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I would like to know if there is any formula for

$ \prod_{x<p\leq y}\left(1+\frac1p\right) $ in terms of $\theta$ or $\psi$ functions $ \theta(x)=\sum_{p\leq x}\log p $ and $ \psi(x)=\sum_{p^\nu\leq x}\log p. $

More precisely, I need to know if we can write that product as some thing similar to the following

$ \frac{\log\theta(y)}{\log\theta(x)}+\epsilon(x,y) $

or

$ \frac{\log\theta(y+\epsilon_1(x,y))}{\log\theta(x+\epsilon_2(x,y))} $

which is equality or very sharp inequality.

avoiding the terms include

$ \frac{\log y}{\log x} $

thanks

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I missed that you restrict the range of $p$ on both sides and thus the answer I gave does perhaps not make too much sense, so I deleted it for the moment. However, could you perhaps elabortae why/if you want the expression specifically in terms of theta or psi. Also, is there some relation on the size of $x$ and $y$. Just the product p <= y can be asymptotically avanluated to c log y with c= 6 e^(gamma) /pi^2 –  quid Mar 30 '13 at 1:36
    
Asymptotically, your product is $\log \theta(y)/\log\theta(x)$, by my answer below. But this won't have that good an error term. Taking $\log\theta(\cdot)$ seems like a pretty unnatural operation, since $\theta(\cdot)$ is a sum. –  Greg Martin Apr 2 '13 at 20:08
    
@Greg Martin, you are right. Asymptotically the product is logθ(y)/logθ(x). but I would like more sharper, since if I write it as the formula in your Answer, I lose some accuracy to invert to terms including only $\theta$ and it is not appropriate for my question. thanks! –  asd Apr 4 '13 at 23:07
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2 Answers

If you really wanted to, you could use partial summation to write $$ \log \prod_{x < p\le y} \bigg( 1+\frac1p \bigg) = \theta(y)f(y) - \theta(x)f(x) - \int_x^y \theta(t)f'(t)\ dt, $$ where $f(t) = \log(1+1/t)/\log t$. But I suspect the most useful thing one can say about your product follows from what quid said: $$ \prod_{x < p\le y} \bigg( 1+\frac1p \bigg) = \frac{\log y}{\log x} \bigg( 1 + O\bigg( \frac1{\log x} \bigg) \bigg) $$

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@Greg Martin. Thank you very much. Actually, I should have clarified my question. So I did now. Sorry I am late. –  asd Apr 2 '13 at 18:53
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[I undelete my (non-)answer (slightly edited), since Greg Martin's answer refers to part of my comment, and the answer contains some details related to this, which thus are perhaps interesting to have around.]

One has $$ \prod_{p\le x} (1+ 1/p) = \prod_{p\le x} (1- 1/p^2)/(1-1/p) = (\zeta(2)+o(1))^{-1} \prod_{p\le x} 1/(1-1/p) $$ and (essentially Mertens' 3rd Theorem) $$\prod_{p\le x} 1/(1-1/p)= (e^{\gamma} +o(1)) \log x.$$ So $$\prod_{p\le x} (1+ 1/p)$$ is asymtotically $$ \frac{6 e^{\gamma}}{\pi^2 } \log x . $$

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