The discussion revolves around the convergence of the improper integral ∫sin(ax) / xp dx over the interval [0, ∞], specifically under the condition that 0 < p < 2.
Participants are actively engaging with the problem, examining the implications of their choices for g(x) and discussing the boundedness of the integral of sin(x). Some guidance has been offered regarding the behavior of g'(x) and the conditions for convergence, but no consensus has been reached on the best approach.
There is an ongoing examination of the conditions under which the integral converges, particularly focusing on the limits and behavior of the functions involved as x approaches infinity.
Yes. sin(ax) is not a good choice for g(x) since it doesn't approach 0 as x approaches infinity.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 ?
Mark44 said:Yes. sin(ax) is not a good choice for g(x) since it doesn't approach 0 as x approaches infinity.
No, not right. For p > 0,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?).
Jamin2112 said:Obviously ∫sin(x) dx is bounded, since the area goes 1, 0, 1, 0, ... as x goes π/2, π, 3π/2, ...