Question

${\int }_1^{\infty }\frac{\log (t-1)}{t^2\log t+\log \left(\frac{t}{t-1}\right)}dt$ equals

• A. $\frac{1}{2}$
• B. $\frac{1}{3}$
• C. $\frac{2}{3}$
• D. None of these

Answer 1

The correct option is D None of these

${\int }_1^{\infty }\frac{\log (t-1)}{t^2\log t\log \frac{t}{(t-1)}}dt={\int }_0^1\frac{\log \left(\frac{1}{x}-1\right)}{\log x\log (1-x)}dx$
(By substituting $t=\frac{1}{x}dt=\frac{1}{x^{2}dx}\text{and}t=1\iff x=1\text{and}t=\infty \iff x=0$)
$\therefore I={\int }_0^1\frac{\log (1-x)-\log x}{\log x\log (1-x)}dx$
$={\int }_0^1\left(\frac{1}{\log x}-\frac{1}{\log (1-x)}\right)dx$
$={\int }_0^1\left(\frac{1}{\log (1-x)}-\frac{1}{\log x}\right)dx=-I\therefore I=0$
$\left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx\right]$