $$f(x) =1+2\sum_{n=1}^{\infty}\frac{q^{n}}{1+q^{2n}}=\frac{1+\vartheta_{3}^{2}(q)}{2}\tag{1}$$ where $\vartheta_{3}(q)$ is one of Jacobi's theta functions. It is known from the theory of theta functions and elliptic integrals that $$\vartheta_{3}^{2}(q)=\frac{2K(k) }{\pi}\tag{2}$$ where $q=e^{-\pi K'(k) /K(k) } $ so that $K'(k) /K(k) =x/\pi$ and hence we can see that $$f(x) =\frac{1}{2}+\frac{K(k) }{\pi}=\frac{1}{2}+\frac{K'(k) }{x}$$ Now as $x\to 0^{+}$ the variable $q\to 1^{-}$ and hence the elliptic modulus $k\to 1^{-}$ so that $K'(k) \to \pi/2$ and we obtain $$f(x) =\frac{1}{2}+\frac{\pi}{2x}+o(1)$$

You can avoid elliptic integrals by using transformation formula $$\vartheta_{3}(q)=\vartheta_{3}(e^{-x})=\sqrt{\frac{\pi} {x}} \vartheta_{3}(e^{-\pi^{2}/x})\tag{3}$$ which gives us $$f(x) =\frac{1}{2}+\frac{\pi}{2x}\vartheta_{3}^{2}(e^{-\pi^{2}/x})\tag{4}$$ Since $$\vartheta_{3}(q)=1+2\sum_{n=1}^{\infty}q^{n^{2}}\tag{5}$$ the desired result is obtained easily.

Equation $(5)$ is the usual definition of Jacobi theta function $\vartheta_{3}$ and equation $(3)$ is the transformation formula for this function and it can be proved using Poisson summation formula. Second equality in equation $(1)$ is another deep relation satisfied by Jacobi theta function $\vartheta_{3}$ and it is related to the Fourier series of elliptic function $\operatorname {dn} (u, k) $.

Paramanand Singh
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  • Again I don't understand what you write. When you are defining something you are supposed to make it clear, when you are using a theorem, too. How do you define $K,K';k$ ? They are functions of what ? You are using 5 more different special functions that you are not even capable to define.. This is totally wrong. – reuns Dec 07 '17 at 14:31
  • @reuns : We have discussed this in past and the best would be to add a link to one of my answers which is having all the definitions. – Paramanand Singh Dec 07 '17 at 15:08
  • No. Adding more special functions, more theorems, more complicated properties won't help... The goal in maths is to make things as simple as possible. – reuns Dec 07 '17 at 15:14
  • @reuns : I myself believe in this philosophy and avoid any unnecessary abstractions. However prefacing every answer with definitions of elliptic integrals and theta functions is bit repetitive. And I am not adding any extra theorems, just linking an answer which contains the definitions. – Paramanand Singh Dec 07 '17 at 15:17
  • Good luck to explain what means and proves $k=\frac{\vartheta_{2}^{2}(q)}{\vartheta_{3}^{2}(q)}$ to you... Honnestly your elliptic theory is just a huge mess. – reuns Dec 07 '17 at 16:24
  • @reuns: that's a pretty strong feedback. But I will try to take it in positive manner and try to improve as best as I can. – Paramanand Singh Dec 07 '17 at 18:17
  • Let us continue this discussion in chat. – Paramanand Singh Dec 08 '17 at 08:54
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Since $\frac{1}{\cosh(x)}$ belongs to $\mathcal{S}(\mathbb{R})$ and $\int_{-\infty}^{+\infty}\frac{e^{-2\pi i n u}}{\cosh(n x)}\,dn = \frac{\pi}{x\cosh\left(\frac{\pi^2 u}{x}\right) } $, the Poisson summation formula ensures

$$ \sum_{n\in\mathbb{Z}}\frac{1}{\cosh(nx)} = \frac{\pi}{x}\sum_{u\in\mathbb{Z}}\frac{1}{\cosh\frac{\pi^2 u}{x}} $$ for any $x>0$. It can be rearranged as

$$ \sum_{n\geq 0}\frac{1}{\cosh(nx)} = \color{green}{\frac{1}{2}+\frac{\pi}{2x}}+\sum_{u\geq 1}\frac{\pi}{x\cosh\frac{\pi^2 u}{x}} $$ where for $x\to 0^+$ we have $$\begin{eqnarray*} \sum_{u\geq 1}\frac{\pi}{x\cosh\frac{\pi^2 u}{x}} &\approx& \frac{\pi}{2x\cosh\frac{\pi^2}{x}}+\int_{1}^{+\infty}\frac{\pi\,du}{x\cosh\frac{\pi^2 u}{x}}\\&=&\frac{\pi}{2x\cosh\frac{\pi^2}{x}}+\frac{\pi-4\arctan\tanh\frac{\pi^2}{2x}}{2\pi}\end{eqnarray*}$$ by the Euler-MacLaurin formula.

Jack D'Aurizio
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