integration - How to solve $int_{-infty}^{infty} frac {sin(t)}{t^2+1} dt$?

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?

Answer

The integrand is and odd function of $t$ and the integral is absolutely convergent, so
$$\int_{-\infty}^{\infty}\frac{\sin t}{t^2+1}dt=0$$
EDIT: Writing out the theorem longhand,

$$\begin{align}\int_{-\infty}^{\infty}\frac{\sin t}{t^2+1}dt&=\lim_{M\rightarrow-\infty}\int_M^0\frac{\sin t}{t^2+1}dt+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin t}{t^2+1}dt\\ &=\lim_{M\rightarrow-\infty}\int_{-M}^0\frac{\sin(-u)}{u^2+1}(-du)+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin u}{u^2+1}du\\ &=\lim_{M\rightarrow-\infty}-\int_0^{-M}\frac{\sin u}{u^2+1}du+\lim_{N\rightarrow\infty}\int_0^N\frac{\sin u}{u^2+1}du\\ &=-\int_0^{\infty}\frac{\sin u}{u^2+1}du+\int_0^{\infty}\frac{\sin u}{u^2+1}du=0\end{align}$$
Where we have used the substitution $u=-t$ in the left integral and $u=t$ in the right integral. Notice that all that was required for this proof to work was that the integrand was an odd function, the interval of integration was $(-\infty,\infty)$, although an interval of $(-a,a)$ would have worked and that the function was Riemann integrable over the interval. This theorem can reduce some perfectly horrible-looking integrals to zero with relatively little effort.