# Mittag Leffler-like expansion of $\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}$

Let $$x\in \mathbb{R^{+}}$$ and $$s\in\mathbb{C}\;\;\Re(s)>0$$. The function in question has the well-known Mittag Leffler expansion: $$\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}=2xs^{2}\sum_{n=1}^{\infty}\frac{1}{4\pi^{2}n^{2}+(sx)^{2}}$$ But is it possible for our function to have an expansion of the form : $$\frac{s}{e^{sx}-1}-\frac{1}{x}+\frac{s}{2}=F(s,x)+\sum_{n=0}^{\infty}\frac{A_{n}(s)}{e^{x}-Z_{n}(s)}$$ Where $$A_{n}(s)$$, $$Z_{n}(s)$$ are appropriate functions in $$s$$, and $$F(s,x)$$ is entire in $$x$$.

I doubt your condition $$x>0$$ makes much sense, the continuation to $$x$$ complex should be implied by your expansions.
In that case fix $$s=\pi$$.
$$F(s,x)$$ analytic in $$x$$ and the $$2i\pi$$-periodicity in $$x$$ of $$\sum_{n=0}^{\infty}\frac{A_{n}(s)}{e^{x}-Z_{n}(s)}$$ would give that
$$\frac{s}{e^{sx}-1}-\frac{1}{x}-\frac{s}{e^{s(x+2i\pi)}-1}+\frac{1}{x+2i\pi}$$ is analytic in $$x$$ which is absurd.
• @renus. i was hoping the condition $x$ is real and positive would solve this problem. At any rate, i am after an expansion where $e^{x}$ would appear, but not $e^{sx}$. – Mohammad Al Jamal Mar 17 at 16:34