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Chapter 8: Problem 22

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

Short Answer

Expert verified
The integral evaluates to -2.

Step by step solution

01

Identify the Integral Type

The given integral \( \int_{0}^{\infty} 2 e^{-\theta} \sin \theta \, d\theta \) is an improper integral due to the upper limit being infinity. The integrand has the form \( e^{-\theta} \sin \theta \), which suggests using integration by parts.
02

Choose Functions for Integration by Parts

Recall that integration by parts is based on the formula \( \int u \, dv = uv - \int v \, du \). Choose \( u = \sin \theta \) and \( dv = 2 e^{-\theta} \, d\theta \). This implies that \( du = \cos \theta \, d\theta \) and \( v = -2 e^{-\theta} \).
03

Apply Integration by Parts

Using \( u = \sin \theta \) and \( v = -2 e^{-\theta} \), the integration by parts formula becomes:\[\int 2e^{-\theta} \sin\theta \, d\theta = -2e^{-\theta} \sin\theta - \int -2e^{-\theta} \cos\theta \, d\theta.\] Simplify to rewrite the integral as:\[= -2e^{-\theta} \sin\theta + 2 \int e^{-\theta} \cos\theta \, d\theta.\]
04

Integrate \( \int e^{-\theta} \cos \theta d\theta \) Again by Parts

For \( \int e^{-\theta} \cos \theta \, d\theta \), again use integration by parts.- Let \( u = \cos \theta \), \( dv = e^{-\theta} \, d\theta \) so \( du = -\sin \theta \, d\theta \), and \( v = -e^{-\theta} \).Apply integration by parts:\[\int e^{-\theta} \cos \theta \, d\theta = -e^{-\theta} \cos \theta - \int -e^{-\theta} (-\sin \theta) \, d\theta.\] This simplifies to:\[= -e^{-\theta} \cos \theta + \int e^{-\theta} \sin \theta \, d\theta.\]
05

Solve the System of Equations

We observe that our original integral appears again: \( \int e^{-\theta} \sin\theta \, d\theta \).Let \( I = \int e^{-\theta} \sin \theta \, d\theta \).Then \[ I = -e^{-\theta} \cos \theta + I.\] Isolate \( I \):\[ 2I = -2e^{-\theta} \cos \theta. \]
06

Evaluate the Definite Integral

After integration and rearranging, we evaluate the limits: \[ -2e^{-\theta} \sin\theta \bigg|_0^{\infty} + 2(-e^{-\theta} \cos\theta )\bigg|_0^{\infty}. \]At \( \theta = \infty \), the exponential terms vanish because \( e^{-\infty} = 0 \).At \( \theta = 0 \), evaluate:\[-2e^0 \sin 0 + 2(-e^0 \cos 0) = 0 - 2(-1) = -2. \]
07

Sum the Results

Since the terms involving \( e^{-\infty} \) go to zero and we confirmed our solution values:\[ I = -2e^{-\theta} \cos \theta + e^{-\theta} \sin \theta. \]Therefore, the value of the integral evaluated from 0 to infinity is \(-2\).

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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 handy technique in calculus for solving certain integrals. The method derives from the product rule for differentiation and is especially useful when the integrand is a product of two types of functions. You apply the integration by parts formula:\[\int u \, dv = uv - \int v \, du.\]Here's the step-by-step approach:
  • Choose the function \(u\) such that its derivative \(du\) is simpler than \(u\).
  • Choose \(dv\) so it's easy to integrate to get \(v\).
  • Apply the formula and simplify the resulting expression.
For example, in the exercise, setting \(u = \sin \theta\) and \(dv = 2e^{-\theta} \, d\theta\) simplifies the process because the derivative of \(\sin \theta\) is straightforward and integrating \(e^{-\theta}\) is also simple.
Exponential Functions
Exponential functions are functions where the variable is an exponent. They are written in the form \(e^x\), where \(e\) is approximately equal to 2.718. These functions are unique as they model growth and decay processes and have a constant rate of change relative to their value.
Key properties:
  • Derivative: \(\frac{d}{dx}e^x = e^x\).
  • Integral: \(\int e^x \, dx = e^x + C\).
They often appear in integrals due to their simple differentiation and integration properties. In the improper integral from the original exercise, the term \(e^{-\theta}\) represents exponential decay. Being mindful of how exponential terms behave at infinity helps in evaluating limits when dealing with improper integrals.
Trigonometric Functions
Trigonometric functions like \(\sin\theta\) and \(\cos\theta\) are fundamental in mathematics and are defined using ratios from a right-angled triangle. Here's a brief understanding:
  • \(\sin\theta\) relates to the ratio of the opposite side to the hypotenuse.
  • \(\cos\theta\) is the ratio of the adjacent side to the hypotenuse.
These functions are periodic, meaning they repeat values over specific intervals, and have derivatives that easily swap between sine and cosine:
  • \(\frac{d}{d\theta}\sin\theta = \cos\theta\)
  • \(\frac{d}{d\theta}\cos\theta = -\sin\theta\)
In the integration process, these properties are critical in selecting \(u\) and \(dv\) effectively when each appears as part of the product in the integrand.
Evaluating Integrals
Evaluating an integral means computing the area under the curve described by the function. In the context of an improper integral, such as the one spanning from 0 to \(\infty\), the challenge is often about managing the limits鈥攊nfinity in this case.
Here's how you can think about the evaluation process:
  • Substitute the limits after integrating the function.
  • Evaluate the behavior of terms like \(e^{-\theta}\) as \(\theta\) approaches infinity, which tends to zero.
  • Simplify the results by subtracting values of the function evaluated at each limit.
In the original exercise, after integrating through parts, the behavior of the exponential ensures terms vanish as theta approaches infinity, simplifying the final computation considerably. Understanding these steps well enables handling more complex integrals smoothly.

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