Find the value of $$\int_{[0,\infty)}\biggl(\int_{[0,\infty)}2x\sqrt{y}e^{-x^2\sqrt{y}-y}dy\biggl)dx$$
My attept:
Let $f(x,y)=2x\sqrt{y}e^{-x^2\sqrt{y}-y}$ to apply Tonelli theorem I did the following
$$|f(x,y)|\leq 2x\sqrt{y}$$ But $\int_0^{\infty}\int_0^{\infty} 2x\sqrt{y}dxdy$ is not finite. What function $g$ should I use to bound $f(x,y)$ such that $\int_0^{\infty}\int_0^{\infty}<\infty$?
You ask about a bound. I think in this case, when you want to calculate the integral, you should not try to find a bound. Instead, argue as follows:
Call your integrand $f$. We note that $f$ is measurable since it is continuous. By the Tonelli theorem, to guarantee that $f$ is integrable over $[0,+\infty)^2$, it suffices to check that one of $$ \int_0^{+\infty}\Bigl(\int_0^{+\infty}|f(x,y)|\,dx\Bigr)\,dy \quad \text{and} \quad \int_0^{+\infty}\Bigl(\int_0^{+\infty}|f(x,y)|\,dy\Bigr)\,dx\tag{$*$} $$ exists. In your case the first one is in fact easily calculated (note that $f\geq 0$), as noticed in the answer by @felix-marin. Hence $f$ is integrable.
Next, by the Fubini theorem, since $f$ is integrable, the integrals in $(*)$ above are equal, and the value of the integral you want to calculate is $1$, as @felix-marin notice.
Finally, I cannot resist to suggest the book Introduction to integration by Hilary Priestley. I wish I would have had it by my side when I first learned the theory of integration.