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Given the complex variable $x$, complex constant $c$, and integer number $r$. I want to solve the equation: $\sum_{k=1}^{\infty}\frac{e^{kx}}{k^r}=c$. I was thinking that if there is a formula or special function that can simplify the left side of the equation, then I may be able to find the analytical solution of $x$. So my question is: is there any formula for the sum of infinite series in the form of $\sum_{k=1}^{\infty}\frac{e^{kx}}{k}$ or $\sum_{k=1}^{\infty}\frac{e^{kx}}{k^2}$?

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If you don't have to be rigorous just take the derivatives until you get a geometric series. – 36min Nov 17 '12 at 1:46

According to Mathematica, $$\sum_{k=1}^\infty \frac{e^{k x}}{k^r} = Li_r(e^x),$$ which is not much of a revelation since $Li$ is the polylogarithm function, defined as $$Li_n(z) =\sum_{k=1}^\infty \frac{z^k}{k^n}.$$ But at least you know now that it is the polylogarithm that you are looking for. Wikipedia seems to have a decent description of it.

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Given the popular inverse function of the logarithm, you would think the polylog's inverse would have a name (the polyexp?) – Igor Rivin Nov 17 '12 at 3:28
Thanks Andrej & Kevin, that was really helpful. – spherical chicken Nov 17 '12 at 17:18

Where did you look? What did you try? $$ \sum_{k = 1}^{\infty} \frac{\operatorname{e} ^{k x}}{k} = -\operatorname{ln} \bigl(1 - \operatorname{e} ^{x}\bigr) $$ (the derivative is a geometric series, whose sum is known.)

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