How to show that $\int_1^\infty a^{-\lfloor{\log_b x\rfloor}}dx=\frac{a(b-1)}{a-b}$?

Question

How to show that $\int_1^\infty a^{-\lfloor{\log_b x\rfloor}}dx=\frac{a(b-1)}{a-b}$?

This was a question in the MIT integration Bee 2015 at 59:01. I thought of doing a Riemann sum but I don't know how to calculate the length of the rectangles. Any idea how to solve this problem?

Answer 1

Assuming $b>1$, here is a hint: $$\int_1^\infty a^{-\lfloor\log_b x\rfloor}\,\mathrm{d}x=\sum_{n=0}^\infty\int_{b^n}^{b^{n+1}}a^{-n }\,\mathrm{d}x$$