# unusual integral involving ceiling and floor functions

• Apr 17th 2011, 10:52 AM
Random Variable
unusual integral involving ceiling and floor functions
http://quicklatex.com/cache3/ql_a52c...3079a83_l3.png

where b is an integer greater than 1.

If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.
• Apr 17th 2011, 11:51 AM
ivalmian
Quote:

Originally Posted by Random Variable
http://quicklatex.com/cache3/ql_a52c...3079a83_l3.png

where b is an integer greater than 1.

If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.

Plugging into mathematica with specific values for b you get

b = 1.1, integral = 8.04619
b = 2, integral = 0.989315
b = 10, integral = 0.101846
b = 100, integral = 0.00999388

This seems close to 1/(b-1) and I wouldn't be too suprised if the discrepancies are due to the fact this integral is poorly behaved.
• Apr 18th 2011, 01:05 AM
Opalg
Quote:

Originally Posted by Random Variable
http://quicklatex.com/cache3/ql_a52c...3079a83_l3.png

where b is an integer greater than 1.

If I was confident in my answer 1/(b-1) I would have posted this in the Math Challenge forum. Wolfram Alpha and Maple have no idea what to do with it, even for specific values of b and for finite intervals.

I agree with the answer 1/(b-1) in the case where b is an integer. For non-integer values of b I'm not so happy with it. I think that if no power of b is an integer then the answer will be \sum_{k=1}^{\infty} \frac{1}{\lceil b^k \rceil}. If it were not for the ceiling function in the denominator, that would agree with your answer.

If b is a number like $\displaystyle \sqrt 2$ with some powers integral and some powers non-integral, the answer would be very hard to compute.

Without LaTeX being available at present, I couldn't face trying to give a detailed solution. Basically, you need to notice that the integrand is zero when x>1, so you only need to look at the intervals 1/(n+1) < x < 1/n, where the integrand is equal to the integer part of \log_{b}n. The difficulties arise from handling the cases where \log_{b}n is already an integer.

Edit. As usual, I didn't read the problem carefully enough, where it stipulates that b is an integer. So yes, I agree with the given solution!
• Apr 18th 2011, 10:36 AM
ivalmian
Quote:

Originally Posted by Opalg
Without LaTeX being available at present, I couldn't face trying to give a detailed solution.

You can use http://www.quicklatex.com/ to render latex and then imbed them as images. It's very quick and simple.