Let I = \(\int\limits^{\infty}_{0} \frac{logx}{1+x^2}\,dx\)
Let x = tanθ
⇒ dx = sec2θdθ
\(\therefore I = \int\limits _{0}^{\frac{\pi}2} \frac{log(tan\theta )\,sec^2\theta d\theta}{1+ tan^2\theta}\)
\(= \int\limits _{0}^{\frac{\pi}2} log(tan\theta )d\theta\) ......(i) (∵ 1 + tan2θ = sec2θ)
\(\therefore I = \int\limits _{0}^{\frac{\pi}2} log(tan(\frac{\pi}2-\theta))d\theta\)
\(= \int\limits _{0}^{\frac{\pi}2} log(cot \theta )d\theta\) ......(ii)
On adding (i) & (ii), we got
\(2 I = \int\limits _{0}^{\frac{\pi}2} (log\,tan\theta + log\,cot\theta )d\theta\)
\(= \int\limits _{0}^{\frac{\pi}2} log(tan\theta -cot \theta )d\theta\)
\(= \int\limits _{0}^{\frac{\pi}2} log1d\theta =0\) (∵ log 1 = 0)
\(\therefore I = 0\)
⇒ \(\int\limits^{\infty}_{0} \frac{logx}{1+x^2}\,dx = 0\)