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 $$ B(n, m)=\frac{(n-1) !(m-1) !}{(n+m-1) !} $$

Step-by-Step Solution

Answer

Change variables, integrate by parts twice, and use formulas to derive Beta function's factorial form.

Step 1: Establish the Relation for Change of Variables

To show that \(B(\alpha, \beta) = B(\beta, \alpha)\), we perform a change of variables. Consider the substitution \(u = 1-x\), which implies \(du = -dx\). Then, when \(x = 0\), \(u = 1\) and when \(x = 1\), \(u = 0\). Substituting into the Beta function, we get:\[ B(\alpha, \beta) = \int_{1}^{0} (1-u)^{\alpha-1} u^{\beta-1} (-du) = \int_{0}^{1} u^{\beta-1} (1-u)^{\alpha-1} du = B(\beta, \alpha) \]This completes the first part of the problem.

Step 2: Integration by Parts Formula

Integration by parts is based on the formula \(\int u dv = uv - \int v du\). For \(B(\alpha, \beta)\), let \(u = x^{\alpha-1}\) and \(dv = (1-x)^{\beta-1}dx\). Then, \(du = (\alpha-1)x^{\alpha-2}dx\) and \(v = -\frac{1}{\beta}(1-x)^{\beta}\). Perform integration by parts on \(B(\alpha, \beta)\):\[ B(\alpha, \beta) = \left[-\frac{1}{\beta}x^{\alpha-1}(1-x)^\beta \right]_{0}^{1} + \frac{\alpha-1}{\beta} \int_{0}^{1} x^{\alpha-2}(1-x)^{\beta} dx \]Assuming regular boundary conditions, the boundary terms vanish, so:\[ B(\alpha, \beta) = \frac{\alpha-1}{\beta} B(\alpha-1, \beta+1) \]

Step 3: An Alternative Integration by Parts Step

To derive the second formula \(B(\alpha, \beta) = \frac{\beta-1}{\alpha} B(\alpha+1, \beta-1)\), we again integrate by parts but this time let \(u = (1-x)^{\alpha-1}\) and \(dv = x^{\beta-1}dx\). Then, \(du = -(\alpha-1)(1-x)^{\alpha-2}dx\) and \(v = \frac{1}{\beta}x^\beta\). Substituting these into the integration by parts formula:\[ B(\alpha, \beta) = \left[\frac{1}{\beta}x^\beta(1-x)^{\alpha-1}\right]_{0}^{1} + \frac{\beta-1}{\alpha} \int_{0}^{1} (1-x)^{\alpha}(x)^{\beta-2} dx \]After canceling out boundary terms at 0 and 1:\[ B(\alpha, \beta) = \frac{\beta-1}{\alpha} B(\alpha+1, \beta-1) \]

Step 4: Using Induction to Reach Factorial Expression

Assume \(\alpha = n\) and \(\beta = m\). We use the recursive relation from Step 2 iteratively as follows:\[ B(n, m) = \frac{n-1}{m} B(n-1, m+1) \]Repeating this, each time reducing \(n\) and increasing \(m\) (and vice versa), eventually we reach:\[ B(1, n+m-1) = \int_{0}^{1} x^{0}(1-x)^{n+m-2} dx = \frac{1}{n+m-1} \]Thus:\[ B(n, m) = \frac{(n-1)(n-2)...1}{(n+m-1)(n+m-2)...(m)} \]Simultaneously using both reductions leads to \(B(n, m) = \frac{(n-1)!(m-1)!}{(n+m-1)!}\).

Key Concepts

Change of Variables

To solve mathematical problems involving integrals, sometimes we need to alter the variable of integration, using a method called "change of variables." This approach is particularly useful for symmetry arguments, as seen with the Beta function. In our original task, we start with the function:\[ B(\alpha, \beta) = \int_{0}^{1} x^{\alpha-1}(1-x)^{\beta-1} \ dx \]Our goal is to show that this function is symmetric, meaning it remains unchanged if we swap \( \alpha \) and \( \beta \). We use the substitution \( u = 1-x \), which gives us \( du = -dx \). Here's how it simplifies our integral:

• When \( x = 0 \), \( u = 1 \).
• When \( x = 1 \), \( u = 0 \).

Substitute these into the integral:\[ B(\alpha, \beta) = \int_{1}^{0} (1-u)^{\alpha-1} u^{\beta-1} (-du) \ = \int_{0}^{1} u^{\beta-1} (1-u)^{\alpha-1} \ du \ = B(\beta, \alpha) \]The negative sign from \( du \) flips the integration limits, allowing us to write it as an integral from 0 to 1 again. This shows the symmetry \( B(\alpha, \beta) = B(\beta, \alpha) \), illustrating the beautiful property that this substitution reveals.

Integration by Parts

Integration by parts is a fundamental technique, grounded in calculus, that breaks down complex integrals into simpler parts. The formula is given by:\[ \int u \, dv = uv - \int v \, du \]Let's apply this concept to the Beta function to derive useful relations. We need to calculate:\[ B(\alpha, \beta) = \int_{0}^{1} x^{\alpha-1}(1-x)^{\beta-1} \, dx \]First, choose \( u = x^{\alpha-1} \) and \( dv = (1-x)^{\beta-1} \, dx \), making \( du = (\alpha-1)x^{\alpha-2} \, dx \) and \( v = -\frac{1}{\beta}(1-x)^{\beta} \). Substituting these into the formula gives:\[ B(\alpha, \beta) = \left[-\frac{1}{\beta}x^{\alpha-1}(1-x)^\beta \right]_{0}^{1} + \frac{\alpha-1}{\beta} \int_{0}^{1} x^{\alpha-2}(1-x)^{\beta} \, dx \]The boundary terms vanish because the expression evaluates to 0 at both limits, so we are left with:\[ B(\alpha, \beta) = \frac{\alpha-1}{\beta} B(\alpha-1, \beta+1) \]We can also perform integration by parts a second time using different substitutions, leading to:\[ B(\alpha, \beta) = \frac{\beta-1}{\alpha} B(\alpha+1, \beta-1) \]These derivations are powerful as they provide recursive relations, which are crucial in understanding the structure of functions like the Beta function.

Recursive Relationship

Recursive relationships are repetitive processes where the outcome in each step is built upon previous results. For the Beta function problem, recursion significantly simplifies calculations when \( \alpha \) and \( \beta \) are integers. Using the relationships derived from integration by parts, let's assume \( \alpha = n \) and \( \beta = m \) with both as positive integers.We start with the formula:\[ B(n, m) = \frac{n-1}{m} B(n-1, m+1) \]We continue applying this recursively, repeatedly adjusting \( n \) and \( m \). The recursion progresses until we reach manageable integrals like:\[ B(1, n+m-1) = \int_{0}^{1} (1-x)^{n+m-2} \, dx = \frac{1}{n+m-1} \]Through recursion, step by step, the Beta function can be expressed entirely in terms of factorials:\[ B(n, m) = \frac{(n-1)! (m-1)!}{(n+m-1)!} \]

• This transformation is particularly neat because it relates the Beta function to factorials, which are easier to handle and deeply connected to combinatorial mathematics.
• Recursive relationship techniques are highly useful in calculations involving sequences or functions that have a self-referential structure.

This approach illustrates how powerful recursive relationships can be in simplifying and solving problems effectively.