The Euler Beta function is defined by $$B(\alpha, \beta) := \int_0^1 (1-t)^{\alpha -1} t^{\beta -1}dt$$ for $\alpha, \beta \in \mathbb C$ with $\Re(\alpha), \Re(\beta) >0$. I don't see how this is well defined. For example, take $\alpha , \beta = 1/2$. We then have the integrand $(1-t)^{-1/2} t^{-1/2}$ which is not bounded for $t\to0$ nor for $t\to 1$.
Where is my mistake?
The integrand needn't be bounded. Note that:
$$0<t<1/2\implies0<(1-t)^{-1/2}t^{-1/2}<\sqrt2t^{-1/2}$$
And further,
$$\underbrace{\int_0^{1/2}t^{-1/2}~\mathrm dt=2t^{1/2}\bigg|_0^{1/2}=\frac1{\sqrt2}}_{\text{Unbounded integrand, but the integral is finite}}$$
Thus, we find that
$$0<\int_0^{1/2}(1-t)^{-1/2}t^{-1/2}~\mathrm dt<1$$
And by symmetry,
$$0<B(1/2,1/2)<2$$