20110906, 19:21  #12 
May 2003
7·13·17 Posts 
William,
By prime density I am assuming you simply mean . If you mean something else, please let me know. If that is the case, then yes there are some improvements available. My favorite source currently for clean inequalities is in the work of Pierre Dusart. From his paper "Sharper bounds for " we have for and for . Better bounds than these are also available (which are asymptotically equal to ), using zerofree regions for the zeta function, but they are not nearly as clean. In other words, you can get for where f(x) is an explicit function which is better than for any k, and x_0 is also explicit. Last fiddled with by ZetaFlux on 20110906 at 20:07 
20110906, 20:44  #13 
May 2003
7×13×17 Posts 
On a second reading, I'm not entirely sure what you are asking. Let me ask some followup questions.
It is known that has an error term which is for all k. Under the Riemann hypothesis, the error term is . Is your question whether or not we can replace inside the integrand with another function, and get an even better error term? If so, the answer is yes. You just replace it with the step function which when integrated matches the prime counting function. On the other hand, if you are asking whether or not there is a simple smooth function which improves the integrand, I think the answer is unknown. At any rate, the difference between that function and would be small, and up to (after integration) would depend only on the zerofree region of the zetafunction. Last fiddled with by ZetaFlux on 20110906 at 20:45 
20110906, 21:02  #14 
May 2003
7·13·17 Posts 
In particular, note that if you replace the integrand by a better approximation to the prime density, like , then you do not necessarily get a "better" approximation to the prime counting function.
In this particular case, if I plugged things into Mathematica correctly, I think that the error term got worse. The error term is now asymptotically . 
20110906, 23:39  #15  
Aug 2006
5,987 Posts 
Quote:


20110907, 00:05  #16 
May 2003
11000001011_{2} Posts 
CRGreathouse,
What is the definition of "instantaneous density" in this context? 
20110907, 00:19  #17 
Aug 2006
5,987 Posts 

20110907, 00:30  #18 
May 2003
7×13×17 Posts 
Can you give an example of a suitable function e?

20110907, 01:42  #19 
Aug 2006
5,987 Posts 

20110907, 02:42  #20 
May 2003
7·13·17 Posts 
I'm having a hard time understanding your definition. First, if I take e(x)=1/sqrt{x}, and plot the function inside your limit, it heads off to infinity. So you don't get a function with growth 1/ln(x).
Second, if you actually do the limit then you get a constant (which in our case is infinity) and not a function (which in our case we want 1/ln(x)). 
20110907, 21:45  #21  
Aug 2006
5,987 Posts 
Quote:


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