Prove a relation between Theta functions and their derivatives to prove $\sum_{n=-\infty}^\infty \frac{1}{\cosh^2(\pi n )}$

Question

In an attempt of proving

\begin{align*} \sum_{n=-\infty}^\infty \frac{1}{\cosh^2(\pi n )}=\frac{1}{\pi}+\frac{\Gamma^4\left( \frac14\right)}{8 \pi^3} \tag{1} \end{align*}

I found the following relationship between theta functions and their derivatives which could help me evaluate $(1)$, but I don´t know how to prove it.

\begin{align*} \frac{\vartheta_3^\prime(e^{-s \pi})}{\vartheta_3(e^{-s \pi})}-\frac{\vartheta_2^\prime(e^{-s \pi})}{\vartheta_2(e^{-s \pi})}=\frac{\pi}{4}\vartheta_4^4(e^{-s \pi}) \end{align*}

Backgroud: To prove $(1)$ I started by the following equality:

\begin{align*} \vartheta_2(q)=2 q^{1/4}\prod_{n=1}^\infty (1-q^{2n})(1+q^{2n})^2 \tag{2} \end{align*}

Proof:

Recall Jacobi´s triple product

\begin{align*} \sum_{n=-\infty}^{\infty} q^{n^{2}} z^{n}&=\prod_{n=1}^{\infty}\left(1+z q^{2 n-1}\right)\left(1+z^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right) \end{align*}

The letting $z=q$ in the equation above we get

\begin{align*} \sum_{n=-\infty}^{\infty} q^{n^{2}} q^{n}&=\sum_{n=-\infty}^{\infty} q^{n^{2}+n+\frac{1}{4}-\frac{1}{4}}\\ &=q^{-1/4}\sum_{n=-\infty}^{\infty} q^{(n+1/2)^2}\\ &= \prod_{n=1}^{\infty}\left(1+q q^{2 n-1}\right)\left(1+q^{-1} q^{2 n-1}\right)\left(1-q^{2 n}\right)\\ &= \prod_{n=1}^{\infty}\left(1+ q^{2 n}\right)\left(1+ q^{2 n-2}\right)\left(1-q^{2 n}\right)\\ &= \prod_{n=1}^{\infty}\left(1+ q^{2 n-2}\right)\prod_{n=1}^{\infty}\left(1+ q^{2 n}\right)\left(1-q^{2 n}\right)\\ &= 2\left(1+ q^{2}\right)\left(1+ q^{4}\right)\left(1+ q^{6}\right)\cdots\prod_{n=1}^{\infty}\left(1+ q^{2 n}\right)\left(1-q^{2 n}\right)\\ &=2 \prod_{n=1}^{\infty}\left(1+ q^{2 n}\right)\prod_{n=1}^{\infty}\left(1+ q^{2 n}\right)\left(1-q^{2 n}\right)\\ &=2 \prod_{n=1}^\infty (1-q^{2n})(1+q^{2n})^2 \end{align*}

Now rewrite $(2)$ as

\begin{align*} \frac{\vartheta_2(q)}{2 q^{1/4}}&=\prod_{n=1}^\infty (1-q^{2n})(1+q^{2n})^2 \\ &=\prod_{n=1}^\infty (1-q^{4n})(1+q^{2n})\\ &=\prod_{n=1}^\infty (1-q^{4n})(1+q^{2n})\cdot \frac{(1-q^{2n})}{(1-q^{2n})}\\ &=\prod_{n=1}^\infty\frac{(1-q^{4n})^2}{(1-q^{2n})} \tag{3} \end{align*}

Thus taking logarithms on both sides of $(3)$ gives

\begin{align*} \ln \vartheta_2(q)-\ln 2 ~\frac14\ln q &=2\sum_{n=1}^\infty\ln (1-q^{4n}) -\sum_{n=1}^\infty\ln(1-q^{2n})\\ &=\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{q^{2nk}}{k}-2 \sum_{n=1}^\infty \sum_{k=1}^\infty \frac{q^{4nk}}{k} \\ &=\sum_{k=1}^\infty \sum_{n=0}^\infty \frac{q^{2k(n+1)}}{k}-2 \sum_{k=1}^\infty \sum_{n=0}^\infty \frac{q^{4k(n+1)}}{k} \\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{q^{2k}}{1-q^{2k}}-2\sum_{k=1}^\infty \frac{1}{k}\frac{q^{4k}}{1-q^{4k}}\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{q^{2k}}{1-q^{2k}}-2\sum_{k=1}^\infty \frac{1}{k}\frac{q^{4k}}{(1-q^{2k})(1+q^{2k})}\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{1}{1-q^{2k}}\left[q^{2k}-\frac{2q^{4k}}{1+q^{2k}} \right]\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{1}{1-q^{2k}}\left[\frac{q^{2k}+q^{4k}-2q^{4k}}{1+q^{2k}} \right]\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{1}{1-q^{2k}}\left[\frac{q^{2k}}{1+q^{2k}(1-q^{2k})} \right]\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{q^{2k}}{1+q^{2k}}\\ &=\sum_{k=1}^\infty \frac{1}{k}\frac{1}{1+q^{-2k}} \end{align*}

