0

Why is the Beta function only defined for x,y > 0 ? $$B(x,y)=\int_0^1t^{x−1}(1−t)^{y−1}dt $$

SAJ
  • 11
  • 2
    Because otherwise you would have singularities at the endpoints that are not integrable. Why are you asking basically the same question as before? (https://math.stackexchange.com/q/4054950) – Gary Mar 09 '21 at 14:27
  • Actually it's defined for $\text{Re}(x), \text{Re}(y) > 0$. – Robert Israel Mar 09 '21 at 15:51
  • ... and by analytic continuation (or using the expression in terms of the Gamma function) for all complex $x,y$ except nonpositive integers. – Robert Israel Mar 09 '21 at 15:58

3 Answers3

1

set them $=0$ and verify that

$$B(0;0)= \int_0^1 \frac{1}{t(1-t)}dt$$

diverges

The same happens for any values of $a,b\leq 0$

tommik
  • 33,201
  • 4
  • 17
  • 35
0

You have

$$ B(x,y)\geq\int_0^{1/2}t^{x−1}(1−t)^{y−1}dt \geq \frac{1}{2^{y-1}} \int_0^{1/2}t^{x−1} \, dt. $$

Now the last integral is finite if and only if $x>0$. Similarly for $y$.

Suman Chakraborty
  • 3,014
0

Just to add to what @tommik and @Suman Chakraborty have written above.

Beta function $B(x,y)$ can be analytically continued to all complex numbers $x,y$, including negative ones, but in this case it is not presented by $\int_0^1t^{x−1}(1−t)^{y−1}dt$.

Analytical continuation of Beta function for all $x,y$ has the form $$B(x,y)=\frac{1}{(1-e^{2 \pi i{x}} )(1-e^{2 \pi i{y}} )}\int_P t^{x-1} (1-t)^{y-1} {\rm d}x$$

where the integration goes along the Pochhammer contour (for example, look here: Confused about Pochhammer contour?)

It can be shown that analytically continued Beta function is related to the analytically continued Gamma function in its usual way: $B(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

For $x,y>0$ both views on Beta-function are identical.

It can be show, for example, through integration along Pochhammer contour that for $x\in(-1,0)$ and $y>0$ $B(x,y)=\lim_{r\to0}(\int_r^1t^{x-1}(1-t)^{y-1}dt+\frac{r^x}{x})$

These representations can be useful for integrals evaluation.

Svyatoslav
  • 20,502