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Math Help - Improper integral with log, floor and ceil function

  1. #1
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    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}.
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    very hard problem!! could u post a solution?
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  3. #3
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    I can do that, within few hours.
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    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
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  5. #5
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    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 <br />
\int_{0}^{1}{\left\lfloor \log _{\alpha }\left\lfloor \frac{1}{x} \right\rfloor \right\rfloor \,dx}<br />
?
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  6. #6
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    \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.
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