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**Laurent** The only fact to use is that $\displaystyle \int_0^1 t^\alpha dt$ is convergent if, and only if $\displaystyle \alpha>-1$.

Indeed, as you said, the problem lies at endpoints, and we can split the integral in two parts.

In the first part, for $\displaystyle 0\leq t\leq \frac{1}{2}$, we have $\displaystyle 0\leq t^{m-1}(1-t)^{n-1} \leq t^{m-1}$, hence it converges if $\displaystyle m-1>-1$ (i.e. if $\displaystyle m>0$). On the other hand, we have, for the same $\displaystyle 0\leq t \leq \frac{1}{2}$, $\displaystyle 0<t^{m-1}\left(\frac{1}{2}\right)^{n-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1>0$ and $\displaystyle 0<t^{m-1}\leq t^{m-1}(1-t)^{n-1}$ if $\displaystyle n-1\leq 0$, and in both cases this shows that the integral diverges if $\displaystyle m-1\leq -1$. Hence this part converges iff $\displaystyle m>0$.

You can do the same on the interval $\displaystyle \left[\frac{1}{2},1\right]$ by letting $\displaystyle u=1-t$. This part converges iff $\displaystyle n>0$.

Finally, the integral converges iff $\displaystyle m>0$ and $\displaystyle n>0$.