Is Lerch's transcendent a multi-valued function?

Lerch’s Transcendent is defined by $${\Phi\left(z,s,a\right)=\sum_{n=0}^{\infty}\frac{z^{n}}{(a+n)^{s}}},$$ when $$|z|<1$$ or $$\Re s>1,|z|=1$$. If $$s$$ is not an integer then $$|\mathrm{ph}(a)|<\pi$$; if $$s$$ is a positive integer then $$a \ne 0,−1,−2,\dots$$; if s is a non-positive integer then a can be any complex number. For other values of $$z$$, $$\Phi(z,s,a)$$ is defined by analytic continuation.

My question is, is Lerch's Transcendent considered a multi-valued function?

I guess it depends on how the term $$(a+n)^s$$ is interpreted. For example if $$s = 3/2$$ and $$a = 1$$. We can take $$(a+n)^s$$ to be either the positive root or the negative root, which will give different value to the sum.

What is the traditional interpretation of such definitions in complex analysis?

• For $\alpha >0, s$ complex, $\alpha ^s= \exp{(s\log {\alpha})}$ is (conventionally taken to be) uniquely defined, where $\log {\alpha}$ is the unique real variable logarithm and $\exp$ is the usual exponential function with Taylor series $\Sigma{\frac{z^n}{n!}}$ so again uniquely defined – Conrad May 14 at 12:16
• @Conrad So by the same idea we should also just take $\Phi$ as single valued? – ablmf May 14 at 12:38
• yes, in the same way RZ is taken to be single valued – Conrad May 14 at 12:41

Suppose the function of the variable $$z$$ is $$\sum_{n \geq 0} z^n/(n + 1) = -\ln(1 - z)/z, \, |z| < 1$$, where $$\ln$$ is the principal value of the logarithm. Looping once around $$z = 1$$ gives $$-\ln(1 - z)/z \pm 2 \pi i/z$$. $$\Phi(z, 1, 1)$$ is multi-valued in this sense.