integrate of functions involving floor - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T00:30:05Z http://mathoverflow.net/feeds/question/110679 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/110679/integrate-of-functions-involving-floor integrate of functions involving floor asd 2012-10-25T15:19:50Z 2012-10-25T21:50:28Z <p>Is there any exact formula or at least exact inequalities for the following intehral</p> <p>$$\int_2^x\frac{dt}{[\frac{\log x}{\log t}]\log t}$$</p> <p>where [x] is the greatest integer less than or equal to x.</p> <p>added:</p> <p>When I use </p> <p>$$x-1&lt;[x]\le x$$ I get $$\frac{x-2}{\log x}=\int_2^x\frac{dt}{\log x}\leq \int_2^x\frac{dt}{[\frac{\log x}{\log t}]\log t}\le \int_2^x\frac{dt}{\log x-\log t}$$ but they are not exact enough. I need more closer bounds.</p> http://mathoverflow.net/questions/110679/integrate-of-functions-involving-floor/110704#110704 Answer by Robert Israel for integrate of functions involving floor Robert Israel 2012-10-25T21:35:06Z 2012-10-25T21:50:28Z <p>Note that $\left\lfloor \dfrac{\log x}{\log t} \right\rfloor = n$ for $x^{1/(n+1)} &lt; t &lt; x^{1/n}$, so if $m = \left\lceil \dfrac{\log x}{\log 2} \right \rceil$</p> <p>\eqalign{\int_{2}^x &amp;\frac{dt}{\left\lfloor \frac{\log x}{\log t}\right\rfloor \log t} = \int_2^{x^{1/m}} \dfrac{dt}{m \log t} + \sum_{j=1}^{m-1} \int_{x^{1/(j+1)}}^{x^{1/j}} \dfrac{dt}{j \log t} \cr &amp;= \frac{1}{m} \text{Ei}(1,-\log 2) - \text{Ei}(1,-\log x)+ \sum_{j=2}^{m} \frac{1}{j(j-1)} \text{Ei}(1,-\frac{1}{j} \log x) \cr}</p> <p>You might also note that $$\frac{x^{1/j} - x^{1/(j+1)}}{\log x} \le \int_{x^{1/(j+1)}}^{x^{1/j}} \dfrac{dt}{j \log t} \le \frac{j+1}{j} \frac{x^{1/j} - x^{1/(j+1)}}{\log x}$$</p> <p>so using these bounds for all $j > N$ and the exact values for $j \le N$ will give you approximations with relative error at most $1/N$. </p>