Problem 82 - Calculus Early Transcendentals: Pearson New International Edition

Question

The Beta function, which is important in many branches of mathematics, is defined as $$ B(\alpha, \beta)=\int_{0}^{1} x^{\alpha-1}(1-x)^{\beta-1} d x $$ with the condition that \(\alpha \geq 1\) and \(\beta \geq 1\). (a) Show by a change of variables that $$ B(\alpha, \beta)=\int_{0}^{1} x^{\beta-1}(1-x)^{\alpha-1} d x=B(\beta, \alpha) $$ (b) Integrate by parts to show that \(B(\alpha, \beta)=\frac{\alpha-1}{\beta} B(\alpha-1, \beta+1)=\frac{\beta-1}{\alpha} B(\alpha+1, \beta-1)\) (c) Assume now that \(\alpha=n\) and \(\beta=m\), and that \(n\) and \(m\) are positive integers. By using the result in part (b) repeatedly, show that $$ B(n, m)=\frac{(n-1) !(m-1) !}{(n+m-1) !} $$ This result is valid even in the case where \(n\) and \(m\) are not integers, provided that we can give meaning to \((n-1) !\), \((m-1)\) !, and \((n+m-1) !\).

Step-by-Step Solution

Answer

(a) Symmetry shown. (b) Integration by parts holds. (c) Result as factorials confirmed.

Step 1: Change of Variables

Let's start by demonstrating part (a). We need to show that \(B(\alpha, \beta) = B(\beta, \alpha)\) by changing the variable of integration. Consider the substitution \(x = 1-u\), hence \(dx = -du\). This changes the limits from \(x = 0\) to \(x = 1\) to \(u = 1\) to \(u = 0\) respectively. Substituting into the integral, we have:\\[ B(\alpha, \beta) = \int_{0}^{1} (1-u)^{\alpha-1}u^{\beta-1} (-du) = \int_{1}^{0} (1-u)^{\alpha-1} u^{\beta-1} du \] Reversing the limits will remove the negative sign: \\[ = \int_{0}^{1} u^{\beta-1} (1-u)^{\alpha-1} du = B(\beta, \alpha) \]. This shows the symmetry of the Beta function.

Step 2: Integration by Parts for Beta Function

Next, for part (b), applying integration by parts, we want to integrate \(B(\alpha, \beta)\). Set \(u = (1-x)^{\beta-1}\) and \(dv = x^{\alpha-1} dx\); then \(du = -(\beta-1)(1-x)^{\beta-2} dx\) and \(v = \frac{x^\alpha}{\alpha}\). Integration by parts formula, \(\int u \, dv = uv - \int v \, du\), gives: \\[ B(\alpha, \beta) = \left[ \frac{x^\alpha}{\alpha} (1-x)^{\beta-1} \right]_0^1 - \int_0^1 \frac{x^\alpha}{\alpha}(-(\beta-1)(1-x)^{\beta-2}) dx \] The boundary terms vanish, leaving: \\[ \frac{\beta-1}{\alpha} \int_0^1 x^{\alpha} (1-x)^{\beta-2} dx = \frac{\beta-1}{\alpha} B(\alpha+1, \beta-1) \]. Reversing \(\alpha \) and \(\beta-1\) for completeness: \\(B(\alpha, \beta)=\frac{\alpha-1}{\beta} B(\alpha-1, \beta+1)\). Thus, both are true symmetrically.

Step 3: Applying Result for Integer \\(\alpha, \\beta\\)

For part (c), assume \(\alpha = n\) and \(\beta = m\), with both being positive integers. Using the recursive formula \(B(n, m)=\frac{n-1}{m} B(n-1, m+1)\) from Step 2 repeatedly, reduce \(n\) until it hits 1, while increasing \(\beta\) correspondingly: \\[ B(n, m) = \frac{n-1}{m} \frac{n-2}{m+1} \cdots \frac{1}{m+(n-2)} B(1, m+n-1) \]. Since \(B(1, k) = \frac{1}{k}\), simplify: \\[ B(n, m) = \frac{(n-1)!}{m(m+1)\cdots(m+n-1)} \cdot \frac{1}{m+n-1} = \frac{(n-1)!(m-1)!}{(n+m-1)!} \].

Key Concepts

Change of Variables

In mathematics, the change of variables is a technique where we substitute one variable for another to simplify an integral or equation. In the context of the Beta function, we aim to show its symmetry: specifically, that \( B(\alpha, \beta) = B(\beta, \alpha) \).To affect this change, we use the substitution \( x = 1-u \). With this substitution, \( dx = -du \). When \( x \) ranges from 0 to 1, \( u \) ranges from 1 to 0. This flips the limits of integration, which necessitates changing the sign of the integral.The process involves:

• Substituting the entire integral: \( x^{\alpha-1}(1-x)^{\beta-1} \) becomes \( (1-u)^{\alpha-1} u^{\beta-1} \).
• Adjusting the integral's limits and sign: the negative sign from \( dx = -du \) is canceled by reversing the integration limits.

This manipulation results in \( B(\beta, \alpha) \), demonstrating the inherent symmetry of the Beta function.

Integration by Parts

Integration by parts is a versatile technique in calculus, akin to the product rule for differentiation. It allows us to integrate products of functions by differentiating one and integrating another. The formula is \( \int u \, dv = uv - \int v \, du \).For the Beta function, we apply this technique to show the recursion relation:1. Choose functions: set \( u = (1-x)^{\beta-1} \), and \( dv = x^{\alpha-1} \, dx \).

• This means \( du = -(\beta-1)(1-x)^{\beta-2} \, dx \)
• And \( v = \frac{x^\alpha}{\alpha} \)

2. Applying the integration by parts formula:

• The first term \( uv \) evaluated at the boundaries is zero, because either factor has a boundary term, becoming zero when \( x = 0 \) or \( x = 1 \).
• For \( \int v \, du \), it simplifies the original function allowing a recursive step.

Ultimately, integration by parts here provides insight into the relation \( B(\alpha, \beta) = \frac{\beta-1}{\alpha} B(\alpha+1, \beta-1) \) and a reversed version \( B(\alpha, \beta)=\frac{\alpha-1}{\beta} B(\alpha-1, \beta+1) \). Each showcases how the function can be broken down recursively.

Factorial Properties

Factorials are essential in combinatorics and various branches of mathematics, including calculus. A factorial, denoted as \( n! \), is the product of all positive integers up to \( n \). With the Gamma function, factorials extend to real and complex numbers. This property is vital when evaluating functions like the Beta function with integer and non-integer inputs.In the context of the Beta function for integer \( \alpha = n \) and \( \beta = m \), we employ the recursive result from integration by parts repeatedly. This reduces the function down to known factorial terms:

• Use the relation \[ B(n, m) = \frac{n-1}{m} B(n-1, m+1) \] repeatedly until \( n = 1 \)
• This continues building a factorial in the numerator for \( n \) while managing the multiplicative terms in the denominator

Upon collapsing the terms, we reach \[ B(n, m) = \frac{(n-1)!(m-1)!}{(n+m-1)!} \]. This expression intertwines factorial properties, providing a bridge between discrete mathematics and continuous functions, allowing the application even when \( n \) and \( m \) are real numbers.