$\displaystyle\int_0^{\infty} \dfrac{\ln x}{1 + x^2}\, dx = ?$
- Diverges
- $0$
- $\pi/4 \cdot \ln 2$
- $\ln(\pi/2)$
Show answer and solution
Correct answer: B. $0$
Split at $x = 1$ and substitute $x = 1/u$ in the part from $1$ to $\infty$: $\int_1^\infty \ln x/(1+x^2)\,dx = \int_0^1 -\ln u/(1+u^2)\,du$.
**So total = $\int_0^1 \ln x/(1+x^2)dx + \int_0^1 -\ln u/(1+u^2)du = 0$.**