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Magazine Chapter 8: Problem 23
Evaluate each improper integral or show that it diverges. \(\int_{0}^{\infty} e^{-x} \cos x d x\)
Short Answer
Expert verified
The integral converges to \( \frac{1}{2} \).
Step by step solution
01
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 \( b \to \infty \).
02
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.
03
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 \( \int u \, dv = uv - \int v \, du \).
04
Apply Integration by Parts Formula
Substitute the chosen \( u \), \( dv \), \( du \), and \( v \) into the formula: \[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x + \int e^{-x} \sin x \, dx \].
05
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 \( v = -e^{-x} \).
06
Apply Second Integration Formula
Substitute again: \( \int e^{-x} \sin x \, dx = -e^{-x} \sin x - \int (-e^{-x}) \cos x \, dx \), which becomes \[-e^{-x} \sin x + \int e^{-x} \cos x \, dx \].
07
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: \( \int e^{-x} \cos x \, dx = \frac{-e^{-x}(\cos x + \sin x)}{2} \).
08
Evaluate the Definite Integral
Evaluate \( \lim_{b \to \infty} \left[ \frac{-e^{-x}(\cos x + \sin x)}{2} \right]_{0}^{b} \), where \( \int_{0}^{b} = \frac{-e^{-b}(\cos b + \sin b)}{2} - \frac{(\cos 0 + \sin 0)}{2} \).
09
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 \( \frac{1}{2} \).
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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 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: \( \int u \, dv = uv - \int v \, du \).
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:
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.
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:
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.
