= Solution
For $0<\varepsilon<e^{-1}$, set $u=\log(1/x)$. The ordinary <change of variables formula> gives
$$
\int_\varepsilon^{e^{-1}}
\frac{dx}{x(\log(1/x))^\theta}
=\int_1^{\log(1/\varepsilon)}u^{-\theta}\,du.
$$
Taking $\varepsilon\downarrow0$ is exactly the defining <limit> of the <improper integral>. By the <improper power integral>, it converges for $\theta>1$, and
$$
\boxed{\int_0^{e^{-1}}
\frac{dx}{x(\log(1/x))^\theta}
=\int_1^\infty u^{-\theta}\,du
=\frac1{\theta-1}}.
$$
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