For $q=e^{-s \pi}$ we obtain

\begin{align*} \ln \vartheta_2\left(e^{-s \pi}\right)-\ln 2 +\frac{\pi s}{4}&=\sum_{n=1}^\infty \frac{1}{n}\frac{1}{e^{2 \pi s n}+1} \nonumber\\ &=\frac{1}{2}\sum_{n=1}^\infty \frac{1-\tanh(n \pi s)}{n} \tag{4} \end{align*}

Now differentiate $(4)$ w.r.t. to $s$ to obtain

\begin{align*} \sum_{n=1}^\infty \frac{1}{\cosh^2(n \pi s)} &=-\frac{1}{2}-\frac{2}{\pi}\frac{\vartheta_2^\prime\left(e^{-s \pi}\right)}{\vartheta_2\left(e^{-s \pi}\right)} \end{align*}

By differentiating both sides of the functional equation of $\vartheta_3$ w.r.t. $t$

\begin{align*} \sqrt{t} \sum_{n=-\infty}^\infty e^{-\pi t n^2}&=\sum_{n=-\infty}^\infty e^{-\frac{\pi n^2}{t} } \end{align*}

I obtained

\begin{align*} \frac{1}{2\sqrt{t}}\sum_{n=-\infty}^\infty e^{-\pi t n^2}-\sqrt{t}\sum_{n=-\infty}^\infty n^2 e^{-\pi t n^2}&= \frac{\pi}{t^2}\sum_{n=-\infty}^\infty n^2e^{-\frac{\pi n^2}{t} } \end{align*}

Letting $t \to 1$

\begin{align*} \sum_{n=-\infty}^\infty n^2 e^{-\pi n^2 } &=\frac{1}{4 \pi} \sum_{n=-\infty}^\infty e^{-\pi n^2 } \tag{5} \end{align*}

Now recall that

\begin{align*} \sum_{n=-\infty}^\infty e^{-\pi n^2 } &=\frac{\sqrt[4]{\pi}}{\Gamma\left(\frac34 \right)} \end{align*}

Plugging these result in $(5)$ we conclude that

\begin{align*} \sum_{n=-\infty}^\infty n^2 e^{-\pi n^2 } &=\frac{1}{4\pi^{3/4}\Gamma\left(\frac34 \right)} \end{align*}

But the same technique fails when trying to evaluate the derivative of $\vartheta_2$. The functional equation

\begin{align*} \sqrt{t} \sum_{n=-\infty}^\infty (-1)^n e^{-\pi t n^2}&=\sum_{n=-\infty}^\infty e^{-\frac{\pi }{t}\left(n+\frac12\right)^2 } \end{align*}

when differentaited w.r.t. to $t$ ends up having a relation between $\vartheta_2$ and the derivatives of $\vartheta_2$ and $\vartheta_4$, and I dont know how to evaluate neither derivatives.

NOTE: Equation $(4)$ can be written in the alternative form

\begin{align*} \frac{1}{2}\sum_{n=1}^\infty \frac{1-\tanh(n \pi s)}{n}&=\ln \vartheta_2\left(e^{-s \pi}\right)-\ln 2 +\frac{\pi s}{4} \\ &=\ln \frac{\vartheta_2\left(e^{-s \pi}\right)\vartheta_3\left(e^{-s \pi}\right)}{\vartheta_3\left(e^{-s \pi}\right)}-\ln 2 +\frac{\pi s}{4}\\ &=\ln \left(\sqrt{k}\sqrt{\frac{2 \operatorname{K}(k)}{\pi}}\right)-\ln 2 +\frac{\pi s}{4} \\ &=\frac12\ln \left(\frac{2 k \operatorname{K}(k)}{2\pi}\right) +\frac{\pi s}{4} \end{align*}

