We consider the function $f(z)=\frac{e^{iaz}}{(z^2+1)(z^2+4)}$.
$(z^2+1)(z^2+4)=0\Rightarrow$$z=\pm i,$$\pm 2i.$ Only i, 2i are in the upper-half plane.
$Res(f,i)=\frac{e^{iai}}{2i(-1+4)}=\frac{e^{-a}}{6i}$ $Res(f,2i)=\frac{e^{ia2i}}{(-4+1)4i}=\frac{e^{-2a}}{-12i}$. $I=2\pi i(\frac{e^{-a}}{6i}-\frac{e^{-2a}}{12i})=\pi(\frac{e^{-a}}{3}-\frac{e^{-2a}}{6})$
