Problem 12 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Show that \(\Gamma(1)=1\), and use integration by parts to show that $$ \Gamma(x+1)=x \Gamma(x) \quad \text { for } x>0 . $$ Use this last expression to show for \(n=1,2, \ldots\) that $$ \Gamma(n)=(n-1) ! $$

Step-by-Step Solution

Answer

\( \Gamma(1) = 1 \) and \( \Gamma(n) = (n-1)! \) for integers \( n \geq 1 \).

Step 1: Definition of Gamma Function

The Gamma function is defined as \( \Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \, dt \). To find \( \Gamma(1) \), substitute \( x = 1 \) into the definition to get \( \Gamma(1) = \int_0^\infty e^{-t} \, dt \).

Step 2: Evaluate \( \Gamma(1) \)

Evaluate \( \Gamma(1) = \int_0^\infty e^{-t} \, dt \). This is a standard integral that equals 1, verifying that \( \Gamma(1) = 1 \).

Step 3: Integration by Parts Setup

Set up integration by parts to show \( \Gamma(x+1) = x \Gamma(x) \). Choose \( u = t^x \), \( dv = e^{-t} \, dt \), leading to \( du = x t^{x-1} \, dt \) and \( v = -e^{-t} \).

Step 4: Integration by Parts Formula

Applying integration by parts: \[ \Gamma(x+1) = \int_0^\infty t^x e^{-t} \, dt = \left[ -t^x e^{-t} \right]_0^\infty + x \int_0^\infty t^{x-1} e^{-t} \, dt = 0 + x\Gamma(x). \] The boundary term vanishes, leaving \( \Gamma(x+1) = x\Gamma(x) \).

Step 5: Use Recurrence to Prove \( \Gamma(n) = (n-1)! \) for Natural Numbers

The recurrence relation \( \Gamma(x+1) = x \Gamma(x) \) can be used iteratively starting from \( \Gamma(1) = 1 \). For \( n = 2 \), \( \Gamma(2) = 1 \Gamma(1) = 1! \). For \( n = 3 \), \( \Gamma(3) = 2 \Gamma(2) = 2! \), and so on. By induction, for any integer \( n \), \( \Gamma(n) = (n-1)! \).

Key Concepts

Integration by Parts

Integration by parts is a fundamental technique in calculus, often used to integrate products of functions. It is based on the product rule for differentiation and is a helpful method for solving integrals that are not straightforward. The formula used in integration by parts is:

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

To apply integration by parts for the Gamma function, consider:

• \( u = t^x \), then \( du = x t^{x-1} \, dt \)
• \( dv = e^{-t} \, dt \), then \( v = -e^{-t} \)

Putting these into the integration by parts formula, the original integral \( \int_0^\infty t^x e^{-t} \, dt \) becomes:

• \( \Gamma(x+1) = \left[ -t^x e^{-t} \right]_0^\infty + x \int_0^\infty t^{x-1} e^{-t} \, dt \)
• The boundary terms cancel out, as -t^x e^{-t} evaluates to zero at both limits, leaving: \( \Gamma(x+1) = x \Gamma(x) \)

This result is crucial as it sets up the recurrence relation for the Gamma function.

Factorial

A factorial, denoted by \( n! \), is the product of all positive integers up to a given number n. It is used extensively in permutations, combinations, and other areas of discrete mathematics.

• For example, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)

Factorials have a close relationship with the Gamma function, as the Gamma function extends the concept of factorials to non-integer and complex values. Specifically, for natural numbers, the relationship is as follows:

• \( \Gamma(n) = (n-1)! \)

This formula shows that the Gamma function is a continuous extension of the factorial function where the argument is decreased by 1. Hence, while the factorial function is defined only for non-negative integers, the Gamma function broadens this concept to real and complex numbers.

Recurrence Relation

A recurrence relation is a mathematical relation that defines each term of a sequence with its preceding terms. In the context of the Gamma function, the recurrence relation is expressed as:

• \( \Gamma(x+1) = x \Gamma(x) \)

This formula is instrumental in bridging the gap between the Gamma function and factorials. Starting from the known value \( \Gamma(1) = 1 \), it can be applied iteratively:

• For \( x = 1 \), the relation gives \( \Gamma(2) = 1 \Gamma(1) = 1 \times 1 = 1 \), which matches 1!
• For \( x = 2 \), \( \Gamma(3) = 2 \Gamma(2) = 2 \times 1 = 2 \), which equals 2!
• The process can be repeated for higher integer values of n.

Thus, for any natural number n, the formula confirms \( \Gamma(n) = (n-1)! \). This relationship shows the Gamma function as an eloquent generalization of factorials beyond the realm of natural numbers.