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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?2012-02-06
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    @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](http://math.stackexchange.com/q/49383/19341). 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.2012-02-07
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    @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}))$2012-02-07
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    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.2012-02-07
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    I have to thank.2012-02-07
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    (1) Correct, you can't change the order. (2) The other "variable" here is the index $i$. Two-variable change might work if it was an inner integration instead of a summation. (3) You can put $t$ inside the sum at least, just like $n^2=n\sum_{k=1}^n 1 =\sum_{k=1}^n n$.2012-02-07
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    @anon (2) My eyes must have been blind to the $i$. (3) isn't $\sum_1^n n = \frac{n(n+1)}{2}$?2012-02-07
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    No, you're thinking of $\sum_{k=1}^n k =\frac{n(n+1)}{2}$. Don't confuse the index variable with one from the outside.2012-02-08

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