Evaluate $int_{ - infty}^{infty} frac{dx}{1+x^2}$ using complex integration

I'm trying to evaluate the real integral $$\int_{ - \infty}^{\infty} \frac{dx}{1+x^2}$$

Denote $\Gamma_{1}=\left[-R,R\right]\ \Gamma_{2}=Re^{it}$, for $t\in\left[0,\pi\right]$,
and let $\gamma$ be a small circle around $i$ so $\gamma$ is inside
the area bounded by $\Gamma_{1}\cup\Gamma_{2}$. By Cauchy's theorem:
$$
\int_{\Gamma_{1}}f\left(z\right)dz+\int_{\Gamma_{2}}f\left(z\right)dz=\int_{\gamma}f\left(z\right)dz
$$
And calculating $\int_{\gamma}f\left(z\right)dz$ gives us $\pi$

(operating Cauchy's formula on the function $\frac{1}{z+i}$). so
we got
$$\int_{\Gamma_{1}}f\left(z\right)dz+\int_{\Gamma_{2}}f\left(z\right)dz=\pi$$
now I need to show that
$$\lim_{R\rightarrow\infty}\int_{\Gamma_{2}}f\left(z\right)dz=0$$
and I'm stuck.

Answer

You can apply Estimation lemma. Since
$$
\left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right| \le \frac{\pi R}{R^2 -1}

$$
for large $R$,
$$
\lim_{R\to\infty}\left|\int_{\Gamma_2} \frac{1}{1+z^2}dz\right|=0.
$$
Then you can get what you want.