convergence of $\int_{0}^{\infty}{\frac{x^2-1}{x^2+1}\frac{\sin(2x)}{x} dx}$

Question

How can I show that $\int_{0}^{\infty}{\frac{x^2-1}{x^2+1}\frac{\sin(2x)}{x} dx}$ converges? I tried the criteria listed in my book but none seem to apply

Answer 1

Note that $${\frac{x^2-1}{x^2+1}\frac{\sin(2x)}{x}}=\frac{\sin(2x)}{x} -\frac 2 {x^2+1}\frac{\sin(2x)}{x}$$ Also, it's a known fact that $$\int_{0}^{+\infty} \frac{\sin(u)}{u}du = \int_{0}^{+\infty} \frac{\sin(2x)}{x}dx$$ converges. So all we have to prove is that

$$\int_0^{+\infty}\frac 2 {x^2+1}\frac{\sin(2x)}{x}dx$$ converges. This is because the integrand converges to 4 as $x\rightarrow 0$, and its absolute value is bounded by $\frac 4 {x^3}$ as $x\rightarrow +\infty$.