= Solution
For $\alpha>0$, <integration by parts> on $[0,R]$ gives
$$
\int_0^R xe^{-\alpha x}\,dx
=\frac1{\alpha^2}-e^{-\alpha R}\left(\frac R\alpha+\frac1{\alpha^2}\right).
$$
The exponential remainder tends to zero. Since the integrand is nonnegative, the <monotone convergence theorem> applied to truncations identifies the improper <integral> with the Lebesgue <integral>. Therefore
$$
\boxed{\int_0^\infty xe^{-\alpha x}\,dx=\alpha^{-2}.}
$$
Its finite value proves integrability as well as evaluating it.
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