Question

$\underset{0}{\overset{\infty }{\int }}\frac{\log x}{1+x^2}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\underset{0}{\overset{\infty }{\int }}\frac{\log x}{1+x^2}dx \\ \mathrm{Putting}x=\tan \theta \\ \Rightarrow dx={\sec }^2\theta d\theta \\ \mathrm{When}x\to 0;\theta \to 0 \\ \mathrm{and}x\to \infty ;\theta \to \frac{\mathrm{\pi }}{2} \\ \mathrm{Now},\mathrm{integral}\mathrm{becomes}, \end{gathered}$$

$$\begin{gathered} I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\frac{\log \left(\tan \theta \right)}{1+{\tan }^2\theta }{\sec }^2\theta d\theta \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\log \left(\tan \theta \right)d\theta .....\left(1\right) \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\log \left[\tan \left(\frac{\mathrm{\pi }}{2}-\theta \right)\right]d\theta \left[\because {\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f\left(a-x\right)dx\right] \\ \Rightarrow I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\log \left(\cot \theta \right)d\theta .....\left(2\right) \\ \mathrm{Adding}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get} \end{gathered}$$

$$\begin{gathered} 2I=\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\log \left(\tan \mathit{}\theta \right)d\theta +\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\log \left(\cot \theta \right)d\theta \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left[\log \left(\tan \theta \right)+\log \left(\cot \theta \right)\right]d\theta \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left[\log \left(\tan \theta \times \cot \mathit{}\theta \right)\right]\mathit{}d\theta \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(\log 1\right)\mathrm{d}\mathrm{\theta } \\ =\underset{0}{\overset{\frac{\mathrm{\pi }}{2}}{\int }}\left(0\right)\mathrm{d}\mathrm{\theta } \\ \Rightarrow 2I=0 \\ \Rightarrow I=0 \\ \therefore \underset{0}{\overset{\infty }{\int }}\frac{\log \mathit{}x}{1+x^{\mathit{2}}}dx=0 \end{gathered}$$