Show that for $f(s,t)=(s^2+t)^{-a}$ the function $f^b$ is integrable over $(0,1)\times(0,T)$ This should be a rather simple question, but I wasn't able to figure out what's the trick to find a suitable upper bound: Given $a\in(1/2,3/2)$, $b\in[1,1/(a-1/2))$ and $T>0$, how can we show that for $f(s,t):=(s^2+t)^{-a}$, $s\in(0,1)$, $t\in(0,T)$, the function $f^b$ is integrable over $(0,1)\times(0,T)$?
 A: Here we assume $a,b>0$. For convergence purposes, it is enough to consider the integral over the cube $(0,1]^2$. By Fubini-Tonelli's theorem:

*

*If $ab\neq1$
$$\int^1_0\int^1_0\frac{1}{(s^2+t)^{ab}}\,dtds=\int^1_0\frac{1}{1-ab}\big((s^2+1)^{1-ab} - s^{2(1-ab)}\big)\,ds$$
The integral $$\int^1_0\frac{1}{s^{2(ab-1)}}\,ds$$ converges if $2(ab-1)<1$ and diverges if $2(ab-1)\geq1$. The integral $$\int^1_0(s^2+1)^{1-ab}\,ds$$ is convergent since $1\leq s^2 +1\leq 2$ for $0\leq s\leq 1$.

*If $ab=1$
$$\int^1_0\int^1_0\frac{1}{(s^2+t)}\,dtds=\int^1_0\log(s^2+1)-2\log(s)\,ds$$
The integral is convergent since $\int^1_0|\log s|\,ds =-(s\log(s)-s)|^1_0=1<\infty$.

Thus, the integral in question converges iff $2(ab-1)<1$, i. e., $ab<\tfrac32$.
Here is a plot of the function $\phi(a)=\frac{3}{2a}$ (in black), $\psi(a)=\frac{1}{a-1/2}$ (in blue) and $\xi(a)\equiv1$ (in red) for $\frac12 <a\leq \frac32$. Notice that convergence in the region of interest for the OP occurs between the black and the red plots, that is $\{(a,b): \frac12<a<\frac32,\, 1\leq b\leq \frac{3}{2a}\}$.

