Question

Solve: $\stackrel{\infty }{\underset{0}{\int }}\frac{logx}{1+x^2}dx$

Answer 1

$I={\int }_0^{\infty }\frac{\log x}{1+x^2}dx$

Let $x=\frac{1}{t}$

$dx=\frac{-1}{t^2}dt$

$I={\int }_{infty}^0\frac{\log t^{-1}}{1+{\left(\frac{1}{t}\right)}^2}\left(\frac{-1}{t^2}\right)dt$

$I={\int }_{\infty }^0\frac{-\log t}{t^2+1}\left(\frac{-t^2}{t^2}\right)dt$

$I={\int }_{\infty }^0\frac{\log t}{1+t^2}dt$

$I=-{\int }_0^{\infty }\frac{\log t}{1+t^2}dt$

$I=-I$

$2I=0$

$I=0$

$\therefore {\int }_0^{\infty }\frac{\log x}{1+x^2}dx=0$.