Solution
= Solution
Set $w=i$, which satisfies $\Re w=0$, and take <real parts> of the absolutely convergent integral. Since $x^i=\cos(\log x)+i\sin(\log x)$ and $\cos(i\pi/2)=\cosh(\pi/2)$,
$$
\boxed{\int_0^\infty\frac{\cos(\log x)}{1+x^2}\,dx=\frac\pi{2\cosh(\pi/2)}.}
$$
Absolute integrability justifies moving the real-part operation through the integral.