Answer 1

Following @reuns comment I got the following:

We have the q-series expansion of $\vartheta_4^4$

\begin{align*} \vartheta_4^4(q)=1+8\sum_{n=1}^\infty \frac{(-1)^n n q^n}{1+q^n} \tag{1} \end{align*}

Using the relation between series

\begin{align*} \sum_{n=1}^\infty (-1)^n a_n=\sum_{n=1}^\infty a_{2n}-\sum_{n=1}^\infty a_{2n-1} \end{align*}

we can rewrite $(1)$ as

\begin{align*} \vartheta_4^4(q)=1+8\left[\sum_{n=1}^\infty \frac{(2n)q^{2n}}{1+q^{2n}}-\sum_{n=1}^\infty \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}} \right] \end{align*}

or

\begin{align*} \frac{\vartheta_4^4(q)-1}{8}=\left[\sum_{n=1}^\infty \frac{(2n)q^{2n}}{1+q^{2n}}-\sum_{n=1}^\infty \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}} \right] \tag{2} \end{align*}

Now, we have already established in the question that

\begin{align*} \vartheta_2(q)=2 q^{1/4}\prod_{n=1}^\infty (1-q^{2n})(1+q^{2n})^2 \tag{3} \end{align*}

setting $z=1$ in the Jacobi´s triple product formula we obtain

\begin{align*} \vartheta_3(q)=\prod_{n=1}^\infty (1-q^{2n})(1+q^{2n-1})^2 \tag{4} \end{align*}

Dividing $(4)$ by $(3)$ we get

\begin{align*} \frac{ \vartheta_3(q)}{\vartheta_2(q)}=\frac{1}{2 q^{1/4}}\prod_{n=1}^\infty \frac{(1+q^{2n-1})^2}{(1+q^{2n})^2} \tag{5} \end{align*}

Taking logarithms of both sides of $(5)$ and then differentiating w.r.t. q we get

\begin{align*} \frac{ \vartheta_3^\prime(q)}{\vartheta_3(q)}-\frac{ \vartheta_2^\prime(q)}{\vartheta_2(q)}=-\frac{1}{4q}-\frac{2}{q}\left[\sum_{n=1}^\infty \frac{(2n)q^{2n}}{1+q^{2n}}-\sum_{n=1}^\infty \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}} \right] \tag{6} \end{align*}

plugging $(2)$ in $(6)$ we obtain

\begin{align*} \frac{ \vartheta_3^\prime(q)}{\vartheta_3(q)}-\frac{ \vartheta_2^\prime(q)}{\vartheta_2(q)}&=-\frac{1}{4q}-\frac{2}{q}\left[\frac{\vartheta_4^4(q)-1}{8} \right] \\ &=-\frac{\vartheta_4^4(q)}{4q} \tag{7} \end{align*}

Being more explicit, we let $q=e^{-s \pi}$, then $(7)$ becomes

\begin{align*} \frac{\pi\sum_{n=-\infty}^\infty\left(n+\frac12\right)^2e^{-\pi s \left(n+\frac12\right)^2}}{\sum_{n=-\infty}^\infty e^{-\pi s \left(n+\frac12\right)^2}}&= \frac{\pi\sum_{n=-\infty}^\infty n^2e^{-\pi s n^2}}{\sum_{n=-\infty}^\infty e^{-\pi s n^2}}+\frac{\pi \vartheta_4^4(e^{-s \pi})}{4} \tag{8} \end{align*}

and our series is equal to

\begin{align*} -\frac{\pi}{2}\sum_{n=1}^\infty \frac{1}{\cosh^2(n \pi s)} &=\frac{\pi}{4}-\pi\frac{\sum_{n=-\infty}^\infty\left(n+\frac12\right)^2e^{-\pi s \left(n+\frac12\right)^2}}{\sum_{n=-\infty}^\infty e^{-\pi s \left(n+\frac12\right)^2}} \tag{9} \end{align*}

Plugging $(8)$ in $(9)$

\begin{align*} \sum_{n=1}^\infty \frac{1}{\cosh^2(n \pi s)} &=-\frac{1}{2}+2\left[ \frac{\sum_{n=-\infty}^\infty n^2e^{-\pi s n^2}}{\sum_{n=-\infty}^\infty e^{-\pi s n^2}}+\frac{ \vartheta_4^4(e^{-s \pi})}{4} \right]\\ &=-\frac{1}{2}+2\left[ \frac{1}{4 \pi}+\frac{1 }{4}\frac{\Gamma^4\left( \frac14\right)}{8 \pi^3} \right] \end{align*}

and since the series is even we get our final result

\begin{align*} \sum_{n=-\infty}^\infty \frac{1}{\cosh^2(\pi n )}=\frac{1}{\pi}+\frac{\Gamma^4\left( \frac14\right)}{8 \pi^3} \end{align*}

