Solving Convergence Problem: Integrals with Lebesgue Measure

  • Thread starter Thread starter jeterfan
  • Start date Start date
  • Tags Tags
    Convergence
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
jeterfan
Messages
2
Reaction score
0

Homework Statement



Consider the integrals [tex]\int_1^\infty \frac{k}{x^2+k^p\cos^2x}dm(x)[/tex], where m is the Lebesgue measure. For what p do the integrands have an integrable majorant? For what p do the integrals tend to 0?


Homework Equations





The Attempt at a Solution



Pick some large constant C. For [tex]x> C k^{p/2}[/tex], the denominator is approximately x2, so the integral is at least as big as

[tex]k\int_{Ck^{p/2}}^\infty \frac{dx}{x^2} = \frac{1}{C k^{p/2-1}}[/tex].
So, when p/2<1, (so p<2) the integral diverges.

When p≥2, that is not a problem, so we need to look at
[tex]\int_1^{Ck^{p/2}} \frac{k}{x^2+k^p \cos^2 x} dx[/tex].

Now, substitute x=[tex]k^{p/2}u[/tex]. The integral becomes

[tex]k^{1+p/2} \int_{k^{-p/2}}^1 \frac{du}{k^p u^2 + k^p \cos^2 k^{p/2} u} = k^{1-p/2}<br /> \int_{k^{-p/2}}^1 \frac{du}{u^2 + \cos^2 k^{p/2} u}.[/tex]
Now, the integral is has no singularity at 0, but I'm not sure where to go from here.
 
Last edited:
Physics news on Phys.org
There is no problem with the integral from ##1## to ##C k^{p/2}##:
$$\int_{1}^{C k^{p/2}} \frac{k}{x^2 + k^p \cos^2 x} dx \leq \int_{1}^{C k^{p/2}} \frac{k}{x^2} dx$$
So you just need a bound for ##1/x^2## on the interval ##[1, C k^{p/2}]##.
 
Ok, I think I can manage that. Thanks!