Show that the Beta function B(p, q) = integral from 0

Show that the Beta function $$ B(p, q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx $$ converges for $p > 0$ and $q > 0$.

The Beta function is potentially improper at both endpoints $x = 0$ and $x = 1$. We must analyze convergence at both points.

Step 1: Split the integral

We split at $x = \frac{1}{2}$ to separate the potential singularities:

$$B(p, q) = \int_0^{1/2} x^{p-1}(1-x)^{q-1} \, dx + \int_{1/2}^1 x^{p-1}(1-x)^{q-1} \, dx$$

For convergence, BOTH integrals must converge.

Step 2: Analyze convergence near $x = 0$

Consider $\int_0^{1/2} x^{p-1}(1-x)^{q-1} \, dx$.

As $x \to 0^+$:

• $x^{p-1} \to \begin{cases} 0 & \text{if } p > 1 \\ 1 & \text{if } p = 1 \\ +\infty & \text{if } p < 1 \end{cases}$
• $(1-x)^{q-1} \to 1^{q-1} = 1$ (bounded)

Therefore, near $x = 0$:

$$x^{p-1}(1-x)^{q-1} \approx x^{p-1} \cdot 1 = x^{p-1}$$

Convergence criterion at $x = 0$:

The integral $\int_0^{1/2} x^{p-1} \, dx$ is of the form $\int_0^a x^\alpha \, dx$.

We know this converges if and only if $\alpha > -1$.

In our case, $\alpha = p - 1$, so we need:

$$p - 1 > -1 \implies p > 0$$

Therefore: The integral converges at $x = 0$ if and only if **$p > 0$**. ✓

Step 3: Analyze convergence near $x = 1$

Consider $\int_{1/2}^1 x^{p-1}(1-x)^{q-1} \, dx$.

As $x \to 1^-$:

• $x^{p-1} \to 1^{p-1} = 1$ (bounded)
• $(1-x)^{q-1} \to 0^{q-1} = \begin{cases} 0 & \text{if } q > 1 \\ 1 & \text{if } q = 1 \\ +\infty & \text{if } q < 1 \end{cases}$

Therefore, near $x = 1$:

$$x^{p-1}(1-x)^{q-1} \approx 1 \cdot (1-x)^{q-1} = (1-x)^{q-1}$$

Let $u = 1 - x$, so as $x \to 1^-$, we have $u \to 0^+$.

The integral becomes:

$$\int_{1/2}^1 (1-x)^{q-1} \, dx \sim \int_0^{1/2} u^{q-1} \, du$$

Convergence criterion at $x = 1$:

This is of the form $\int_0^b u^\beta \, du$ with $\beta = q - 1$.

This converges if and only if $\beta > -1$, i.e.:

$$q - 1 > -1 \implies q > 0$$

Therefore: The integral converges at $x = 1$ if and only if **$q > 0$**. ✓

Step 4: Conclusion

Combining both conditions:

• Convergence at $x = 0$: requires $p > 0$
• Convergence at $x = 1$: requires $q > 0$

Convergence of Beta Function

The Beta function

$$B(p, q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx$$

converges if and only if $p > 0$ and $q > 0$.

$$\boxed{\text{For } p > 0 \text{ and } q > 0, \text{ the Beta function converges}}$$

Rigorous proof using comparison

Near $x = 0$ (for $x \in (0, 1/2]$):

Since $(1-x)^{q-1}$ is bounded on $[0, 1/2]$, there exists $M_1 > 0$ such that:

$$0 \leq (1-x)^{q-1} \leq M_1 \quad \text{for } x \in [0, 1/2]$$

Therefore:

$$0 \leq x^{p-1}(1-x)^{q-1} \leq M_1 x^{p-1} \quad \text{for } x \in (0, 1/2]$$

Since $\int_0^{1/2} x^{p-1} \, dx$ converges for $p > 0$, by the comparison test:

$$\int_0^{1/2} x^{p-1}(1-x)^{q-1} \, dx \quad \text{converges for } p > 0$$

Near $x = 1$ (for $x \in [1/2, 1)$):

Since $x^{p-1}$ is bounded on $[1/2, 1]$, there exists $M_2 > 0$ such that:

$$0 \leq x^{p-1} \leq M_2 \quad \text{for } x \in [1/2, 1]$$

Therefore:

$$0 \leq x^{p-1}(1-x)^{q-1} \leq M_2 (1-x)^{q-1} \quad \text{for } x \in [1/2, 1)$$

Since $\int_{1/2}^1 (1-x)^{q-1} \, dx$ converges for $q > 0$, by the comparison test:

$$\int_{1/2}^1 x^{p-1}(1-x)^{q-1} \, dx \quad \text{converges for } q > 0$$

Special cases

Case 1: $p = q = 1$

$$B(1, 1) = \int_0^1 x^0(1-x)^0 \, dx = \int_0^1 1 \, dx = 1$$

Case 2: $p = 2, q = 1$

$$B(2, 1) = \int_0^1 x \, dx = \frac{1}{2}$$

Case 3: $p = 1, q = 2$

$$B(1, 2) = \int_0^1 (1-x) \, dx = \frac{1}{2}$$

Case 4: $p = q = 2$

$$B(2, 2) = \int_0^1 x(1-x) \, dx = \int_0^1 (x - x^2) \, dx = \frac{1}{2} - \frac{1}{3} = \frac{1}{6}$$

Symmetry property

The Beta function is symmetric:

$$B(p, q) = B(q, p)$$

Proof: Let $u = 1 - x$, so $du = -dx$.

When $x = 0$: $u = 1$; when $x = 1$: $u = 0$.

$$B(p, q) = \int_0^1 x^{p-1}(1-x)^{q-1} \, dx = \int_1^0 (1-u)^{p-1}u^{q-1} \cdot (-du)$$

$$= \int_0^1 u^{q-1}(1-u)^{p-1} \, du = B(q, p)$$

Relationship to Gamma function

The Beta function is related to the Gamma function by:

$$B(p, q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$

where $\Gamma(p) = \int_0^\infty t^{p-1}e^{-t} \, dt$ is the Gamma function.

This provides an alternative way to compute Beta function values.

Applications

The Beta function appears in:

• Probability theory: Beta distribution
• Statistics: Bayesian analysis
• Combinatorics: Related to binomial coefficients
• Physics: Various integrals in quantum mechanics
• Special functions: Hypergeometric functions

Summary

1. The Beta function has potential singularities at $x = 0$ and $x = 1$
2. Near $x = 0$: integrand $\sim x^{p-1}$, converges if $p > 0$
3. Near $x = 1$: integrand $\sim (1-x)^{q-1}$, converges if $q > 0$
4. Both conditions must hold for convergence
5. The function is well-defined for all $p > 0$ and $q > 0$
6. The Beta function is symmetric: $B(p,q) = B(q,p)$

Key lessons:

• Integrals on $[0,1]$ may have singularities at both endpoints
• Each endpoint must be analyzed independently
• The comparison test with power functions is essential
• Boundedness of one factor allows comparison with the other
• The conditions $p > 0$ and $q > 0$ are both necessary and sufficient
• The Beta function is a fundamental special function in mathematics