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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?

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  • $\begingroup$ 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 $\endgroup$ – Conrad May 14 at 12:16
  • $\begingroup$ @Conrad So by the same idea we should also just take $\Phi$ as single valued? $\endgroup$ – ablmf May 14 at 12:38
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    $\begingroup$ yes, in the same way RZ is taken to be single valued $\endgroup$ – Conrad May 14 at 12:41
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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.

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