Question about poles of the Lerch Transcendent

Question

The Lerch Transcendent is defined as the analytic continutation of the sum $$ \Phi(z,s,a)=\sum_{k=0}^\infty(k+a)^{-s}z^k. $$ According to Wolfram functions, for fixed $s$, $a$, the function $\Phi(z,s,a)$ has no poles or essential singularities. But consider, $$ \Phi(z,-1,1)=\frac{1}{(1-z)^2}, $$ which clearly has a pole at $z=1$. Is the information on the Wolfram functions site incorrect or am I missing something?

Answer 1

According to Maple documentation:

• If $v$ and $a$ are positive integers, $\Phi(z, a, v)$ has a branch cut in the $z$-plane along the real axis to the right of $z=1$, with a branch point at $z=1$.
• If $a$ is a non-positive integer, $\Phi(z, a, v)$ is a rational function of $z$ with a pole of order $1−a$ at $z=1$.

It doesn't say what happens for non-integer $a$, but I suspect a branch point at $z=1$.