Show that $\beta_{x,y}:(0,1) \to \mathbb{C}, \beta_{x,y}(t)=(1-t)^{x-1}t^{y-1}$ is integrable

Question

Let $\mathbb{C}_{Re>0}=\{x \in \mathbb{C} \space | \space Re(x)>0 \}$.

For all $x,y \in \mathbb{C}_{Re>0}$ the function $\beta_{x,y}:(0,1) \to \mathbb{C}, \beta_{x,y}(t)=(1-t)^{x-1}t^{y-1}$ is integrable.

How do I show this?

Answer 1

First, note that for all $x,a,b\in\mathbb{R}:x^{a+ib} = x^ae^{ib\ln(x)}$.

Now we can use this because $$\left|\int_0^1(1-t)^{x-1}t^{y-1}dt \right|\leq\int_0^1|(1-t)^{x-1}t^{y-1}|dt,$$

and for all $t\in\mathbb{R}:|t^x|=|t^a|$, where $x=a+ib$. Therefore it is sufficient to show that the function $$f:t\mapsto (1-t)^{\alpha-1}t^{\beta-1}$$ is integrable on $(0,1)$ when $\alpha,\beta>0$.

To do this, first note that $f$ is bounded on any closed interval $[a,b]\subseteq (0,1)$ because $f$ is continuous. Furthermore, note that $$\lim_{t\to 0}\dfrac{f(t)}{t^{\beta-1}}=1 \hspace{0.5cm} \text{and} \hspace{0.5cm} \lim_{t\to 1}\dfrac{f(t)}{(1-t)^{\alpha-1}}=1$$ so $$f(t) = \Theta\left(\frac1{t^{1-\beta}}\right)\hspace{0.5cm}\text{when}\hspace{0.5cm} x\to 0$$ and $$f(t) = \Theta\left(\frac1{(1-t)^{1-\alpha}}\right)\hspace{0.5cm}\text{when}\hspace{0.5cm} x\to 1.$$ So, since $\alpha,\beta>0$, $f$ is integrable on $(0,1)$.