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