= Solution
The integrand is positive and <continuous> on the interval in question. Substitute $t=\log(1+x)$ to obtain
$$
\int_1^\infty f(x)\,dx=\int_{\log2}^\infty\frac{e^{(1-\alpha)t}}{t^2}\,dt.
$$
If $\alpha\geq1$, the exponential factor is at most one and the comparison integral of $t^{-2}$ converges. If $\alpha<1$, exponential growth dominates $t^2$, so the integrand eventually exceeds one and the integral diverges. Thus
$$
\boxed{\alpha\geq1}.
$$
The logarithmic square is what makes the boundary value $\alpha=1$ integrable.
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