Question

\[\int\limits_0^\infty \frac{\log x}{1 + x^2} dx\]

Answer 1

\[\text{We have}, \]

\[I = \int\limits_0^\infty \frac{\log x}{1 + x^2} dx\]

\[Putting\ x = \tan \theta\]

\[ \Rightarrow dx = \sec^2 \theta d\theta\]

\[When\ x \to 0 ; \theta \to 0\]

\[and\ x \to \infty ; \theta \to \frac{\pi}{2}\]

\[\text{Now, integral becomes},\]

\[I = \int\limits_0^\frac{\pi}{2} \frac{\log \left( \tan \theta \right)}{1 + \tan^2 \theta} \sec^2 \theta d\theta\]
\[ \Rightarrow I = \int\limits_0^\frac{\pi}{2} \log \left( \tan \theta \right) d\theta ...............\left( 1 \right)\]
\[ \Rightarrow I = \int\limits_0^\frac{\pi}{2} \log\left[ \tan \left( \frac{\pi}{2} - \theta \right) \right] d\theta .................\left[ \because \int_0^a f\left( x \right)dx = \int_0^a f\left( a - x \right)dx \right]\]
\[ \Rightarrow I = \int\limits_0^\frac{\pi}{2} \log \left( \cot \theta \right) d\theta ..................\left( 2 \right)\]
\[\text{Adding} \left( 1 \right)and \left( 2 \right), \text{we get}\]

\[2I = \int\limits_0^\frac{\pi}{2} \log \left( \tan \theta \right) d\theta + \int\limits_0^\frac{\pi}{2} \log \left( \cot \theta \right) d\theta\]

\[ = \int\limits_0^\frac{\pi}{2} \left[ \log \left( \tan \theta \right) + \log \left( \cot \theta \right) \right] d\theta\]

\[ = \int\limits_0^\frac{\pi}{2} \left[ \log \left( \tan \theta \times \cot \theta \right) \right] d\theta\]

\[ = \int\limits_0^\frac{\pi}{2} \left( \log 1 \right) d\theta\]

\[ = \int\limits_0^\frac{\pi}{2} \left( 0 \right) d\theta\]

\[ \Rightarrow 2I = 0\]

\[ \Rightarrow I = 0\]

\[ \therefore \int\limits_0^\infty \frac{\log x}{1 + x^2} dx = 0\]