sequences and series - Does $sum_{n=1}^infty frac{1}{phi(n)^s}$ have a euler product?

Does $$ \sum_{n=1}^\infty \frac{1}{\phi(n)^s}$$

have a euler product and functional equation? $\phi(n)$ is the euler phi function.

Since $\phi(n)$ is multiplicative I think the series could have a euler product and functional equation.

$$ \sum_{n=1}^\infty \frac{1}{\phi(n)^s}= \sum_{n=1}^\infty \frac{a(n)}{n^s}$$

where $a(n)$ is sequence https://oeis.org/A058277 in the OEIS.

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I did some research and found many relations between the zeta function and other special functions such as:

$$ \sum_{n=1}^\infty \frac{\phi(n)}{n^s}=\frac{\zeta(s-1)}{\zeta(s)}. $$

Answer

Firstly, $$\phi(p^k)=p^k\left(1-\frac1p\right)\qquad{\text{$k\ge1$, prime $p$}}$$

Define $f(k)=\phi(k)^{-s}$.

For the moment, consider the set of integers $S_N=\{k\,|\,k=p_1^{a_1}p_2^{a_2}\cdots p_N^{a_N},a_{(\cdot)}\ge1\}$.

Then,
$$\begin{align}
\sum_{k\in S_N}f(k)
&=\sum^\infty_{a_N=1}\cdots\sum^\infty_{a_1=1}
\left[p_1^{a_1}\cdots p_N^{a_N}
\left(1-\frac1{p_1}\right)\cdots\left(1-\frac1{p_1}\right)\right]^{-s} \\
&=\left(\prod^N_{i=1}\frac1{1-1/p_i}\right)^s\cdot\prod^N_{j=1}\sum^\infty_{a_j=1}p_j^{-a_js} \\

&=\prod^N_{i=1}\frac{(1-1/p_i)^{-s}}{p_i^s-1}
\end{align}
$$

Define $$f_i:=\frac{(1-1/p_i)^{-s}}{p_i^s-1}$$

Now we want to find
$\displaystyle{\sum_{k\in S^*_N}f(k)}$ where $S^*_N=\{k\,|\,k=p_1^{a_1}\cdots p_N^{a_N},a_{(\cdot)}\color{red}{\ge0}\}$

Summing $f(k)$ over all the elements in $S_N^*$ with $a_\alpha=0$ and other indexes non-zero gives $\displaystyle{\frac1{f_\alpha}\prod^N_{i=1}f_i}$.

How about having two zero indexes $a_\alpha,a_\beta$? $\displaystyle{\frac1{f_\alpha f_\beta}\prod^N_{i=1}f_i}$.
Three?
$\displaystyle{\frac1{f_\alpha f_\beta f_\gamma}\prod^N_{i=1}f_i}$.

Summing all these and factorizing naturally give
$$\sum_{k\in S^*_N}f(k)=\left(1+\frac1{f_1}\right)\left(1+\frac1{f_2}\right)\cdots\left(1+\frac1{f_N}\right)\cdot\prod^N_{i=1}f_i=\prod^N_{i=1}(1+f_i)$$

Taking the limit $N\to\infty$, we find that
$$\sum^\infty_{n=1}\frac1{\phi(n)^s}=\prod_{\text{prime }p}\left(1+\frac{(1-1/p)^{-s}}{p^s-1}\right)$$

I am still working on the functional equation. It is clear that the function is holomorphic on $\text{Re }s>1$, and is likely to have a pole at $s=1$, as plugging in $s=1$ gives $\prod_p\left(1+\frac1p\right)=\infty$.

A few more words on analytic continuation:

Obviously,
$$\begin{align}
F(s):=\sum^\infty_{n=1}\frac1{\phi(n)^s}
&=\prod_{\text{prime }p}\left(1+\frac{(1-1/p)^{-s}}{p^s-1}\right)\\
&=\prod_{p}\frac1{1-p^{-s}}\cdot\prod_p[1-p^{-s}+(p-1)^{-s}] \\

&=\zeta(s)\cdot \underbrace{\prod_p[1-p^{-s}+(p-1)^{-s}]}_{G(s)}
\end{align}
$$

1. $G(s)$ converges for $\text{Re }s>0$, as $1-p^{-s}+(p-1)^{-s}=1+sp^{-s-1}+O(p^{-s-2})$.


