3

I am trying to find if there is a simple expression for the following series $$ S=\frac{1}{2}\sum_{\ell=1}^\infty\frac{\coth(\pi\ell)-1}{\ell}=\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi \ell}-1)}\,. $$ In principle I think it's a good idea to get rid of the $\ell$ in the denominator, for instance consindering $$ S(a)=\sum_{\ell=1}^\infty \frac{1}{\ell(e^{2\pi a \ell}-1)}\,,\qquad S'(a)=-\frac{\pi}{2} \sum_{\ell=1}^\infty \frac{1}{\sinh^2(\pi a\ell)}\,, $$ but this doesn't seem to help much. I also tried comparing with an integral, writing $$ S = \sum_{\ell=0}^\infty \frac{1}{(\ell+1)(e^{2\pi(\ell+1)}-1)} = \int_0^\infty \frac{d\ell}{(\ell+1)(e^{2\pi(\ell+1)}-1)} + \frac{1}{2(e^{2\pi}-1)} + C $$ where $C$ can be in principle calculated with the Euler-MacLaurin or Abel-Plana formulas. Unfortunately, also these integrals seem pretty hard... Any suggestion?

Paramanand Singh
  • 87,309
Brightsun
  • 6,693
  • 1
    See https://math.stackexchange.com/a/938644/72031 Equation $(2)$ of that answer gives a closed form for your sum as $(1/2)\log(\pi/K)-\pi/12$ where $K=\Gamma^2(1/4)/4\sqrt{\pi}$. – Paramanand Singh Apr 05 '21 at 11:11
  • @ParamanandSingh Quite sure you can't beat my answer, all we need is to prove that $\Delta(z)$ is a weight 12 cusp form (to get $\Delta(z)=\frac{E_4(z)^3-E_6(z)^2}{1728}$) and to find the $\Gamma(1/4)$ closed-form of $G_4(i)$ from the $z\to (\wp_i(z),\wp_i'(z))$ isomorphism $\Bbb{C/(Z+iZ)}\to y^2=4x^3-g_2(i)x$. But perhaps you don't care of the proofs? – reuns Apr 05 '21 at 13:55
  • @reuns: I do care about proofs! I don't know why you have such misunderstanding about me. I admit my tools are a bit different from yours. – Paramanand Singh Apr 05 '21 at 14:17
  • 1
    @reuns : And I don't think I am trying to beat your answer. Please don't get me wrong. I aspire to learn your techniques someday. – Paramanand Singh Apr 05 '21 at 14:23
  • @ParamanandSingh How do we get the relation between the Dedekind function and the elliptic integral $K$? Or could you point me to a place where I can look up these things? Thanks. – Brightsun Apr 13 '21 at 09:05
  • You will need to study about theta functions and their relationship with elliptic integrals. Another key ingredient is Jacobi triple product. See this and also this. – Paramanand Singh Apr 13 '21 at 10:34
  • For a thorough understanding you may have to read posts related to "elliptic integrals, elliptic functions, theta functions, AGM" from this archives page. – Paramanand Singh Apr 13 '21 at 10:36

1 Answers1

3

Theory of modular forms and elliptic curves. For $\Im(z)>0$

$$f(z)=\sum_{l=1}^\infty \frac1{l (e^{-2i\pi lz}-1)}=\sum_{n=1}^\infty \sigma_{-1}(n) e^{2i\pi nz}= -\sum_{m\ge 1} \log(1-e^{2i\pi mz})= -\frac1{24}\log \left(\frac{\Delta(z)}{e^{2i\pi z}}\right)$$

From that $\Delta(z)$ is a weight $12$ cusp form we get the relation with Eisenstein series

$$\Delta(z)=\frac{E_4(z)^3-E_6(z)^2}{1728}$$

$E_6(i)=0$ and $E_4(i)$ has a closed-form in term of $\Gamma(1/4)$.

This closed-form follows from $1=\int_0^1 dz=\int_0^1 \frac{d\wp_i(z)}{\wp_i'(z)}=\int_C \frac{dx}{\sqrt{4x^3-g_2(i) x}}$ which reduces to $\frac{A}{E_4(i)}\int_0^1 \frac{dx}{x^3-x}=\frac{A}{E_4(i)}\ B(1/2,1/4)$ (the beta function)

reuns
  • 77,999
  • OK, I see, so the first step is to reduce this to the logarithm of the Euler function $\varphi(q)=\prod_n (1-q^n)$ or the Dedekind function (I think there is a slight confusion between $l$ and $n$ in the summation indices). I am unfamiliar with the function $\Delta(z)$ and its properties, though, so I cannot really follow the rest... – Brightsun Apr 13 '21 at 09:00
  • 1
    @Brightsun Yes sorry for the $l,n$ typo. There are a few different ways to deal with those special functions / special values, but the idea is to reduce it to an elliptic integral which reduces to the beta function $B(1/2,1/4)$, this reduction follows from that your series is the ($\log $ of the) special value of a modular form, and each point of the modular curve corresponds to an elliptic curve. This is a deep and rich theory, a bit hopeless to introduce it in a few lines. – reuns Apr 13 '21 at 11:30
  • 1
    I have to agree with your assessment that "this is a deep and rich theory, a bit hopeless to introduce it in a few lines". – Paramanand Singh Apr 17 '21 at 07:29