 Solution

For positive bounds, integration by parts with $u=\log x$ and $dv=x^{2}dx$ givesFor negative bounds, specify the complex logarithm on the negative real axis. With argument $\pi$, namely the upper-bank boundary value of the principal logarithm, $\log x=\log (-x)+i\pi$. Substitute $t=-x$ in the real part:More generally, a continuous branch taking argument $(2k+1)\pi$ replaces $i\pi$ by $(2k+1)i\pi$. The imaginary part depends on that choice. The analytic principal branch excludes the negative axis itself, so the boundary-value convention is part of this interpretation.