Help with second integral in a Cauchy's integral formula problem. I have been trying to do this problem for a while:

Use Cauchy's integral formula to evaluate $$\int_{-\infty}^\infty \frac{t\operatorname{sin}(\pi t)}{t^2+4}dt.$$

I have factored it into $$\int_{-\infty}^\infty \frac{t\operatorname{sin}(\pi t)}{t^2+4}dt=\frac{1}{2i}\left(\int_{-\infty}^\infty \frac{te^{i\pi t}}{t^2+4}dt-\int_{-\infty}^\infty \frac{te^{-i\pi t}}{t^2+4}dt\right).$$
So for first integral I am supposed to split it up into $\oint f dz - \int_{\gamma}f dz$ where $f$ is the integrand above and $\Gamma$ is a circle of radius $R$ in the upper half plane (ie $\gamma(t)=Re^{i\theta}:0\leq\theta<\pi$).  But I can't seem to evaluate the second integral in this formula - the $\int_{\gamma}f dz$.
I'm sure this is obvious but I could use some help.
 A: $$
\begin{align}
\int_{-\infty}^\infty\frac{t\sin(\pi t)}{t^2+4}dt
&=\frac1{2i}\int_{-\infty}^\infty\frac{te^{i\pi t}}{t^2+4}\mathrm{d}t
-\frac1{2i}\int_{-\infty}^\infty\frac{te^{-i\pi t}}{t^2+4}\mathrm{d}t\tag{1}\\
&=\frac1{2i}\int_{\gamma^+}\frac{te^{i\pi t}}{t^2+4}\mathrm{d}t
-\frac1{2i}\int_{\gamma^-}\frac{te^{-i\pi t}}{t^2+4}\mathrm{d}t\tag{2}\\
&=\frac{2\pi i}{2i}\left(\frac{2ie^{-2\pi}}{4i}\right)
+\frac{2\pi i}{2i}\left(\frac{-2ie^{-2\pi}}{-4i}\right)\tag{3}\\
&=\pi e^{-2\pi}\tag{4}
\end{align}
$$
$(1)\quad\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$
$(2)\quad\gamma^+$ follows the real axis and circles counterclockwise back through the upper half plane.
$\hphantom{(2)}\quad\gamma^-$ follows the real axis and circles clockwise back through the lower half plane.
$(3)\quad$evaluate the residues at $+2i$ and $-2i$ and note that $\gamma^-$ is clockwise.
We choose $\gamma^+$ for $e^{i\pi t}$ since $e^{i\pi t}$ decays in the upper half plane. Similarly, $e^{-i\pi t}$ decays in the lower half plane.
