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Magazine Chapter 7: Problem 26
Decide if the improper integral converges or diverges. $$\int_{1}^{\infty} \frac{2+\cos \phi}{\phi^{2}} d \phi$$
Short Answer
Expert verified
The improper integral converges.
Step by step solution
01
Recognizing the Improper Integral
We are dealing with the integral \( \int_{1}^{\infty} \frac{2+\cos \phi}{\phi^{2}} d \phi \). This is an improper integral because the upper limit of integration is infinite.
02
Simplifying the Integrand
Notice that the integrand can be separated into two parts: \( \frac{2}{\phi^2} + \frac{\cos \phi}{\phi^2} \). This will help in determining convergence.
03
Evaluating \( \int_{1}^{\infty} \frac{2}{\phi^2} d \phi \)
We know \( \int_{1}^{\infty} \frac{2}{\phi^2} d \phi \) converges, as \( \frac{2}{\phi^2} \) behaves like \( k x^{-2} \), a \( p\)-integral with \( p = 2 > 1 \). Evaluating, we get: \( \int_{1}^{\infty} \frac{2}{\phi^2} d \phi = \lim_{b \to \infty} \left[ -\frac{2}{\phi} \right]_{1}^{b} = 2 \).
04
Evaluating \( \int_{1}^{\infty} \frac{\cos \phi}{\phi^2} d \phi \)
The term \( \frac{\cos \phi}{\phi^2} \) converges absolutely. By comparison, \( \left| \frac{\cos \phi}{\phi^2} \right| \leq \frac{1}{\phi^2} \). Since \( \int_{1}^{\infty} \frac{1}{\phi^2} d \phi \) converges, by the comparison test, \( \int_{1}^{\infty} \frac{\cos \phi}{\phi^2} d \phi \) converges.
05
Conclusion of Convergence
Since both \( \int_{1}^{\infty} \frac{2}{\phi^2} d \phi \) and \( \int_{1}^{\infty} \frac{\cos \phi}{\phi^2} d \phi \) converge, the original integral \( \int_{1}^{\infty} \frac{2+\cos \phi}{\phi^{2}} d \phi \) also converges.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Comparison Test
The comparison test is a valuable tool in determining the convergence of integrals and series by comparing them with another function that has a known behavior. In essence, if you can find a function that is either always greater than or equal to your given function, and it converges, then your original function must also converge.
Here's how it generally works:
Here's how it generally works:
- Compare the function you are interested in with another function that you know converges or diverges.
- If your function is smaller and the comparison function converges, then your function also converges.
- Conversely, if your function is larger and the comparison function diverges, then your function also diverges.
- The term \( \frac{\cos \phi}{\phi^2} \) was compared to \( \frac{1}{\phi^2} \), a known convergent function.
- Since \( \left| \frac{\cos \phi}{\phi^2} \right| \leq \frac{1}{\phi^2} \), and \( \int_{1}^{\infty} \frac{1}{\phi^2} d\phi \) converges, \( \int_{1}^{\infty} \frac{\cos \phi}{\phi^2} d\phi \) converges by the comparison test.
p-integrals
P-integrals are a type of integrals of the form \( \int_{a}^{b} \frac{1}{x^p} \, dx \), where \( p \) is a constant exponent. These are well-known benchmark integrals used to evaluate whether a given improper integral converges or diverges based on the value of \( p \).
The rule for p-integrals is quite simple:
The rule for p-integrals is quite simple:
- If \( p > 1 \), the integral converges.
- If \( p \leq 1 \), the integral diverges.
- The expression \( \frac{2}{\phi^2} \) behaves like \( \frac{k}{\phi^p} \) with \( p = 2 \) and \( k = 2 \).
- Since \( p = 2 \) is greater than 1, we know \( \int_{1}^{\infty} \frac{2}{\phi^2} d\phi \) converges.
- This property was essential for concluding the convergence of the original integral.
Absolute Convergence
Absolute convergence is a stronger form of convergence which implies that an integral or series converges when the absolute value of the function or terms are considered. If a function converges absolutely, then it converges normally too, but the reverse is not always true.
Here's why absolute convergence matters:
Here's why absolute convergence matters:
- Absolute convergence ensures that irrespective of the signs of the function elements involved, the integral or series will still converge.
- In mathematical terms, if \( \int_{a}^{b} |f(x)| \, dx \) converges, then \( \int_{a}^{b} f(x) \, dx \) converges as well.
- The term \( \frac{\cos \phi}{\phi^2} \) has an oscillating numerator \( \cos \phi \) which might seem problematic.
- By considering its absolute value \( \left| \frac{\cos \phi}{\phi^2} \right| \leq \frac{1}{\phi^2} \), we ensure convergence through the comparison test.
- Thus, absolute convergence provides an additional layer of confirmation that \( \int_{1}^{\infty} \frac{\cos \phi}{\phi^2} d\phi \) converges.
