Where is $\varphi(z)=0?$ $$ \varphi(z)=\sum_{n=1}^\infty e^{-n^z} $$
I plotted $\varphi(z)$ for $z\in\Bbb C$ to get an idea of what the function looked like and it seemed like there could be roots. Also looking for references about this specific function and if it goes by a name, since I don't really know how to solve the equation.
I experimented with the function $f(x)=1/x$ and summed its values, $\sum_{n\ge1} 1/n,$ but it fails to converge. But if you do $v(x)=\sum_{n\ge1}n^{-x}$ then it does converge for $\Re(x)>1.$ And you can analytically continue this function. In the complex plane, this function now has some roots.
As for $\varphi(z)$ relating to this, $\varphi(z),$ is the summatory function of the nonlinear image of $f(x).$ What this means is that you take $f(x)$ and define a map $\lambda:\Bbb R^2\to\Bbb R^{2+}$ with $\lambda(x,y)=(e^x,e^y)$ which is clearly nonlinear. $f$ transforms from $1/x$ into $h(x)=e^{\frac{1}{\log(x)}}$Then you take the sum of, $h$ which is the image of $f.$ The sum is of $h$ is $\sum_{n\ge1}e^{-1/n}.$ Then consider adding a power of $-z$ to $n$ to get $\varphi(z).$
Whereas $v(x)$ admits an analytic continuation, it is unknown whether $\varphi(z)$ admits an analytic continuation. Note that if instead after the mapping $\lambda$ we multiply points on $h,$ i.e. $\prod_{n\ge1}e^{-n^z}$ the result would be $e^{-v(-x)}$ and the analytic continuation would coincide with the analytic continuation of $v(x).$ So from my viewpoint it seems as though multiplying these points on $h$ is more natural because then both analytic continuations coincide.
I think that for some reason unbeknownst to me, but probably obvious to someone else, by way of $\lambda,$ the summation of $f$ is converted to the product of $h$. And that by imposing the summation on $h$ there is an inconsistency.
I am confused because it would seem to me that $\lambda$ should change multiplication into addition not the other way around.
So I think, that the possible underlying inconsistency (using a sum instead of a product) is really what makes the analytic continuation and finding the possible zeros of $\varphi$ difficult.
That being said, I'm still interested in the question because if the assumption that a product is more natural and one still imposes a sum, then what are the consequences for the analytic continuation and other important properties of the function? That is, does the imposition of the sum directly imply that the analytic continuation fails to exist in all cases?
