= Solution
Use the <complex exponential> representation of <cosine>. For real $\alpha$ and $\beta>0$, exponential decay makes the <improper integral> absolutely convergent, and
$$
I(\alpha,\beta)=\operatorname{Re}\int_0^\infty e^{-(\beta-i\alpha)x}\,dx=\operatorname{Re}\frac1{\beta-i\alpha}=\boxed{\frac\beta{\beta^2+\alpha^2}}.
$$
For the second <integral>, use <differentiation under the integral sign> with respect to $\beta$. The minus sign is essential: $\partial_\beta e^{-\beta x}=-xe^{-\beta x}$. Hence
$$
\boxed{\int_0^\infty x\cos(\alpha x)e^{-\beta x}\,dx=-\frac{\partial I}{\partial\beta}=\frac{\beta^2-\alpha^2}{(\beta^2+\alpha^2)^2}.}
$$
To justify this step near any fixed $\beta_0>0$, restrict $\beta\geq\beta_0/2$. The absolute value of the differentiated integrand is at most $xe^{-\beta_0x/2}$, an integrable bound, so <dominated convergence> permits the differentiation. At $\alpha=0$ the answers reduce to $1/\beta$ and $1/\beta^2$, respectively.
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