I'm considering here the fact that $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {e^{iz}}{z^2+1} dz=0$$ , where $\Gamma$ is a contour defined as a semicircle centred about the origin, of radius $R>1$, in the upper half-plane of $\mathbb{C}$, positively oriented.
Now, I think, one could express $\int\limits_{-\infty}^{\infty} \frac {sin(t)}{t^2+1} dt$ as $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {sin(t)}{t^2+1} dt$$ But here's where I don't know what to do next. Would appreciate a hint.
On a side note, $$\int\limits_{-\infty}^\infty \frac {e^{iz}}{z^2+1} dz\ne 0$$ , so I'm not sure what I'm not seeing here because $$\lim\limits_{R\to\infty} \int_{\Gamma_R} \frac {e^{iz}}{z^2+1} dz=0$$, so where's the analogy?