calculus - Evaluate the following improper integral with bounds.

I need ideas for solve this improper integral, i know is hard and is a bonus for my analysis course, so i would really appreciate your help, thanks

$$\int_{1}^{\infty}\dfrac{x\sin(2x)}{x^2+3}dx$$

Hint: $$\begin{cases} \sin(\theta)\geq \dfrac{2\theta}{\pi},& 0\leq\theta\leq \dfrac{\pi}{2}\\\\\sin(\theta)\geq \dfrac{-2\theta}{\pi}+2,& \dfrac{\pi}{2}\leq\theta\leq{\pi}\end{cases}$$

In order to bound the integral

$$\int_{0}^{\pi} e^{-2R\sin(\theta)}d\theta$$

I don't know a nice and beauty approach in order to attack this properly to obtain a closed answer, so....

Answer

Hint: Use
$$\int_1^\infty \frac{x \sin(2x)}{x^2+3} \mathrm{d} x = \Im \int_1^\infty \frac{x \mathrm{e}^{2 i x} }{x^2+3} \mathrm{d} x = \frac{1}{2} \Im \int_1^\infty \mathrm{e}^{2 i x} \left( \frac{1}{x - i \sqrt{3}} - \frac{1}{x + i \sqrt{3}} \right) \mathrm{d} x$$
Then check out the definition of the exponential integral.