# Improper integral with log, floor and ceil function

• February 17th 2009, 01:56 PM
Krizalid
Improper integral with log, floor and ceil function
Compute for all integer $\alpha>1$ the value of $\int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}.$
• February 18th 2009, 07:27 PM
Ajreb
very hard problem!! could u post a solution?
• February 18th 2009, 07:28 PM
Krizalid
I can do that, within few hours.
• February 19th 2009, 11:08 AM
Krizalid
For each $x>1$ it's $x\le \left\lceil x \right\rceil <2x$ and the integrand is $\left\lfloor \log _{\alpha }1 \right\rfloor =0,$ hence the integral becomes $\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}.$ Since $\alpha$ is an integer, then the inner floor function is reduntant so the integral is $\int_{0}^{1}{\left\lfloor -\log _{\alpha }x \right\rfloor \,dx}=\sum\limits_{k=0}^{\infty }{\int_{\alpha ^{-k-1}}^{\alpha ^{-k}}{k\,dx}}=\sum\limits_{k=0}^{\infty }{k\left( \frac{1}{\alpha ^{k}}-\frac{1}{\alpha ^{k+1}} \right)}=\frac{1}{\alpha -1}$ and we're done. $\blacksquare$
• February 21st 2009, 06:09 PM
Ajreb
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 $
\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}
$
?
• February 21st 2009, 07:23 PM
Krizalid
$\int_{0}^{\infty }{\left\lfloor \log _{\alpha }\left\lfloor \frac{\left\lceil x \right\rceil }{x} \right\rfloor \right\rfloor \,dx}=\sum\limits_{n=1}^{\infty }{\int_{n-1}^{n}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x} \right\rfloor \right\rfloor \,dx}}=\sum\limits_{n=1}^{\infty }{\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{n}{x+n-1} \right\rfloor \right\rfloor \,dx}}.$

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