Problem 22 - Thomas Calculus

Question

Evaluate the integrals in Exercises \(1-34\) without using tables. $$ \int_{0}^{\infty} 2 e^{-\theta} \sin \theta d \theta $$

Step-by-Step Solution

Answer

The integral evaluates to \(2\).

Step 1: Recognize the Integral Type

The integral given is \(\int_{0}^{\infty} 2 e^{-\theta} \sin \theta \, d\theta\). This is an improper integral as it is an integral from 0 to infinity.

Step 2: Set Up the Integral for Evaluation

To evaluate this integral, we first need to set it up using limits: \( \lim_{b \to \infty} \int_{0}^{b} 2 e^{-\theta} \sin \theta \, d\theta \).

Step 3: Use Integration by Parts

Apply integration by parts to \( \int e^{-\theta} \sin \theta \, d\theta \), where \( u = \sin \theta \) and \( dv = 2e^{-\theta} \, d\theta \). Then, \( du = \cos \theta \, d\theta \) and \( v = -2e^{-\theta} \). Recall the formula: \( \int u \, dv = uv - \int v \, du \).

Step 4: Compute the Components

Compute \( uv = -2e^{-\theta} \sin \theta \). Now find the integral of \( \int v \, du = -2 \int e^{-\theta} \cos \theta \, d\theta \).

Step 5: Second Integration by Parts

For \( \int e^{-\theta} \cos \theta \, d\theta \), apply integration by parts again with \( u = \cos \theta \) and \( dv = e^{-\theta} \, d\theta \). Then, \( du = -\sin \theta \, d\theta \) and \( v = -e^{-\theta} \).

Step 6: Combine the Parts

After the second part of integration by parts, complete the formula: \(-2 \int e^{-\theta} \cos \theta \, d\theta = -2(-e^{-\theta} \cos \theta + \int e^{-\theta} \sin \theta \, d\theta)\).

Step 7: Solve the Equations

Solve these integrals by combining like terms and using substitutions. You end up with a repeating pattern that enables you to solve for the original integral.

Step 8: Calculate the Limit

Evaluate the limit: \(\lim_{b \to \infty} [-2e^{- heta} \sin \theta - 2e^{-\theta} \cos \theta]_{0}^{b}\). Simplify this limit to obtain the result.

Key Concepts

Improper Integrals

Improper integrals are integrals that describe areas or values over an infinite interval or involve an integrand with infinite discontinuities. In our exercise, the integral is improper because the upper limit is infinity, specifically going from 0 to \( \infty \). This requires special techniques since you can't directly compute an integral with infinity as a limit. The conventional method to approach this is to replace the infinite limit with a variable, such as \( b \), and then take the limit as this variable approaches infinity: \[ \lim_{b \to \infty} \int_{0}^{b} 2 e^{-\theta} \sin \theta \, d\theta \].This technique essentially "slices" the infinite integral into a finite one, making it solvable using familiar calculus methods. Once evaluated, if the result tends to a finite number as the limit extends to infinity, the original improper integral converges.

Exponential Functions

Exponential functions feature prominently in various calculus problems due to their unique properties. Here, the function \( e^{-\theta} \) plays a crucial role. Exponential functions have some attractive properties:

• They always have a constant base, for example, \( e \), raised to the power of a variable.
• They exhibit rapid growth or decay depending on whether the exponent is positive or negative.

In our integral, \( e^{-\theta} \) serves as a decaying factor. As \( \theta \) increases, \( e^{-\theta} \) decreases exponentially towards zero. This decay significantly impacts the convergence of the integral over the infinite interval and is essential for the convergence of many improper integrals. Because such functions have simple derivatives and integrals, they often simplify complex integration problems when combined with other functions such as trigonometric functions.

Trigonometric Integrals

In calculus, trigonometric integrals involve integrating products or powers of trigonometric functions. In this exercise, the function \( \sin \theta \) is present in the integrand.Trigonometric functions have oscillating patterns with values ranging between -1 and 1. These inherent properties:

• Make them periodic and cyclic, useful in a wide range of mathematical and physical problems.
• They often appear in integrals combined with other function types, like exponentials, requiring integration techniques like integration by parts or substitutions.

In the integral \( \int_{0}^{b} 2 e^{-\theta} \sin \theta \, d\theta \), the presence of \( \sin \theta \) multiplies the complexity. Its periodicity means that although \( \sin \theta \) oscillates, when combined with the decaying \( e^{-\theta} \), it contributes to a finite value over the infinite range. Understanding how these trigonometric integrals behave is crucial for solving them effectively using suitable integration methods.