Integral with floor and ceil function

[Krizalid]

Here's an integral I love:

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

 

[Random Variable]

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} $

 

[Krizalid]

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