Solving Convergence Problem: Integrals with Lebesgue Measure

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SUMMARY

The discussion focuses on the convergence of the integral \(\int_1^\infty \frac{k}{x^2+k^p\cos^2x}dm(x)\) using Lebesgue measure. It concludes that for \(p < 2\), the integral diverges, while for \(p \geq 2\), convergence is achievable. The substitution \(x=k^{p/2}u\) is utilized to analyze the integral's behavior, leading to the expression \(k^{1-p/2}\int_{k^{-p/2}}^1 \frac{du}{u^2 + \cos^2 k^{p/2} u}\), which indicates no singularity at zero. The discussion emphasizes the importance of bounding the integral from 1 to \(C k^{p/2}\) for convergence analysis.

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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?


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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.
 
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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!
 

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