Integral with floor and ceil function

[Krizalid1]

Here's an integral I love:

For each integer $\alpha>1,$ compute $\displaystyle \int_0^\infty {\left\lfloor {{{\log }_\alpha }\left\lfloor {\frac{{\left\lceil x \right\rceil }}{x}} \right\rfloor } \right\rfloor \,dx} .$

 

[polygamma]

For $x>1$, $\displaystyle 1 \le \frac{\lceil x \rceil}{x} < 2 \implies \left\lfloor \frac{\lceil x \rceil}{x} \right\rfloor = 1 \implies \log_{a} \left\lfloor \frac{\lceil x \rceil}{x} \right\rfloor =0$

So $\displaystyle \int_{0}^{\infty} \left\lfloor \log_{a} \Big\lfloor \frac{\lceil x \rceil}{x} \Big\rfloor \right\rfloor \ dx = \int_{0}^{1} \left\lfloor \log_{a} \Big\lfloor \frac{\lceil x \rceil}{x} \Big\rfloor \right\rfloor \ dx = \int_{0}^{1} \left\lfloor \log_{a} \Big\lfloor \frac{1}{x} \Big\rfloor \right\rfloor \ dx$

Now find where the integrand is constant.

$\displaystyle \left\lfloor \log_{a} \Big\lfloor \frac{1}{x} \Big\rfloor \right\rfloor = k $

$ \displaystyle k \le \log_{a} \Big\lfloor \frac{1}{x} \Big\rfloor < k+1 $

$ \displaystyle a^{k} \le \Big\lfloor \frac{1}{x} \Big\rfloor < a^{k+1} $

$ \displaystyle \displaystyle a^{k} \le \frac{1}{x} < a^{k+1} $ since $a$ is a positive integer

$\displaystyle \implies \frac{1}{a^{k+1}} < x \le \frac{1}{a^{k}} $

$ \displaystyle \int_{0}^{1} \left\lfloor \log_{a} \Big\lfloor \frac{1}{x} \Big\rfloor \right\rfloor \ dx = \sum_{k=1}^{\infty} \int_{\frac{1}{a^{k+1}}}^{\frac{1}{a^{k}}} k \ dx $

$ \displaystyle =\sum^{\infty}_{k=1}k\left(\frac{1}{a^{k}}-\frac{1}{a^{k+1}}\right) =\sum^{\infty}_{k=1}k\left(\frac{1}{a}\right)^{k}-\sum^{\infty}_{k=1}k\left(\frac{1}{a}\right)^{k+1} $

$ \displaystyle =\frac{\frac{1}{a}}{(1-\frac{1}{a})^{2}}-\frac{(\frac{1}{a})^{2}}{(1-\frac{1}{a})^{2}}=\frac{1}{a-1} $

 

[Krizalid1]

Yes that's correct.

 

[oasi]

Here's an integral I love:

For each integer $\alpha>1,$ compute $\displaystyle \int_0^\infty {\left\lfloor {{{\log }_\alpha }\left\lfloor {\frac{{\left\lceil x \right\rceil }}{x}} \right\rfloor } \right\rfloor \,dx} .$

this one is very hard

 

[CellCoree]

The integral provided is a fascinating mathematical problem that involves the use of both the floor and ceiling functions. These functions are commonly used in mathematics to round a number down or up to the nearest integer, respectively. In this case, the integral involves taking the logarithm of a fraction and then rounding it down to the nearest integer.

I find this integral to be particularly interesting because it presents a challenge in terms of computation and also has potential real-world applications. The use of the floor and ceiling functions can be seen in various fields such as computer science, engineering, and statistics.

In order to solve this integral, one would need to carefully consider the behavior of the floor and ceiling functions and how they interact with the logarithm. Additionally, the bounds of the integral, from 0 to infinity, also add an interesting element to the problem.

I would be interested to see the results of computing this integral for different values of $\alpha$ and to explore any patterns or relationships that may emerge. This could potentially lead to further understanding of the behavior of these functions and their applications.

Overall, this integral serves as a great example of the complexity and beauty of mathematics and its potential for practical use in various scientific fields.