Absolite convergence of the beta function

[Ted123]

Determine the values of the complex parameters $p$ and $q$ for which the beta function $$\int_0^1 t^{p-1} (1-t)^{q-1}dt$$ converges absolutely.

The solution says:

Split the integral into 2 parts: $$\displaystyle \int_0^1 t^{p-1} (1-t)^{q-1}dt = \int_0^{1/2} t^{p-1} (1-t)^{q-1}dt + \int_{1/2}^1 t^{p-1} (1-t)^{q-1}dt$$

As $t \to 0$ $$|t^{p-1} (1-t)^{q-1}| \sim |t^{p-1}| \Rightarrow \text{Re}(p)>0$$ As $t\to 1$ $$|t^{p-1} (1-t)^{q-1}| \sim |(1-t)^{q-1}| \Rightarrow \text{Re}(q)>0$$

Why are these implications true?

 

[sunjin09]

Determine the values of the complex parameters $p$ and $q$ for which the beta function $$\int_0^1 t^{p-1} (1-t)^{q-1}dt$$ converges absolutely.

The solution says:

Split the integral into 2 parts: $$\displaystyle \int_0^1 t^{p-1} (1-t)^{q-1}dt = \int_0^{1/2} t^{p-1} (1-t)^{q-1}dt + \int_{1/2}^1 t^{p-1} (1-t)^{q-1}dt$$

As $t \to 0$ $$|t^{p-1} (1-t)^{q-1}| \sim |t^{p-1}| \Rightarrow \text{Re}(p)>0$$ As $t\to 1$ $$|t^{p-1} (1-t)^{q-1}| \sim |(1-t)^{q-1}| \Rightarrow \text{Re}(q)>0$$

Why are these implications true?

because otherwise a neighborhood of the endpoints is non-integrable?