Problem 54 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate \(\int_{0}^{\infty} x^{n-1} e^{-x} d x\) for \(n=1,2,3,4\), and 5, thereby confirming Problem \(53(\mathrm{c})\).

Step-by-Step Solution

Answer

The integrals evaluate to \(\Gamma(1) = 1\), \(\Gamma(2) = 1\), \(\Gamma(3) = 2\), \(\Gamma(4) = 6\), \(\Gamma(5) = 24\).

Step 1: Identify the Integral

The exercise requires evaluating the integral \( \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \). This is the definition of the Gamma function, \( \Gamma(n) \), where \( \Gamma(n) = \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \). We need to compute this for \( n = 1, 2, 3, 4, \) and \( 5 \).

Step 2: Evaluate the Integral for n=1

For \( n = 1 \):\[\Gamma(1) = \int_{0}^{\infty} x^{1-1} e^{-x} \, dx = \int_{0}^{\infty} e^{-x} \, dx.\]This simplifies to:\[\left[ -e^{-x} \right]_{0}^{\infty} = 1 - 0 = 1.\]Thus, \( \Gamma(1) = 1 \).

Step 3: Evaluate the Integral for n=2

For \( n = 2 \):\[\Gamma(2) = \int_{0}^{\infty} x^{2-1} e^{-x} \, dx = \int_{0}^{\infty} x e^{-x} \, dx.\]Using integration by parts, let \( u = x \) and \( dv = e^{-x} \, dx \). Then, \( du = dx \) and \( v = -e^{-x} \).\[\Gamma(2) = \left[ -x e^{-x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-x} \, dx = [0] + 1 = 1.\]Thus, \( \Gamma(2) = 1 \).

Step 4: Recursive Property of Gamma Function

Use the recursive formula of the Gamma function: \( \Gamma(n) = (n-1) \Gamma(n-1) \).

Step 5: Evaluate the Integral for n=3

For \( n = 3 \):Using the recursive property:\[\Gamma(3) = (3-1)\Gamma(2) = 2 \times 1 = 2.\]

Step 6: Evaluate the Integral for n=4

For \( n = 4 \):Using the recursive property:\[\Gamma(4) = (4-1)\Gamma(3) = 3 \times 2 = 6.\]

Step 7: Evaluate the Integral for n=5

For \( n = 5 \):Using the recursive property:\[\Gamma(5) = (5-1)\Gamma(4) = 4 \times 6 = 24.\]

Step 8: Confirm with the Known Function

Recall that \( \Gamma(n) = (n-1)\Gamma(n-1) = (n-1)! \). Thus, for \( n = 1, 2, 3, 4, \) and \( 5 \),\( \Gamma(n) = (n-1)! \ = 1, 1, 2, 6, 24 \), which matches our calculated values.

Key Concepts

Integration by Parts

Integration by parts is a technique used to solve integrals where the product of two functions is involved. This technique transforms the original integral into a simpler one that is easier to solve. It is based on the product rule for differentiation and is given by the formula:

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

To apply integration by parts, you need to choose which parts of your integral will be \( u \) and \( dv \). For example, in the integral \( \int_{0}^{\infty} x e^{-x} \, dx \), we set \( u = x \) and \( dv = e^{-x} \, dx \). This means \( du = dx \) and \( v = -e^{-x} \), resulting in the transformed integral:

• \( \int_{0}^{\infty} x e^{-x} \, dx = \left[ -xe^{-x} \right]_{0}^{\infty} + \int_{0}^{\infty} e^{-x} \, dx \)

By solving this, we find that the integral simplifies considerably to 1, confirming the value of \( \Gamma(2) \). Remember, the trickiest part is often selecting appropriate \( u \) and \( dv \). But with practice, it becomes more intuitive.

Recursive Formula

The recursive formula for the Gamma function is an essential tool for solving advanced integrals and understanding factorial-related calculations. The formula is:

• \( \Gamma(n) = (n-1) \Gamma(n-1) \)

This powerful relation ties the Gamma function at \( n \) to its value at \( n-1 \), thus simplifying the computation process. For instance, if you know \( \Gamma(2) = 1 \), you can easily find \( \Gamma(3) \) using the recursive formula:

• \( \Gamma(3) = 2 \times \Gamma(2) = 2 \times 1 = 2 \)

This formula essentially reduces complex computations by utilizing previously solved values, converting a potentially cumbersome process into a series of straightforward multiplications.

Factorial Function

The factorial function, denoted \( n! \), is the product of all positive integers up to \( n \). It plays a central role in permutations, combinations, and a wide range of mathematical formulas.

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

The Gamma function extends the concept of the factorial to non-integer values. In essence, \( \Gamma(n) = (n-1)! \) when \( n \) is a positive integer. This connection between the Gamma function and the factorial is evidence in calculations like \( \Gamma(5) \) equating to \( 4! \).

• That is \( \Gamma(5) = 4 \times 3 \times 2 \times 1 = 24 \), which matches \( 4! \).

Understanding this relationship provides a seamless transition between continuous and discrete approaches to factorial calculations.

Definite Integral

A definite integral is one that evaluates the area under a curve within a given range. It offers a numerical value, in contrast to an indefinite integral which includes a constant term. For example, in the exercise, we evaluate the integral

• \( \int_{0}^{\infty} x^{n-1} e^{-x} \, dx \)

over the range \( 0 \) to \( \infty \). This integral calculates the area from the curve \( x^{n-1} e^{-x} \) to the x-axis, stretching from zero to infinity. By determining these areas for each integer \( n \), we uncovered integral values that relate to the Gamma function.

• This shows the powerful application of definite integrals in evaluating functions over a specified interval and their connection to broader mathematical concepts like the Gamma function.

Mastering definite integrals is fundamental not only in mathematics but also in applied fields like physics and engineering, where they frequently model real-world problems.