Question

Evaluate : ${\int }_0^1\log (\frac{1}{x}-1)dx$

Answer 1

${\int }_0^1\log (\frac{1}{x}-1)dx=I$ (say)

$={\int }_0^1\log \left(\frac{1-x}{x}\right)dx$

Using property, ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$

$I={\int }_0^1\log \left(\frac{1-(1-x)}{1-x}\right)dx$

$={\int }_0^1\log \left(\frac{x}{1-x}\right)dx$

$={\int }_0^1\log {\left(\frac{1-x}{x}\right)}^{-1}dx$

$={\int }_0^1(-1)\log \left(\frac{1-x}{x}\right)dx$

$I=-{\int }_0^1\log \left(\frac{1-x}{x}\right)dx$

$I=-I$

$2I=0$

$\therefore I=0$

$\therefore {\int }_0^1\log (\frac{1}{x}-1)dx=0$