Note: To find $\vartheta_4(e^{-s \pi})$

\begin{align*} \sum_{n=-\infty}^\infty (-1)^n e^{-\pi n^2} &=\frac{\vartheta_4\left(e^{-\pi}\right)}{\vartheta_3\left(e^{-\pi}\right)}\vartheta_3\left(e^{-\pi}\right)\\ &=\vartheta_3\left(e^{-\pi}\right)\sqrt{k^\prime} \\ &=\frac{\vartheta_3\left(e^{-\pi}\right)}{\sqrt[4]{2}}\\ &=\frac{\sqrt[4]{\pi}}{\sqrt[4]{2}\Gamma\left(\frac{3}{4} \right)} \end{align*}

Then we apply the reflection formula for the gamma function to obtain the form

\begin{align*} \sum_{n=-\infty}^\infty (-1)^n e^{-\pi n^2}=\frac{\Gamma\left(\frac{1}{4} \right)}{\pi^{3/4}2^{3/4}} \end{align*}

Answer 2

This is an alternative approach to find sum of your series in terms of elliptic integrals. I have mentioned about the theta function identity in comments and will reiterate the same at the end of the answer as well.

Your series related to $\cosh(n\pi)$ can be handled directly in terms of Plouffe's function $a(q) $ given by $$a(q) =\sum_{n=1}^{\infty}\frac{q^n}{n(1-q^n)}\tag{1}$$ We can observe that $$b(q):=\sum_{n=1}^{\infty}\frac{q^n}{n(1+q^n)}=a(q)-2a(q^2)$$ and differentiating the above we get $$qb'(q) =\sum_{n=1}^{\infty}\frac{q^n }{(1+q^n)^2}=qa'(q)-4q^2a'(q^2)$$ If $q=e^{-\pi} $ we can write the sum in question as $$\sum_{n\in\mathbb{Z}} \frac {1}{\cosh^2(n\pi)}=1+8\sum_{n=1}^{\infty}\frac {q^{2n}}{(1+q^{2n})^2}=1+8q^2b'(q^2)=1+8q^2a'(q^2)-32q^4a'(q^4)\tag{2}$$ We can now use the expressions for $a(q^2),a(q^4)$ from the linked answer $$a(q^2) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\tag{3}$$ and $$a(q^{4}) = -\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\tag{4}$$ And differentiating these equations we get $$q^2a'(q^2)=\frac{q}{2}\frac{dk}{dq}\frac{d}{dk}\left(-\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}\right) $$ and $$4q^4a'(q^4)=q\frac{dk}{dq}\frac{d}{dk}\left(-\frac{1}{12}\log(k^{4}k') + \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{6K}\right) $$ and using $(2)$ we get our desired sum as $$\frac{4kk'^2K^2}{\pi^2}\frac{d}{dk}\log(kK)=\frac{4KE}{\pi^2}\tag{5}$$ The above derivation uses the formulas $$\frac{dK} {dk} =\frac{E-k'^2K}{kk'^2},\,\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{6}$$ The latter one is equivalent to the identity between derivatives of theta functions given in question.

To get the closed form for the sum let us note that if $q=e^{-\pi}$ then $$k=k'=\frac{1}{\sqrt{2}},K=\frac{\Gamma^2(1/4)}{4\sqrt{\pi}}$$ To get the value of $E$ we use the Legendre's identity $$KE'+K'E-KK'=\frac{\pi} {2}$$ Since $k=k'$ we have $K'=K, E'=E$ and then $$4KE=\pi+2K^2$$ and the sum in question equals $$\frac{1}{\pi}+\frac{2K^2}{\pi^2}=\frac{1}{\pi}+\frac{\Gamma^4(1/4)} {8\pi^3}$$

---

The identity $$1+8\sum_{n=1}^{\infty}\frac{q^{2n}}{(1+q^{2n})^2}=\frac{4KE}{\pi^2}$$ was established by Jacobi in his Fundamenta Nova using Fourier series of square of elliptic functions and I have described his approach in this answer (see equation $(13)$ there).