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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?

Andy Tam
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1 Answers1

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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 $0 < t < 1/2$.

Robert Israel
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  • Ok, TY. Although I lack the dexterity to figure it out, at least I wasn't wrong. – Andy Tam Jun 24 '13 at 00:52
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    @AndyTam Very good question Andy regarding the beta function, and I too fell for the trap of concluding that the integral of $\beta(x,y)$ converges when $x,y\geq 1$. Can anyone show the above result by Robert? – dandar Sep 26 '13 at 16:09