Residue Theorem: compute the integral $\int_0^\infty \frac{x \sin x}{x^4+4a^4}$ 
Compute the integral $$\int_0^\infty \frac{x \sin x}{x^4+4a^4}$$

Since, it's an even function I can rewrite the expression as $$\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin x}{x^4+4a^4}$$.
In the previous part I found that the integral along the upper semicirlce as $R \rightarrow \infty$$$\frac{1}{2}\int_{C} \frac{z e^{iz}}{z^4+4a^4}\rightarrow 0$$.
I'm left to evaluate $$\oint \frac{z\sin z}{z^4+4a^4}$$. Should I compute the residues of sin or should I rewrite $$\sin z=\frac{e^{it}-e^{-it}}{2i}$$
I found the roots of $z^4+4a^4$ to be $z=\sqrt2 ae^{\pi/4+2\pi k/4}$ for $k=0,1,2,3$.
But I can't seem to be able to compute the answer.

Write the integral of interest $I(a)$ as 
$$\begin{align}
I(a)&=\frac12\int_{-\infty}^\infty \frac{x\sin(x)}{x^4+4a^4}\,dx\\\\
&=\text{Im}\left(\frac12\int_{-\infty}^\infty \frac{xe^{ix}}{x^4+4a^4}\,dx\right) \tag 1\\\\
&=\lim_{R\to \infty}\text{Im}\left(\frac12\oint_{C_R}\frac{ze^{iz}}{z^4+4a^4}\,dz\right) 
\tag 2\\\\
\end{align}$$
where $C_R$ is the closed contour in the upper-half plane comprised of the line segment from $-R$ to $R$ and the semicircle centered at the origin with radius $R$.  The equivalence of $(1)$ and $(2)$ is guaranteed since, as already determined in the OP, the contribution from the integral over the semi-circle vanishes as $R\to \infty$.  
Therefore, we have 
$$I(a)=\text{Im}\left(\frac12\,(2\pi i) \,\sum \text{Res}_{\text{Im}(z)>0}\left(\frac{ze^{iz}}{z^4+4a^4}\right)\right)$$
where the residues in the upper-half plane are at $z=4^{1/4}ae^{i\pi/4}$ and $z=4^{1/4}ae^{i3\pi/4}$
Can you finish now? 
