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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?

YuiTo Cheng
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LeafGlowPath
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    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 '19 at 12:16
  • @Conrad So by the same idea we should also just take $\Phi$ as single valued? – LeafGlowPath May 14 '19 at 12:38
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    yes, in the same way RZ is taken to be single valued – Conrad May 14 '19 at 12:41

2 Answers2

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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.

Maxim
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Let us check how Lerch transcendent $\Phi$ is multivalued in two of its variables.

In the variable $a$.

At least for $s=1$ integrating with respect to the variable $a$ for each $n$ is

(%i1) integrate(1/(k+a), a);
(%o1)                             log(k + a)

otherwise

(%i2) integrate(1/(k+a)^s, a);
Is - s equal to - 1?

no;
                                        1 - s
                                 (k + a)
(%o2)                            ------------
                                    1 - s

obtaning therefore, after straightforward normalization, the value of lerch zeta function $\Phi(z, s-1,a)$ from the original $\Phi(z, s,a)$, so that for positive integers $s$ would, at some point after iterated integrals and normalizations, cross the logarithm, in which the integration may shift by a constant $2\pi i k$, depending on the chossen branch. This is the way I see why the function is multivalued for the variable $a$: for each natural $s>0$, the Lerch zeta function comes from iterated multiplications by a same constant and differentiations, from the logarithm.

In the variable $z$.

The analytic continuation of $z$ has a branch cut at $[1,\infty)$, this is because the lerch zeta function in some domain equals $$\sum_{j\in\mathbb{N}}\operatorname{Li_{1+j}(z) a^j }$$ interpreted as a Generating function of the polylogarithm.

For $|z|>1$ the all polylogarithms have a branch cut with respect to this variable, located at the corresponding branch cut of the logarihtms $\operatorname{Li}_1(z)$, as discussed in Understanding Polylogarithm's Branch Points, Monodromy and Multivaluedness, Especially Around the Origin

There is where the multivaluedness of lerch zeta fucntion for the first variable $z$ comes from.

As in the other case, these are infinite sums of functions having their branch cut located at the same set (positive naturals $s$ at least and the semi-line $[1,\infty) \ni z$, ) therefore the sum of the functions should have the multiple values.

Dr Potato
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