Problem 53 - Calculus Early Transcendentals: Pearson New International Edition

Question

(Gamma Function) Let \(\Gamma(n)=\int_{0}^{\infty} x^{n-1} e^{-x} d x, n>0\). This integral converges by Problems 51 and 52 . Show each of the following (note that the gamma function is defined for every positive real number \(n\) ): (a) \(\Gamma(1)=1\) (b) \(\Gamma(n+1)=n \Gamma(n)\) (c) \(\Gamma(n+1)=n\) !, if \(n\) is a positive integer.

Step-by-Step Solution

Answer

(a) \(\Gamma(1) = 1\); (b) \(\Gamma(n+1) = n \Gamma(n)\); (c) \(\Gamma(n+1) = n!\).

Step 1: Evaluate \(\Gamma(1)\)

To find \(\Gamma(1)\), we substitute \(n = 1\) into the definition of the Gamma function: \[\Gamma(1) = \int_{0}^{\infty} x^{1-1} e^{-x} \, dx = \int_{0}^{\infty} e^{-x} \, dx.\] This is a standard exponential integral. The integral of \(e^{-x}\) from \(0\) to \(\infty\) equals 1. Therefore, \(\Gamma(1) = 1\).

Step 2: Prove recursive property \(\Gamma(n+1)=n \Gamma(n)\)

To prove \(\Gamma(n+1) = n \Gamma(n)\), use integration by parts:Let \(u = x^n\) and \(dv = e^{-x}\, dx\). This gives \(du = n x^{n-1} \, dx\) and \(v = -e^{-x}\). The integration by parts formula \(\int u \, dv = uv - \int v \, du\) gives:\[\int_{0}^{\infty} x^n e^{-x} \, dx = \left[-x^n e^{-x}\right]_{0}^{\infty} + \int_{0}^{\infty} n x^{n-1} e^{-x} \, dx.\]The boundary term \( \left[-x^n e^{-x}\right]_{0}^{\infty} \) evaluates to 0 because as \(x \to \infty\), \(e^{-x}\) goes to 0 faster than \(x^n\) grows, and at \(x = 0\) the term is also 0. Thus:\[\Gamma(n+1) = n \int_{0}^{\infty} x^{n-1} e^{-x} \, dx = n \Gamma(n).\]

Step 3: Verify \(\Gamma(n+1) = n!\) for positive integer \(n\)

The recursive property \(\Gamma(n+1) = n\Gamma(n)\) implies that if \(\Gamma(1)=1\), then \(\Gamma(n+1)\) can be expressed as:\[\Gamma(n+1) = n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 1 = n!.\]Start from \(\Gamma(1) = 1 != 1\), and notice that \(\Gamma(2) = 1 \times 1 = 1! = 1\), \(\Gamma(3) = 2 \times 1 = 2! = 2\), and so on. This confirms that \(\Gamma(n+1) = n!\) holds for any positive integer \(n\).

Key Concepts

Integration by Parts

Integration by Parts is a powerful technique used to solve integrals that cannot be easily handled by standard methods. It is derived from the product rule for differentiation and is helpful, especially for integrals involving products of algebraic and exponential functions.
The formula for Integration by Parts is:

• \( \int u \, dv = uv - \int v \, du \)

Here, you choose parts of the integrand to be \( u \) and \( dv \). Differentiating \( u \) gives \( du \), and integrating \( dv \) gives \( v \). By applying this formula, tricky integrals become manageable.In the solution, Integration by Parts is used to establish the recursive property of the Gamma Function: \( \Gamma(n+1) = n \Gamma(n) \).
We let \( u = x^n \) and \( dv = e^{-x} \, dx \). Calculating the derivatives and integrals, we get \( du = nx^{n-1} \, dx \) and \( v = -e^{-x} \).
Substituting these into the formula helps simplify the integral, and interestingly, confirms that the boundary term vanishes.

Factorial Function

The Factorial Function, usually denoted as \( n! \), is a fundamental concept in mathematics, particularly in combinatorics and number theory. It's defined for non-negative integers, representing the product of all positive integers up to a given number.

• For example, \( n! = n \times (n-1) \times (n-2) \times \ldots \times 2 \times 1 \).
• By definition, \( 0! = 1 \).

This function grows rapidly with increasing \( n \), and its properties make it useful for counting and probability calculations.
In relation to the Gamma Function, there's an intriguing connection: when \( n \) is a positive integer, the gamma function \( \Gamma(n+1) \) equals \( n! \). This connection is established through the recursive relation \( \Gamma(n+1) = n \Gamma(n) \), showing that these two functions are intrinsically linked for positive integers.
Thus, the Gamma Function can be thought of as a generalized factorial that extends to complex numbers.

Recursive Properties

Recursive Properties in mathematics refer to relationships where a function is defined in terms of itself, usually changing the input slightly with each iteration.
The Gamma Function possesses a recursive property like this. Specifically, it satisfies \( \Gamma(n+1) = n \Gamma(n) \). This unique property allows one to compute \( \Gamma(n+1) \) using the value of \( \Gamma(n) \). To appreciate this property, consider how it works in steps:

• Start with a base case, such as \( \Gamma(1) = 1 \).
• Use the recursive property to compute successive values, like \( \Gamma(2) = 1 \times \Gamma(1) = 1 \), \( \Gamma(3) = 2 \times \Gamma(2) = 2 \times 1 = 2 \), and so on.

This recursive nature simplifies calculations and highlights the Gamma Function's deep connection to factorials. It's a cornerstone feature that makes the Gamma Function so powerful in mathematics.'