= Solution
Again the <function> is continuous on its domain and therefore a <measurable function>. For $0<x<1$, $|(\log x)e^{-x}|\leq-\log x$, and
$$
\int_0^1-\log x\,dx=1.
$$
This standard <improper integral> follows from the <antiderivative> $x-x\log x$ and $x\log x\to0$ at zero. For $x\geq1$, $0\leq\log x\leq x$, and <integration by parts> gives $\int_1^\infty xe^{-x}\,dx=2/e$. Thus
$$
\boxed{\int_0^\infty|(\log x)e^{-x}|\,dx\leq1+\frac2e<\infty.}
$$
The logarithmic singularity is integrable and the exponential factor controls the tail, so this too is a <Lebesgue integrable function>.
Back to article page
