Evaluate \[ \int_0^\infty\frac{\log x}{1+x^2}\,dx. \]
Solution
1. The improper integral converges absolutely.
On \((0,1]\), \[ \frac{|\log x|}{1+x^2} \leq |\log x|, \] and \[ \int_0^1|\log x|\,dx=1. \] On \([1,\infty)\), \[ \frac{\log x}{1+x^2} \leq \frac{\log x}{x^2}, \] whose integral converges by integration by parts. Hence the original integral is absolutely convergent.
2. One has\[ \int_1^\infty\frac{\log x}{1+x^2}\,dx = -\int_0^1\frac{\log t}{1+t^2}\,dt. \]
Use the substitution \[ x=\frac1t, \qquad dx=-\frac{dt}{t^2}. \] Then \[ \begin{aligned} \int_1^\infty\frac{\log x}{1+x^2}\,dx &= \int_1^0 \frac{-\log t}{1+t^{-2}} \left(-\frac{dt}{t^2}\right)\\ &= \int_1^0\frac{\log t}{1+t^2}\,dt\\ &= -\int_0^1\frac{\log t}{1+t^2}\,dt. \end{aligned} \]
3. Therefore\[ \boxed{ \int_0^\infty\frac{\log x}{1+x^2}\,dx=0. } \]
By absolute convergence from step 1, split at \(1\): \[ \int_0^\infty = \int_0^1+\int_1^\infty. \] Step 2 shows that the two contributions cancel.
4. Q.E.D.
Step 3 gives the requested value.