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.

Here is another approach:
$$f(a)=\int_0^{\infty} \frac{x \cdot \sin(ax)}{x^2+3}dx$$ take a Laplace transform with respect to $a$ to obtain
\begin{align}
\mathcal{L}(f(a))&=\int_0^{\infty} \frac{x^2}{(x^2+3)(x^2+s^2)}dx\\
&=\frac{\pi}{2\sqrt3+2s}
\end{align}
now take an inverse Laplace to obtain
$$f(a)=\frac{\pi}{2}   e^{-\sqrt{3} a}$$
Therefore $f(2)=\frac{\pi}{2}   e^{-2\sqrt{3}}$.
