Problem 24 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each improper integral or show that it diverges. \(\int_{0}^{\infty} e^{-x} \sin x d x\)

Step-by-Step Solution

Answer

The integral evaluates to \( \frac{1}{2} \).

Step 1: Identify the type of integral and apply integration by parts

The integral in question is improper because it extends to infinity. To evaluate it, we can use integration by parts since the integrand is a product of two functions, an exponential function and a trigonometric function. Integration by parts is given by the formula \( \int u \, dv = uv - \int v \, du \).

Step 2: Choose functions for integration by parts

Let \( u = \sin x \) and \( dv = e^{-x} \, dx \). Then, \( du = \cos x \, dx \) and \( v = -e^{-x} \).

Step 3: Apply integration by parts

Using integration by parts, the integral becomes:\[\int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int -e^{-x} \cos x \, dx\]which simplifies to:\[-e^{-x} \sin x + \int e^{-x} \cos x \, dx\]

Step 4: Repeat integration by parts for the new integral

Apply integration by parts again for \( \int e^{-x} \cos x \, dx \). Let \( u = \cos x \) and \( dv = e^{-x} \, dx \). Then, \( du = -\sin x \, dx \) and \( v = -e^{-x} \).

Step 5: Apply integration by parts to the new integral

Using integration by parts again, we have:\[\int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int e^{-x} \sin x \, dx\]

Step 6: Combine results to form an equation

The original integral is now expressed as:\[\int e^{-x} \sin x \, dx = -e^{-x} \sin x + (-e^{-x} \cos x) - \int e^{-x} \sin x \, dx\]Let \( I = \int e^{-x} \sin x \, dx \), then:\[ I = -e^{-x} \sin x - e^{-x} \cos x - I \]

Step 7: Solve for the integral

Reorganize the equation to solve for \( I \):\[ 2I = -e^{-x} \sin x - e^{-x} \cos x \]Simplify to:\[ I = \frac{-e^{-x} (\sin x + \cos x)}{2} \]

Step 8: Evaluate the integral from 0 to infinity

Evaluate the limit as \( x \to \infty \):\[ \lim_{{x \to \infty}} \frac{-e^{-x} (\sin x + \cos x)}{2} = \lim_{{x \to \infty}} 0 = 0 \]At \( x = 0 \), the expression evaluates to \( \frac{-1}{2} \).

Step 9: Final result

Combine the limits:\[ \int_{0}^{\infty} e^{-x} \sin x \, dx = 0 + \frac{1}{2} = \frac{1}{2} \]

Key Concepts

Integration by Parts

Integration by parts is a clever technique used to simplify the integration of products of functions, such as our exponential and trigonometric functions in the given problem. The integration by parts formula is expressed as:\[ \int u \, dv = uv - \int v \, du \]This formula essentially swaps the parts of the integral, helping to make otherwise complicated integrals easier to solve.
How did we apply this method here? Let's break it down:

• We identified two functions from the integrand: choose \( u = \sin x \) and \( dv = e^{-x} \, dx \).
• Calculated \( du = \cos x \, dx \) and \( v = -e^{-x} \).
• Substituted into the integration by parts formula to transform the initial integral. This required finding a new integral \( \int e^{-x} \cos x \, dx \), which was then solved using integration by parts again.

This iterative process eventually simplified the original expression into a solvable equation. Repeated integration by parts is common with functions like these that form a cycle.

Limits and Convergence

To evaluate improper integrals that extend to infinity, such as the integral we are tackling here, we must determine their convergence or divergence. Improper integrals are those where at least one of the limits of integration is infinite, or the integrand becomes infinite within the bounds of integration.
Understanding limits is crucial: * As we simplify the integrand through integration by parts, the role of limits becomes more important—specifically evaluating how the behavior of the integral behaves at the boundaries.* For the given problem, we evaluated the limit \( \lim_{{x \to \infty}} \frac{-e^{-x} (\sin x + \cos x)}{2} \), where we assessed the behavior at infinity.* Because the exponential term \( e^{-x} \) diminishes to zero as \( x \to \infty \), this indicated convergence of the integral.
By correctly applying these limits at both \( x = 0 \) and \( x = \infty \), we were able to successfully validate the convergence of the integral to a finite number, confirming it does not diverge.

Trigonometric Integrals

Trigonometric integrals involve sine, cosine, or other trigonometric functions. They often appear in the form of products with other functions like exponentials. Understanding how to simplify these expressions is key.
In the given problem \( \int_{0}^{\infty} e^{-x} \sin x \, dx \):* We effectively utilized properties of both sine and cosine during the integration by parts process.* Trigonometric functions often cycle through their values, contributing periodic characteristics that can be leveraged during calculations.* Notice how both \( \sin x \) and \( \cos x \) appear in our intermediary steps, allowing us to manipulate the integral back and forth until a stable solution is found.
For many integrals containing trigonometric functions, it’s essential to recognize repeating patterns and relationships, which often lead to self-consistent equations, such as occurring cycles helping us solve for the integral. These strategic manipulations are crucial for resolving integrals involving these functions.