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Chapter 4: Problem 81

a. If \(\alpha>0, \Gamma(\alpha)\) is defined by \(\Gamma(\alpha)=\int_{0}^{\infty} y^{\alpha-1} e^{-y} d y,\) show that \(\Gamma(1)=1\) * b. If \(\alpha>1,\) integrate by parts to prove that \(\Gamma(\alpha)=(\alpha-1) \Gamma(\alpha-1)\)

Short Answer

Expert verified
\(\Gamma(1) = 1\) and \(\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)\).

Step by step solution

01

Define the Gamma Function

The Gamma function \( \Gamma(\alpha) \) is defined as \( \Gamma(\alpha) = \int_{0}^{\infty} y^{\alpha-1} e^{-y} \, dy \). This function generalizes the factorial function with \( \Gamma(n) = (n-1)! \) for positive integers \( n \).
02

Prove \(\Gamma(1) = 1\)

Substitute \( \alpha = 1 \) into the definition of the Gamma function: \[ \Gamma(1) = \int_{0}^{\infty} y^{1-1} e^{-y} \; dy = \int_{0}^{\infty} e^{-y} \; dy. \]The integral \( \int_{0}^{\infty} e^{-y} \; dy \) is a standard exponential integral and equals 1. Thus, \( \Gamma(1) = 1 \).
03

Integrate by Parts for \( \Gamma(\alpha) \)

For \( \alpha > 1 \), set \( u = y^{\alpha-1} \) and \( dv = e^{-y} \, dy \). Then, \( du = (\alpha-1)y^{\alpha-2} \, dy \) and \( v = -e^{-y} \).
04

Apply Integration by Parts Formula

Apply the integration by parts formula \( \int u \, dv = uv - \int v \, du \):\[ \Gamma(\alpha) = \left[-y^{\alpha-1} e^{-y}\right]_{0}^{\infty} + \int_{0}^{\infty} e^{-y} (\alpha-1) y^{\alpha-2} \, dy. \]
05

Evaluate Boundary Terms

The boundary term \( \left[-y^{\alpha-1} e^{-y}\right]_{0}^{\infty} \) evaluates to 0 because as \( y \to \infty \), \( e^{-y} \to 0 \) faster than \( y^{\alpha-1} \to \infty \), and at \( y = 0 \), the term is also 0. So we have:\[ \Gamma(\alpha) = (\alpha-1) \int_{0}^{\infty} y^{\alpha-2} e^{-y} \, dy. \]
06

Recognize the Inductive Hypothesis

Notice that \( \int_{0}^{\infty} y^{\alpha-2} e^{-y} \, dy \) is the definition of \( \Gamma(\alpha-1) \), thus:\[ \Gamma(\alpha) = (\alpha-1) \Gamma(\alpha-1). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration by Parts
Integration by parts is a technique used in calculus to simplify certain types of integrals. The approach is derived from the product rule for differentiation and can be extremely helpful for integrating a product of functions.

Here's the basic formula for integration by parts:
  • \( \int u \, dv = uv - \int v \, du \)
To apply this method:
  • First, choose which part of the integrand to assign as \( u \) and \( dv \). This choice can be guided by the type of functions involved; for example, polynomial functions are typically good candidates for \( u \).
  • Next, differentiate \( u \) to get \( du \) and integrate \( dv \) to get \( v \). Substitute these into the integration by parts formula.
  • Finally, compute the remaining integral.
In the context of the Gamma Function, for \( \alpha > 1 \), we selected \( u = y^{\alpha-1} \) and \( dv = e^{-y} \), leading to \( du = (\alpha-1)y^{\alpha-2} \, dy \) and \( v = -e^{-y} \). This choice simplifies the integration process effectively, transforming the problem into a form involving \( \Gamma(\alpha-1) \).
Factorial Function
The factorial function, typically denoted as \( n! \), is the product of all positive integers up to a given number \( n \). It's defined as:
  • \( n! = n \times (n-1) \times \cdots \times 2 \times 1 \)
  • By convention, \( 0! = 1 \)
The relation between the Gamma function and the factorial function is fascinating. For positive integers, the Gamma function generalizes the factorial function with \( \Gamma(n) = (n-1)! \).

This means:
  • The Gamma function provides a way to extend the notion of factorials beyond integers. It handles non-integer and complex number inputs as well.
  • It is a continuous extension of the factorial function, often used in advanced calculus and complex analysis.
Understanding the Gamma function's relation to the factorial function is essential for deeper study in mathematical areas like probability and statistics, where continuous distributions require such extensions.
Exponential Integral
The exponential integral is a category of integrals involving the exponential function, which is a common element in many mathematical problems. In the exercise, the integral \( \int_{0}^{\infty} e^{-y} \, dy \) is an exponential integral that evaluates to 1. This specific integral is essential in demonstrating basic properties of the Gamma function, like showing that \( \Gamma(1) = 1 \).

Key points about the exponential integral:
  • The form \( e^{-y} \) represents a rapidly decreasing function as \( y \) approaches infinity, ensuring convergence of the integral over an infinite domain.
  • Understanding this integral helps develop intuition about the behavior of functions involving exponentials, especially over large intervals.
  • It frequently appears in various fields such as physics, statistics, and engineering, where decay processes or growth models are examined.
Learning how to evaluate such integrals, especially understanding their convergence and properties, is crucial for mastering more advanced topics in calculus and applied mathematics.

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