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.
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)}.
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}
- G(s) converges for \text{Re }s>0, as 1-p^{-s}+(p-1)^{-s}=1+sp^{-s-1}+O(p^{-s-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.)
- \lim_{s\to0^+}G(s)=1 This can be seen by plugging in s=0 into the above expression.
- 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.
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}