Sum of series with exponential denominator

Question

How do I calculate the sum $$ \sum_{j=1}^\infty\frac{j^3}{1+e^{2\pi j z}} $$ where $z\in\mathbb C$ such that $\mathrm{Re}\left(z\right)>0$. Can it be evaluated/expressed in terms of some special functions?

Answer 1

Let $$f(q) =\sum_{j=1}^{\infty}\frac{j^3q^j}{1-q^j},\,|q|<1$$ and $$g(q) =\sum_{j=1}^{\infty} \frac{j^3q^j}{1+q^j},\,|q|<1$$ then we have $$f(q^2)=\frac{f(q)-g(q)}{2}$$ or $$g(q) =f(q) - 2f(q^2)$$ The function $f(q) $ is connected to Ramanujan's $Q(q) $ via $$Q(q) =1+240 f(q) $$ and hence $$g(q) =\frac{1+Q(q) - 2Q(q^2)}{240}$$ The sum in question is $g(e^{-2\pi z}) $ and it is well known from the theory of elliptic integrals/functions and theta functions that $Q(e^{-\pi\sqrt{n}}) $ has a closed form expression in terms of Gamma function values at rational points if $n$ is a positive rational number. Thus your sum can be evaluated in closed form if $z$ is positive and $z^2$ is rational. For other values of $z$ it is not known if there is any closed form expression for the sum in question.

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Here is one evaluation of the sum in question for $z=1/2,q=e^{-\pi}$. We have $$Q(q) =\left(\frac{2K}{\pi}\right)^4(1+14k^2+k^4)$$ and $$Q(q^2)=\left(\frac{2K}{\pi}\right) ^4(1-k^2+k^4)$$ where $k$ is the elliptic modulus corresponding to nome $q$ and $K=K(k) $ is the complete elliptic integral of first kind. When $q=e^{-\pi} $ we have $$k=\frac{1}{\sqrt{2}},K=\frac{\Gamma^{2}(1/4)}{4\sqrt{\pi}}$$ and then \begin{align} g(q) &=\frac{1}{240}+\frac{1}{240}\left(\frac{2K}{\pi}\right)^4(16k^2-k^4-1)\notag\\ &=\frac{1}{240}+\frac{27}{960}\left(\frac {\Gamma ^2(1/4)}{2\pi^{3/2}}\right)^4\notag\\ &=\frac{1}{240}\left(1+\frac{27\Gamma^{8}(1/4)}{64\pi^6}\right)\notag \end{align}