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I came across the following integral, while working with products of $\zeta$ primes function: $$ \int_{1}^{x}t^{-s-1} \sum_{i=1}^{\pi(t^{1/2})}\left[\pi\left(\frac{t}{p_i}\right)-i+1\right] dt, $$ where the inner sum represents $\pi_2(t)$, the number of semi-primes below $t$, and $\pi(t^{1/2})$ gives the number of squares below $t$.

  • Since $t$ also appears inside the integral as limit of the sum, I don't think I could switch summation and integration. Further $t$ doesn't make sense outside the integral at all.

  • Here it was recommended to do a two-variable change. How does that work, when I have only one variable?

  • Does it (somehow) work to move the $t^{-s-1}$ inside the summation? I'm worried, since $t$ also appears in the upper limit $\pi(t^{1/2})$.

Can anybody help me evaluating this integral?

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You should mention that the crazy sum is a formula for $\pi_2(t)$, the semiprime counting function. Note that in the expansion of $P(s)^2$, semiprimes occur twice and squared primes once. I don't see the point in having an $x$ instead of just letting it be $\infty$. Why is the linked question CW anyway? In my opinion it looks like this is pulling away from finding approximations instead of towards. Finally, why do you look at products of prime zeta functions instead of just approximating prime zeta functions themselves well? –  anon Feb 6 '12 at 17:51
    
@anon: Good point with the (semi-) prime counting, I forgot that. Thx for that. It runs to $x$, because I just sum up primes less than $x$, inspired by your own question. Mine's CW due to too many edits. I'll have to approximate somewhen, and I thought the later the better. Your final question I didn't get: Are there prime Zeta functions for semiprimes? ... yeah why not, another good point. Thx a lot. –  draks ... Feb 7 '12 at 8:09
    
@anon: Concerning your first point: Do you agree that $\pi_2$ counts all semiprimes, including squares, and therefore $P(s)^2 = \int_2^x t^{-s} d(2\pi_2(t)-\pi_1(t^{1/2}))$ –  draks ... Feb 7 '12 at 8:18
    
Thanks for the answers. My last question actually should have been posted on the linked question, not this one, and as per your last comment I agree of course. –  anon Feb 7 '12 at 8:22
    
I have to thank. –  draks ... Feb 7 '12 at 12:42

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