2. Therefore, $F(s)$ has a simple pole at $s=1$ due to zeta function. The residue there equals $$\operatorname*{Res}_{s=1}F(s)=\prod_p\left(1+\frac1{p(p-1)}\right)=\frac{315\zeta(3)}{2\pi^4}$$
   (See Landau’s totient constant.)


3. $$\lim_{s\to0^+}G(s)=1$$ This can be seen by plugging in $s=0$ into the above expression.


4. Let's look at $G'(0)$.
   
   $$\frac{G'(s)}{G(s)}=\sum_p\frac{p^{-s}\ln p-(p-1)^{-s}\ln(p-1)}{1-p^{-s}+(p-1)^{-s}}$$
   $$G'(0)=-G(0)\sum_p\ln\left(1-\frac1p\right)=\infty$$
   As $G(0)$ is finite and $G'(0)$ is not, this suggests that $0$ is a branch point of $F(s)$. Thus, meromorphic continuation is not possible.

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Further analysis shows that $\text{Re }s=0$ is the natural boundary of $F(s)$.

Firstly, by means of successive series expansion, we obtain

$$\begin{align}
\ln(1-p^{-s}+(p-1)^{-s})
&=\sum^\infty_{n=1}\frac{\left[p^{-s}-(p-1)^{-s}\right]^n}{n} \\
&=\sum^\infty_{n=1}\frac{1}{n}\sum^n_{r=0}\binom nr p^{-s(n-r)}(-1)^r(p-1)^{-sr} \\
&=\sum^\infty_{n=1}\frac{p^{-ns}}{n}\sum^n_{r=0}\binom nr (-1)^r\left(1-\frac1p\right)^{-sr} \\
&=\sum^\infty_{n=1}\frac{p^{-ns}}{n}\sum^n_{r=0}\binom nr (-1)^r\sum^\infty_{k=0}\binom{-sr}{k}\frac{(-1)^k}{p^k} \\
&=\sum^\infty_{n=1}\sum^\infty_{k=0}\alpha_{n,k}(s)\frac1{p^{k+ns}}
\end{align}
$$

where $$\alpha_{n,k}(s)=\frac{(-1)^k}{n}\sum^n_{r=0}(-1)^r\binom nr \binom{-sr}{k}$$

Furthermore, notice that
$$\alpha_{n,0}(s)=\frac1n\sum^n_{r=0}(-1)^r\binom nr=0$$

Therefore,
$$\ln(1-p^{-s}+(p-1)^{-s})=\sum_{(n,k)\in\mathbb N^2}\alpha_{n,k}(s)\frac1{p^{k+ns}}$$
$$\begin{align}
\implies \ln G(s)
&=\sum_p \ln(1-p^{-s}+(p-1)^{-s}) \\

&=\sum_{(n,k)\in\mathbb N^2}\alpha_{n,k}(s)\zeta_{\mathbb P}(k+ns) \\
\end{align}
$$
where $\zeta_{\mathbb P}$ is the prime zeta function.

It is well known that $\zeta_{\mathbb P}$ has the natural boundary $\text{Re }s=0$, because $\mathcal S$ (the set of singularities of $\zeta_{\mathbb P}$) clusters on the imaginary axis. Hence, obviously $G(s)$ cannot be analytically continued across $\text{Re }s=0$. A functional equation does not exist.

Meanwhile, we obtained a representation of $F(s)$ in terms of well-known functions:

$$\sum^\infty_{n=1}\frac1{\phi(n)^s}
=\zeta(s)\exp\left[\sum_{(n,k)\in\mathbb N^2}\alpha_{n,k}(s)\zeta_{\mathbb P}(k+ns)\right]
$$
$$\alpha_{n,k}(s)=\frac{(-1)^k}{n}\sum^n_{r=0}(-1)^r\binom nr \binom{-sr}{k}$$