Solution
= Solution
For fixed $x>0$, $x\ne1$,
$$
\int_0^1x^y\,dy=\frac{x-1}{\log x},
$$
with limiting value one at $x=1$. Integrating over $0<x<1$ in the opposite order gives
$$
\int_0^1\int_0^1x^y\,dy\,dx
=\int_0^1\frac{dy}{y+1}
=\log2.
$$
On the other hand, after setting $x=e^{-u}$,
$$
\int_0^1\frac{x-1}{\log x}\,dx
=\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du.
$$
Consequently
$$
\boxed{
\int_0^\infty\frac{e^{-u}-e^{-2u}}u\,du=\log2}.
$$