Problem 23 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

Step-by-Step Solution

Answer

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

Step 1: Identify the Type of Integral

The integral \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) is an improper integral because it has an infinite upper limit. First, we need to evaluate the limit of the integral as \( b \to \infty \).

Step 2: Rewrite the Integral with Limits

Rewrite the integral using a limit: \( \lim_{b \to \infty} \int_{0}^{b} e^{-x} \cos x \, dx \). We need to solve this limit to determine if the integral converges or diverges.

Step 3: Integrate by Parts

Use integration by parts: Let \( u = \cos x \) and \( dv = e^{-x} \, dx \). Then, \( du = -\sin x \, dx \) and \( v = -e^{-x} \). Integration by parts formula is \( \int u \, dv = uv - \int v \, du \).

Step 4: Apply Integration by Parts Formula

Substitute the chosen \( u \), \( dv \), \( du \), and \( v \) into the formula: \[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx \].

Step 5: Second Integration by Parts

The integral \( \int e^{-x} \sin x \, dx \) requires integration by parts: Let \( u = \sin x \) and \( dv = e^{-x} \, dx \). Then \( du = \cos x \, dx \) and \( v = -e^{-x} \).

Step 6: Apply Second Integration Formula

Substitute again: \( \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int (-e^{-x}) \cos x \, dx \), which becomes \[-e^{-x} \sin x + \int e^{-x} \cos x \, dx \].

Step 7: Solve the System of Equations

Notice that \( \int e^{-x} \cos x \, dx \) appears on both sides of one equation derived from previous steps: \(-e^{-x} \cos x + e^{-x} \sin x + \int e^{-x} \cos x \, dx = 0 \). Isolate \( \int e^{-x} \cos x \, dx \) to get: \( \int e^{-x} \cos x \, dx = \frac{-e^{-x}(\cos x + \sin x)}{2} \).

Step 8: Evaluate the Definite Integral

Evaluate \( \lim_{b \to \infty} \left[ \frac{-e^{-x}(\cos x + \sin x)}{2} \right]_{0}^{b} \), where \( \int_{0}^{b} = \frac{-e^{-b}(\cos b + \sin b)}{2} - \frac{(\cos 0 + \sin 0)}{2} \).

Step 9: Simplify the Expression

Compute \( \lim_{b \to \infty} \frac{-e^{-b}(\cos b + \sin b)}{2} = 0 \) because \( e^{-b} \to 0 \) as \( b \to \infty \). Thus, this term vanishes. The final result is \( \frac{1}{2} \).

Key Concepts

Integration by Parts

Integration by Parts is a technique employed for solving integrals that are products of two functions, where each function is often more easily differentiated than integrated. The method is a significant tool in calculus and is derived from the product rule for differentiation. Here's a simplified breakdown:

• Choose which function is your "u" (to differentiate), and "dv" (to integrate). For productive choices, "u" is usually a polynomial, and "dv" is typically an exponential or trigonometric function.
• Differentiate "u" to find "du" and integrate "dv" to find "v".
• Substitute these into the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).

In our problem, the original integral function involves an exponential, \( e^{-x} \), and \( \cos x \). We first let \( u = \cos x \) and \( dv = e^{-x} \, dx \), which require repeated application to resolve the converging result of the integral.

Convergence and Divergence

Convergence and Divergence are terms used to determine if an improper integral has a finite value or not. For an improper integral with an infinite limit to be considered convergent, its evaluation must produce a finite number as the upper limit approaches infinity.

In our example, \( \int_{0}^{\infty} e^{-x} \cos x \, dx \) handles convergence by integrating and evaluating through limits:

• First, replace the infinity limit with a variable, usually denoted as \( b \), and later take the limit as \( b \to \infty \).
• Calculate the integral over a finite range, then apply limits to operative calculations.

The solution shows that as \( e^{-x} \to 0 \) when \( b \to \infty \), the oscillatory nature of \( \cos x \) vanishes due to the dominant decay of the exponential term, proving the integral converges to the value \( \frac{1}{2} \).

Limits in Calculus

Limits in calculus are foundational for evaluating expressions as quantities approach a certain value. They are the backbone of defining derivatives and integrals, especially for improper integrals judged on infinite boundaries.

For the integral \( \int_{0}^{\infty} e^{-x} \cos x \, dx \), limits are crucial to verify convergence by:

• Transitioning the problem into a scenario where \( b \to \infty \) for \( \int_{0}^{b} \).
• Applying limits helps compute expressions which, through strategic evaluation, confirm whether the limit results in a finite and constant term.

In this exercise, \( \lim_{b \to \infty} \frac{-e^{-b}(\cos b + \sin b)}{2} \) simplifies to zero, demonstrating the validity of using limits to verify the integral result of \( \frac{1}{2} \). This strategy wrapped the problem to conclude with convergence, supported by the decay to zero where needed.