Let $f: \mathbb R_{\geq 0} \to \mathbb R$ measurable such that there is $\alpha \in (0,1)$ with $|f(t)| \leq \dfrac{t^{\alpha}}{1+t}$ for all $t \geq 0$. Let $G: \mathbb R_{\geq 0} \times \mathbb R_{\geq 0} \to \mathbb R$ be defined as $$G(x,t)=e^{-xt}f(t)$$ Prove that $G$ is integrable.
It is sufficient to show that $|G|$ is integrable. By Fubini-Tonelli's theorem, we have $$\int_{(\mathbb R_{\geq 0})^2}|G(x,t)|=\int_{ \mathbb R_{\geq 0}}(\int_{ \mathbb R_{\geq 0}}|f(t)|e^{-xt}dx)dt$$ $$=\int_{ \mathbb R_{\geq 0}}|f(t)|(\int_{ \mathbb R_{\geq 0}} e^{-xt}dx)dt \space(*)$$
If for each $t>0$ I define $f_{n,t}(x)=e^{-xt}\chi _{[0,n]}$, it is clear that $\lim_n f_{n,t}(x)=f(x)$ with $(f_{n,t})_{n}$ an increasing sequence.
The Riemann-Integral $$\int_0^n f_{n,t}(x)dx=\dfrac{1}{t}-\dfrac{e^{-nt}}{t},$$ which coincides with the Lebesgue Integral. So by this and by the monotone convergence theorem, the last expression becomes $$(*)=\int_{\mathbb R_{\geq 0}}|f(t)|\dfrac{1}{t}dt\leq \int_{\mathbb R_{\geq 0}}\dfrac{t^{\alpha}}{1+t}\dfrac{1}{t}dt$$
I would like to show that this last integral is finite, but I don't know how to prove this. I would appreciate if someone could help me to complete the answer and to correct me if necessary.
So consider the integrals $\int_0^1 t^{- \alpha} \textrm{d}t $ if $\alpha \in (0,1)$ and $\int_1^\infty t^{-\beta} \textrm{d}t $ with $ \beta \in (1 ,2)$.
Then we must bound $\frac{1}{1+t}$ by $1$ on $t \in [0,1]$ and by $\frac{1}{t}$ for $t \in [1, \infty)$.