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Show that this improper integral converges

  • Thread starter Jamin2112
  • Start date
  • #1
986
9

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 ?
 

Answers and Replies

  • #2
33,481
5,169

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
986
9
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
33,481
5,169
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]
Obviously ∫sin(x) dx is bounded, since the area goes 1, 0, 1, 0, ... as x goes π/2, π, 3π/2, ...
 

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