TITLE: Show that the Beta Function $\beta (x,y)$ Converges When $x \gt 0, \space y \gt 0$
QUESTION [12 upvotes]: Given $ \beta (x, y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt$, Show that $\beta (x,y)$ converges when $x \gt 0, \space y \gt 0$.
$$\int_0^1 t^{x-1}(1-t)^{y-1} dt = \int_0^{0.5} t^{x-1}(1-t)^{y-1} dt + \int_{0.5}^1 t^{x-1}(1-t)^{y-1} dt$$
Now, according to the text, for the integral from $0$ to $1 \over 2$, $t^{x-1}(1-t)^{y-1} \le t^{x-1}$. 
$\int_0^{0.5} t^{x-1} dt \lt {\infty}$ entails that $\int_0^{0.5} t^{x-1}(1-t)^{y-1} dt$ converges.

Well, $t^{x-1}(1-t)^{y-1} \le t^{x-1}$ when $y \ge 1$, not for all $y \gt 0$.

Then the text does something similar for the integral from ${1 \over 2}$ to $1$: 
Since $t^{x-1}(1-t)^{y-1} \le (1-t)^{y-1}$, 
$\int_{0.5}^1 (1-t)^{y-1} dt \lt {\infty}$ entails that $\int_{0.5}^1 t^{x-1}(1-t)^{y-1} dt$ converges.

Well, $t^{x-1}(1-t)^{y-1} \le (1-t)^{y-1}$ when $x \ge 1$. 
So the text shows that $\beta (x,y)$ converges when $x, y \ge 1$.

Did I make a mistake somewhere or misunderstand something?

REPLY [7 votes]: It should say that if $y > 0$ there is a constant $c$ such that 
$t^{x-1}(1-t)^{y-1} \le c t^{x-1}$ for $0 < t < 1/2$.