Let $\{\alpha_n\}_{n=1}^\infty$ be a sequence of positive real numbers such that $\sum_{n=1}^\infty \alpha_n<\infty$. In particular, $\lim_n \alpha_n=0$.
Let $\{\varepsilon_n\}_{n=1}^{\infty}\subset (0,1)$ be a decreasing sequence of numbers such that $\lim_n \varepsilon_n = 0$. Define the function
$$f(x) := \sum_{n=1}^{\infty} \alpha_n x^{\varepsilon_n}.$$
The series above converges uniformy on compacts of $\mathbb{R}$.
Consider the integral on an interval $[a,b]\subset [0,\infty)$:
$$\int_a^b f(x)^\beta dx = \int_a^b \left( \sum_{n=1}^{\infty} \alpha_n x^{\varepsilon_n} \right)^\beta dx.$$
I claim that the above integral converges for all $\beta\in \mathbb{R}$.
1) If $\beta\geq 0$ then $f$ is continuous and hence integrable on the compact $[a,b]$.
2) If $\beta:= -\gamma$, $\gamma>0$ then we may use the estimate $$\left( \sum_{n=1}^{\infty} \alpha_n x^{\varepsilon_n}\right)^{-\gamma} \leq (\alpha_{n_0})^{-\gamma} x^{-\gamma \varepsilon_{n_0}}$$ for any $n_0\geq 1$. Hence, $$\int_a^b \left( \sum_{n=1}^{\infty} \alpha_n x^{\varepsilon_n} \right)^{-\gamma} dx\leq \alpha_{n_0}^{-\gamma} \int_a^b x^{-\gamma \varepsilon_{n_0}} dx = \alpha_{n_0}^{-\gamma} \frac{b^{1-\gamma \varepsilon_{n_0}}- a^{1-\gamma \varepsilon_{n_0}}}{1-\gamma \varepsilon_{n_0}},$$ where we choose $n_0\geq 1$ so that $1-\gamma \varepsilon_{n_0}>0$.
Finally taking limit on the quotient we have
$$\int_a^b \left( \sum_{n=1}^{\infty} \alpha_n x^{\varepsilon_n} \right)^{-\gamma} dx\leq \alpha_{n_0}^{-\gamma} (b- a).$$
Are all these arguments correct and is it true that this integral converges always? Of course, as $n_0$ grows, then $\alpha_{n_0}^{-\gamma}$ gets larger, but $n_0$ is fixed.
Thanks a lot for you help! :)
I think you've done a good job with the main ideas. Everything seems to be essentially correct. I would change a few things, add a few things.
Since $x^{\epsilon_n}$ is problematic if $x<0,$ I would restrict the domain to $[0,\infty).$ Then we can say that for $x\in [0,b]$ and $n\in \mathbb N,$
$$|\alpha_nx^{\epsilon_n}| = \alpha_nx^{\epsilon_n}\le \alpha_n+ \alpha_nx^{\epsilon_0} \le \alpha_n+ \alpha_nb^{\epsilon_0}=\alpha_n(1+b^{\epsilon_0}).$$
Because $\sum_n \alpha_n <\infty,$ the series defining $f$ converges uniformly on $[0,b]$ by the Weierstrass M test. Each summand $\alpha_nx^{\epsilon_n}$ is continuous on $[0,b],$ and therefore so is $f.$ Because $b$ is arbitrary, $f$ is continuous on $[0,\infty).$ Note that $f(0)=0$ and $f>0$ on $(0,\infty).$
As for convergence of the integrals $\int_a^bf^\beta,$ note there is nothing to prove if $a>0,$ since $f^\beta$ is continuous on such an interval for any $\beta \in \mathbb R.$ The crux of the matter are the integrals $\int_0^1f^\beta.$ If $\beta \ge 0,$ then $f^\beta$ is continuous on $[0,1]$ and there's no problem. So assume $\beta < 0.$ We'll save work by recalling that $\int_0^1 x^p\,dx$ converges iff $p>-1.$ Now as you observed, we can choose $n_0$ such that $-|\beta|\epsilon_{n_0}>-1.$ Because $f(x) > \alpha_{n_0}x^{\epsilon_{n_0}},$ we have
$$f^\beta(x) = f^{-|\beta|}(x) < (\alpha_{n_0}x^{\epsilon_{n_0}})^{-|\beta|} = \alpha_{n_0}^{-|\beta|}x^{-|\beta|\epsilon_{n_0}}.$$
Because $-|\beta|\epsilon_{n_0}>-1,$ $\int_0^1 x^{-|\beta|\epsilon_{n_0}}\,dx$ converges and we're done.