Let $p$ be a real number. I think: $$q\in \mathbb R^+ \quad \text{iff} \quad \int_0^{\infty} \frac{x^p}{e^{qx}} dx<+\infty$$ because $\frac{x^p}{e^{qx}}$ converges to $0$ very quickly when $q>0$.
Is this right? If so, how can I prove? I appreciate any help, thank you!
On
Near $0$
the integrand function is equivalent to
$$x^p $$ since $e^{qx}$ goes to $1$. thus it converges iff $-p <1$.
Near $+\infty $.
if $q <0$, the integrand goes to $+\infty $ and the integral diverges.
If $q=0$, it diverges cause $-p <1$
If $q>0$, we have
$$\lim_{+\infty}x^{\color {red}{2}}x^pe^{-qx}=0$$
because as you said, the exponential is Faster. thus for $x $ great enough,
$$x^{\color {red}{2}}x^pe^{-qx}<\color {green}{1}$$ and $$\frac {x^p}{e^{qx}}<\frac{\color {green}{1}}{x^{\color {red}{2}} }$$
the integral converges by comparison.
Your generalized integral converges if only if $p>-1$ and $q>0$.
As soon as $p>-1$ and $q>0$ we have $$ \int_{0}^{+\infty}x^p e^{-qx}\,dx = \frac{p!}{q^{p+1}} $$ by the integral definition of the $\Gamma$ function.