Solving Convergence Problem: Integrals with Lebesgue Measure

[jeterfan]

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?

Homework Equations

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

$$k^{1+p/2} \int_{k^{-p/2}}^1 \frac{du}{k^p u^2 + k^p \cos^2 k^{p/2} u} = k^{1-p/2} \int_{k^{-p/2}}^1 \frac{du}{u^2 + \cos^2 k^{p/2} u}.$$
Now, the integral is has no singularity at 0, but I'm not sure where to go from here.

 

[jbunniii]

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 $[1, C k^{p/2}]$.

 

[jeterfan]

Ok, I think I can manage that. Thanks!