# Thread: Improper integral with log, floor and ceil function

1. ## Improper integral with log, floor and ceil function

Compute for all integer $\displaystyle \alpha>1$ the value of $\displaystyle \int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}.$

2. very hard problem!! could u post a solution?

3. I can do that, within few hours.

4. For each $\displaystyle x>1$ it's $\displaystyle x\le \left\lceil x \right\rceil <2x$ and the integrand is $\displaystyle \left\lfloor \log _{\alpha }1 \right\rfloor =0,$ hence the integral becomes $\displaystyle \int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}.$ Since $\displaystyle \alpha$ is an integer, then the inner floor function is reduntant so the integral is $\displaystyle \int_{0}^{1}{\left\lfloor -\log _{\alpha }x \right\rfloor \,dx}=\sum\limits_{k=0}^{\infty }{\int_{\alpha ^{-k-1}}^{\alpha ^{-k}}{k\,dx}}=\sum\limits_{k=0}^{\infty }{k\left( \frac{1}{\alpha ^{k}}-\frac{1}{\alpha ^{k+1}} \right)}=\frac{1}{\alpha -1}$ and we're done. $\displaystyle \blacksquare$

5. thanks a lot for the solution!! but it's kind of hard to me to understand, can you explain any further how did you get that the integral equals $\displaystyle \int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}$?

6. $\displaystyle \int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}=\sum\limits_{n=1}^{\infty }{\int_{n-1}^{n}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x} \right\rfloor \right\rfloor \,dx}}=\sum\limits_{n=1}^{\infty }{\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor \,dx}}.$

Thus, for $\displaystyle 0\le x\le1$ and $\displaystyle n\ge2$ we have $\displaystyle \left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor =0$ hence the integral is $\displaystyle \int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}+\sum\limits_{n=2}^{\infty }{\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor \,dx}}=\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx},$ and we're done.