evaluate the integral Evaluate the integral from: 
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx$$
The way I approach this problem is 
$$\int_0^{\infty} \frac{x \cdot \sin(2x)}{x^2+3}dx  = \frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz$$
and
$$ \text{Res}_{i\sqrt3}(f(z)) = \frac{e ^{-2\sqrt3}}{2\sqrt3}$$
Then, the integral will be: 
$$\frac{1}{2}\int_{-\infty}^{\infty} \frac{z \cdot e^{i2z}}{(z - i\sqrt{3})(i+i\sqrt{3})}dz = \frac{1}{2} \cdot 2\pi i \cdot \frac{e^{-2\sqrt3}}{2\sqrt3} = \frac{\pi i e^{-2\sqrt3}}{2\sqrt3}$$
Is my approach correct? if not, can someone show me? Sorry because I just learn about residue theorem and don't know if my work is correct or not.

First of all, keep in mind that 
$$\sin(2z)=\Im (e^{2iz})$$
Second, the residue is wrong, you've forgotten the $z$ in the numerator:
$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{P(z).e^{i2z}}{Q'(z)}$$
$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{z.e^{i2z}}{2z}$$
$$\text{Res}_{i\sqrt{3}}(f(z))=\frac{e^{i2z}}{2}$$
So the integral $\text{I}$ is:
$$\text{I}=\Im(2i\pi\text{Res}_{i\sqrt{3}}(f(z)))=\Im(2i\pi\frac{e^{-2\sqrt{3}}}{2})=\pi\frac{e^{-2\sqrt{3}}}{2}$$
