Let $\alpha,\beta\in\mathbb{R}$, with $\alpha>-1$. I have shown that the function $(0,1]\to\mathbb{R}:y\mapsto y^\alpha(x+y)^\beta$ is integrable for $x>0$.
We now define $$F:(0,+\infty)\to\mathbb{R}:x\mapsto F(x)=\int_0^1y^\alpha(x+y)^\beta dy.$$
I am trying to show the following:
If $\alpha+\beta>-1$, then $F(x) = \Theta(1)$ as $x\to 0^+$.
$\textbf{My attempt (which I think is wrong)}$: Since $0<y\leq 1$ we have
\begin{align} 0<\lim_{x\to0^+}\int_0^1y^\alpha(x+y)^\beta dy&\leq \lim_{x\to0^+}\int_0^1(x+1)^\beta dy\\ &\leq\lim_{x\to0^+}(x+1)^\beta=1 \end{align}
By the following:
If $h>0$ and $$\lim_{x\to x_0}\dfrac{||f(x)||}{h(x)}\in(0,+\infty),$$ then $f(x)=\Theta(h(x))$ as $x\to x_0$.
Then $F(x) = \Theta(1)$ as $x\to0^+$.
I think the method I use must be wrong because I never make use of the given fact that $\alpha+\beta >-1$. But I don't see where I am wrong... Note that in my inequalities above I make use of the fact that $$\int_A fd\lambda\leq\int_A\left(\sup_A f\right) d\lambda=\sup_A f\int_Ad\lambda.$$ Maybe this is not always true?
I see my mistake now. Indeed, it is true that $$0<\int_0^1y^\alpha(x+y)^\beta dy\leq\int_0^1(x+y)^\beta dy,$$ since $0<y^\alpha\leq1$. But it is not known whether $$(x+y)^\beta\leq (x+1)^\beta$$ is true.