Dirichlet transform of $e^{(2 \pi i / 3) \Omega(n)}$

Question

The Dirichlet transform of the Liouville function $\lambda(n)$ is famously

$$ \sum_{n=1} \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}\tag{1}$$

The Liouville function is defined by $$ \lambda(n) = (-1)^{\Omega(n)}\tag{2} $$ with $\Omega(n)$ being the number of not-necessarily distinct prime factors of the natural number $n$.

I am interested in the Dirichlet transform of a similar (completely multiplicative) function $f$ defined as $$ f(n):= e^{(2 \pi i / 3)\Omega(n)} \tag{3}$$

I have got that far in the evaluation:

$$\begin{eqnarray} \sum_{n=1} \frac{f(n)}{n^s} & = & \sum_{n=1} \frac{e^{(2 \pi i / 3)\Omega(n)}}{n^s} \nonumber \\ & = & \prod_{p\in\Bbb{P}} \left( \sum_{k=0} \frac{e^{(2 \pi i / 3)k}}{p^{ks}} \right) = \prod_{p\in\Bbb{P}} \left( \sum_{k=0} \left(\frac{e^{(2 \pi i / 3)}}{p^{s}}\right)^k \right)\nonumber \\ & = & \prod_{p\in\Bbb{P}} \frac{1}{ 1 - \frac{e^{2 \pi i / 3}}{p^{s}}} \tag{5} \\ \end{eqnarray}$$

at this point I am not completely sure how to continue. In analogy to the case of the Dirichlet transformation of $\lambda(n)$ one could expand these fractions in trinomial fashion possibly by a factor of $ \frac{1}{1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}$ that would yield

$$\begin{eqnarray} \prod_{p\in\Bbb{P}} \frac{1}{ 1 - \frac{e^{2 \pi i / 3}}{p^{s}}} \cdot \frac{{1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}}{{1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}} & = & \prod_{p\in\Bbb{P}} \frac{{1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}}{ 1 - \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^3 } \nonumber \\ & = & \prod_{p\in\Bbb{P}} \frac{{1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}}{ 1 - \frac{1}{p^{3s}} } \nonumber \\ & = & \zeta(3s) \prod_{p\in\Bbb{P}} \left({1 + \frac{e^{2 \pi i / 3}}{p^{s}} + \left(\frac{e^{2 \pi i / 3}}{p^{s}}\right)^2}\right) \nonumber \\ \end{eqnarray}$$

setting $\beta(p)=\frac{e^{2 \pi i / 3}}{p^{s}}$ I obtain

$$\begin{eqnarray} \sum_{n=1} \frac{f(n)}{n^s} & = & \zeta(3s) \prod_{p\in\Bbb{P}} (1 + \beta(p) + \beta^2(p)) & \nonumber \\ & = &\zeta(3s) \prod_{p\in\Bbb{P}} \left(\beta(p) - \frac{i\sqrt{3}-1}{2}\right) \prod_{p\in\Bbb{P}} \left(\beta(p) + \frac{i\sqrt{3}+1}{2}\right) \end{eqnarray}$$

Here I get a bit stuck, the hope would be that one would be able to express the whole thing in terms of elementary functions.

Any help appreciated.

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Edit

From here I can slightly proceed by

$$\begin{eqnarray} \prod_{p\in\Bbb{P}} \left(\beta(p) - \frac{i\sqrt{3}-1}{2}\right) \prod_{p\in\Bbb{P}} \left(\beta(p) + \frac{i\sqrt{3}+1}{2}\right) & = & \prod_{p\in\Bbb{P}} \left(\frac{1-i\sqrt{3}}{2} - \beta(p)\right) \prod_{p\in\Bbb{P}} \left(\frac{i\sqrt{3}+1}{2} + \beta(p)\right) \end{eqnarray}$$

by defining the constants $\alpha_{1/2}:=\frac{2e^{2\pi i / 3}}{(+/-)1-i\sqrt{3}}$ this can be rewritten to

$$\begin{eqnarray} \prod_{p\in\Bbb{P}} \left(\frac{1-i\sqrt{3}}{2} - \beta(p)\right) \prod_{p\in\Bbb{P}} \left(\frac{i\sqrt{3}+1}{2} + \beta(p)\right) & = & \prod_{p\in\Bbb{P}} \left(1-\frac{\alpha_1}{p^s}\right) \prod_{p\in\Bbb{P}} \left(1-\frac{\alpha_2}{p^s}\right) \end{eqnarray}$$

These expressions look up to the factors $\alpha$ very similar to the Euler Product of $\zeta^{-1}$, but I am not aware of if this can be simplified any further.

Answer 1

For your case, it is possible to write the Dirichlet series in the following product. Let $|z|\leq R <2$. Then

$$ \sum_{n=1}^{\infty} z^{\Omega(n)} n^{-s} = \zeta(s)^z F(s,z) $$ where $$ F(s,z) = \prod_p \left( 1- \frac z{p^s}\right)^{-1}\left(1-\frac 1{p^s}\right)^z $$ is absolutely convergent for $\Re(s)>\frac12$.

Thus, you can try putting $z=e^{2\pi i/3}$.

For a general method, you can refer to Montgomery, Vaughan's Multiplicative Number Theory, classical theory I, p.232 Theorem 7.18.