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The problem arose, while I was looking at products of power prime zeta functions $$ P_x(ks)=\sum_{p\,\in\mathrm{\,primes}\leq x} p^{-ks}, $$ with $k\in \mathbb{N}$ and $s=it$ with real $t$. By using (see here) $$ \sum_{p\leq x}p^{-s}= \mathrm{li}(x^{1-s}) + O \left( {2|s|x^{1/2}}\log x \right) $$ I get (omitting the error terms for the moment) $$ \begin{eqnarray*} P_x(ks)P_x(ms) &\sim& \text{li}(x^{1-ks})\text{li}(x^{1-ms}) &=&\int_0^1 \int_0^1 \frac{1}{\ln( x^{1-ks} u_k) \ln(x^{1-ms}u_m)} du_k du_m\\ &=&{\rm Ei}(1-ks){\rm Ei}(1-ms)&&\\ \end{eqnarray*} $$ How can I simplify this expression? Non-trivial approximations are also welcome, when bounds are given.

Convolution: Since $\text{li}(x^{1-s})=\int_0^{x^{1-s}}(\ln u)^{-1}du$, I thought Integration of Convolutions, with something like $$\int(f*g)(x) \, dx=\left(\int f(x) \, dx\right)\left(\int g(x) \, dx\right) $$ might help, since my problem also deals with products of integrals?

Special Case: When I allow negative values for $k$ and $m$ and set $m=-k$ and get the following $$ \begin{eqnarray*} P_x(ks)P_x(-ks)&=&\pi(x) + \sum_{j

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