Solving Convergence Problem: Integrals with Lebesgue Measure

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Homework Statement



Consider the integrals \int_1^\infty \frac{k}{x^2+k^p\cos^2x}dm(x), 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 x> C k^{p/2}, the denominator is approximately x2, so the integral is at least as big as

k\int_{Ck^{p/2}}^\infty \frac{dx}{x^2} = \frac{1}{C k^{p/2-1}}.
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
\int_1^{Ck^{p/2}} \frac{k}{x^2+k^p \cos^2 x} dx.

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

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}.
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!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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