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$\displaystyle\int_0^{\infty} \dfrac{\ln x}{1 + x^2}\, dx = ?$
ADiverges
B$0$
C$\pi/4 \cdot \ln 2$
D$\ln(\pi/2)$
Answer & 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 = $\int_0^1 \ln x/(1+x^2)dx + \int_0^1 -\ln u/(1+u^2)du = 0$.
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