Show that this improper integral converges

In summary, to show that ∫sin(ax) / xp dx converges on the interval [0, ∞] for 0 < p < 2, we can use g(x) = 1 / x^p as it approaches 0 as x approaches infinity. This satisfies the criteria of ∫f(x)g(x)dx converging, as g'(x) = -p * x^(-p-1) and for p > 0, the limit of 1 / x^p is 0 as x approaches infinity. Therefore, the integral converges on the given interval.
  • #1
Jamin2112
986
12

Homework Statement



Show that

∫sin(ax) / xp dx interval: [0, ∞]

converges if 0 < p < 2.

Homework Equations



This is the chapter where we learn that ∫f(x)g(x)dx converges if ∫f(x)dx is bounded and g'(x) is continuous, g'(x) < 0, and g(x) --> 0.

The Attempt at a Solution



I can't figure out where to start. Should I use g(x) = 1 / xp ?
 
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  • #2
Jamin2112 said:

Homework Statement



Show that

∫sin(ax) / xp dx interval: [0, ∞]

converges if 0 < p < 2.

Homework Equations



This is the chapter where we learn that ∫f(x)g(x)dx converges if ∫f(x)dx is bounded and g'(x) is continuous, g'(x) < 0, and g(x) --> 0.

The Attempt at a Solution



I can't figure out where to start. Should I use g(x) = 1 / xp ?
Yes. sin(ax) is not a good choice for g(x) since it doesn't approach 0 as x approaches infinity.
 
  • #3
Mark44 said:
Yes. sin(ax) is not a good choice for g(x) since it doesn't approach 0 as x approaches infinity.



Well, if g(x) = x-p, then g'(x) = -p * x-p-1. We're looking at what happens when x --> ∞. We'll need - p - 1 ≤ 1 (right?).

Obviously ∫sin(x) dx is bounded, since the area goes 1, 0, 1, 0, ... as x goes π/2, π, 3π/2, ...
 
  • #4
Jamin2112 said:
Well, if g(x) = x-p, then g'(x) = -p * x-p-1. We're looking at what happens when x --> ∞. We'll need - p - 1 ≤ 1 (right?).
No, not right. For p > 0,
[tex]\lim_{x \to \infty}\frac{1}{x^p} = 0[/tex]
Jamin2112 said:
Obviously ∫sin(x) dx is bounded, since the area goes 1, 0, 1, 0, ... as x goes π/2, π, 3π/2, ...
 

1. What is an improper integral?

An improper integral is an integral where one or both of the integration limits are infinity, or where the function being integrated is not defined at one or more points within the interval.

2. How do you determine if an improper integral converges?

To determine if an improper integral converges, you can use various tests such as the Comparison Test, the Limit Comparison Test, or the Ratio Test. You can also check for convergence by evaluating the integral using limits.

3. What is the significance of a convergent improper integral?

A convergent improper integral indicates that the function being integrated has a finite area under the curve, even though it may be undefined at certain points. This allows for the use of calculus techniques to analyze and solve problems involving these types of integrals.

4. Can an improper integral diverge?

Yes, an improper integral can diverge if the function being integrated does not approach a finite limit as the integration limits approach infinity, or if the function oscillates infinitely within the interval.

5. How can I use the concept of convergence to solve real-world problems?

The concept of convergence is essential in many scientific fields, such as physics, engineering, and economics. It allows for the accurate calculation of values and the prediction of outcomes in various scenarios, such as calculating the total energy of a system or determining the value of an investment over time.

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