Find every $a>0$ such that $\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges

Question

We know that $\forall x \in [2,+\infty)$,

$$\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$

And it's easy to prove that $\int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx$ converges if and only if $a>3$.

But is it legitimate to conclude that our first integral converges if and only if $a>3$? The Comparison Criterion says that if $0 \leq f\leq g$, then, if $g$ converges, $f$ converges. But the converse isn't valid. Are we losing any possible values for $a$?

For example, the following equivalence holds:

$$\int_2^{\infty} \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} dx = \int_2^{\infty} \frac{2x^6}{(x-1)^2(x+1)^2x^a} dx - \int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$$

Shouldn't we prove that $\int_2^{\infty} \frac{x^6(\sin^2 x)}{(x-1)^2(x+1)^2x^a} dx$ converges to $0$?

I'm confused.

Answer 1

It is true that $\frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$. What is also true is that $\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a}$. Altogether, this inequality is $$\frac{x^6}{(x-1)^2(x+1)^2x^a} \leq \frac{x^6(1+\cos^2 x)}{(x-1)^2(x+1)^2x^a} \leq \frac{2x^6}{(x-1)^2(x+1)^2x^a}$$

The left-hand side diverges for $a \leq 3$. The right-hand side converges for $a > 3$. Therefore, the integral converges if and only if $a > 3$.

Answer 2

Let $f(x)$ be the given integrand and $g(x)=x^{2-a}(1+\cos^{2}x)$. You can easily verify that $\frac {f(x)} {g(x)} \to 1$ as $ x \to \infty$. Hence $\int_2^{\infty} f(x)dx$ is finite iff $\int_2^{\infty} g(x)dx$ is finite. Now using the fact that $x^{2-a} \leq g(x) \leq 2x^{2-a}$ we can conclude that $\int_2^{\infty} g(x)dx$ is finite iff $\int_2^{\infty} x^{2-a}dx$ is finite. Clearly, this is true iff $a >3$.