Integrals of $\int^\infty _2 \frac{x}{x^3 -2\sin x}dx$

Question

How do I calculate the convergence of $\int^\infty_2\frac{x}{x^3 -2\sin x}dx$ ?

I know in generally, integration of (n-dimentional polynomial)/(n+2-dimentional polynomial) will be converge.

Answer 1

$$\left|\int^2_\infty \frac{x}{x^3 -2\sin x}dx\right| \leq \int_2^\infty \left|\frac{x}{x^3 -2\sin x}\right|dx \leq \int_2^\infty \frac{x}{|x^3 -2\sin x|}dx=\int_2^\infty \frac{x}{x^3 -2x}dx=\dfrac{1}{2\sqrt{2}}\ln\dfrac12(2+\sqrt{2})^2$$ Note: $$|x^3 -2\sin x|\geq|x^3| -2|\sin x|\geq|x^3| -2|x|=x^3-2x$$

Answer 2

The numerator $\ge x^3-2\ge\frac{3}{4}x^3$, so the integral's modulus is at most $\frac{4}{3}\int_2^\infty\frac{dx}{x^2}$. This is finite, so the original integral converges.

Answer 3

Your reasoning is correct, but you need to check that the denominator doesn't cancel in the given range. You might also remark that as the sine is bounded, it does not influence the growth of the term $x^3$. Hence the integrand is asymptotic to $x^{-2}$.