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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 $x$.

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Why would you want such an expansion?

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.

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  • $\begingroup$ @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 $e^{sx}$. $\endgroup$ – Mohammad Al Jamal Mar 17 at 16:34

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