Show that this improper integral converges

[Jamin2112]

Homework Statement

Show that

∫sin(ax) / x^p 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 / x^p ?

 

[Mark44]

Homework Statement

Show that

∫sin(ax) / x^p 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 / x^p ?

Yes. sin(ax) is not a good choice for g(x) since it doesn't approach 0 as x approaches infinity.

 

[Jamin2112]

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, ...

 

[Mark44]

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,
\lim_{x \to \infty}\frac{1}{x^p} = 0

Obviously ∫sin(x) dx is bounded, since the area goes 1, 0, 1, 0, ... as x goes π/2, π, 3π/2, ...