Question

If $\underset{1/8}{\overset{1/2}{\int }}\left[\ln \left[\frac{1}{x}\right]\right]dx$ is equal to $\frac{a}{b},$ where $a$ and $b$ are coprime, then the value of $b-4a$ is
($[.]$ denotes the greatest integer function)

Answer 1

$I=\underset{1/8}{\overset{1/2}{\int }}\left[\ln \left[\frac{1}{x}\right]\right]dx=\underset{1/8}{\overset{1/7}{\int }}\left[\ln 7\right]dx+\underset{1/7}{\overset{1/6}{\int }}\left[\ln 6\right]dx+\underset{1/6}{\overset{1/5}{\int }}\left[\ln 5\right]dx+\underset{1/5}{\overset{1/4}{\int }}\left[\ln 4\right]dx+\underset{1/4}{\overset{1/3}{\int }}\left[\ln 3\right]dx+\underset{1/3}{\overset{1/2}{\int }}\left[\ln 2\right]dx$

Since $\left[\ln 2\right]=0$ and $\left[\ln 3\right]=\left[\ln 4\right]=\left[\ln 5\right]=\left[\ln 6\right]=\left[\ln 7\right]=1$
$\therefore I=\underset{1/8}{\overset{1/3}{\int }}1dx=\frac{5}{24}$
$\Rightarrow a=5,b=24\therefore b-4a=4$