$\int^{\infty}_{0} \frac{e^{-\alpha x}}{x^{\beta}} dx < \infty$, where $\alpha > 0$ and $ 0< \beta < 1$. $\int^{\infty}_{0} \frac{e^{-\alpha x}}{x^{\beta}} dx \in L^{1}(0, \infty)$, where $\alpha > 0$ and $ 0< \beta < 1$.
My idea is to use the fact that
$$
\int^{\infty}_{0} \frac{e^{-\alpha x}}{x^{\beta}} dx = \int^{1}_{0} \frac{e^{-\alpha x}}{x^{\beta}} dx + \int^{\infty}_{1} \frac{e^{-\alpha x}}{x^{\beta}} dx
$$
I know
$$
\int^{\infty}_{1} \frac{e^{-\alpha x}}{x^{\beta}} dx < \int^{\infty}_{1} e^{-\alpha x} dx < \infty
$$
My problem is in showing that
$$
\int^{1}_{0} \frac{e^{-\alpha x}}{x^{\beta}} dx < \infty.
$$

You have to use that $\beta\in(0,1)$: If you integrate by parts once, the $x^\beta$ disappears from your denominator:
$$\int\limits_0^1 \frac{e^{-\alpha x}}{x^\beta}dx = \frac{1}{1-\beta}\left([x^{1-\beta}e^{-\alpha x}]_0^1 +\alpha \int\limits_0^1 x^{1-\beta}e^{-\alpha x}dx\right)$$
