# 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

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.

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 $$a > 3$$.
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 $$a >3$$.