In the text, "Complex Analysis 2nd Edition" by Eberhard Freitag and Rolf Busam I'm having difficulties verifying my proof of $\text{Proposition (1.0) - (1.3)}$. Is the following proof correct?
$\text{Proposition (1.0)}$
Let $\gamma_{R}^{1},\gamma_{R}^{2}$ be the curve sketched in the figure with $R>1$ ,and $f(z) = \frac{1}{1+z^{2}}$ . One must show that
$(1.1)$
$$\oint_{\gamma}f(z)dz = \oint_{\gamma_{R}^{1}}f(z)dz + \oint_{\gamma_{R}^{2}}f(z)dz = \pi$$
$(1.2)$
$$\lim_{R \rightarrow \infty} \bigg | \oint_{\gamma_{R}^{2}} f(z) dz \bigg | = 0. $$
$(1.3)$
$$\int_{- \infty}^{\infty} \frac{1}{1+x^{2}}dx = \lim_{R \rightarrow \infty} \int_{-R}^{R} \frac{1}{1+x^{2}}dx = \pi $$
One can pick a Semicircular Contour $\gamma_{R}$ assuming that $R > 1$. Indeed we have
$$\gamma_{R}^{1}(t) = t + i0 \, \, \, \, \, \, \, \, \, \, \, \, \, \, \text{if} \, \, -R \leq t \leq R$$
$$\gamma_{R}^{2}(t) = Re^{it} \, \, \text{if} \, \, \, \, 0\leq t \leq \pi$$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \,\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, ,\, \, \, $ 
$\text{Remark}$
We call these two curves, take together, $\gamma$ or $\gamma_{R}$
To prove $(1.2)$ it's trivial to notation that
$$\oint_{\gamma} \frac{1}{1+z^{2}}dz = \oint_{\gamma_{R}^{1}}\frac{1}{1+z^{2}}dz + \oint_{\gamma_{R}^{2}}\frac{1}{1+z^{2}}dz.$$
Finally we evaluate the integral using the Cauchy Integral Formula
$(**)$
$$\oint_{\gamma_{R}^{1} + \gamma_{R}^{2}} \frac{1}{1+z^{2}}dz = \oint_{\gamma_{R}^{1} + \gamma_{R}^{2}} \frac{f(z)}{(z+1)^{}}dz = (2 \pi i) \cdot f(z) = \pi.$$
To provide a proof for $(1.2)$, one needs to craft the upper bound for $\gamma_{R}^{2}(t)$ hence we will need the Estimation Lemma.
Using the Estimation Lemma one has the following bound,
$(***)$
$$\bigg |\oint_{\gamma_{R}^{2}} \frac{1}{1+z^{2}} dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{1}{1+z^{2}}|\leq \pi R \cdot \frac{1}{R^{2} - 1}.$$
It's trivial to see that,
$$\lim_{R \rightarrow \infty}\bigg | \oint_{\gamma_{R}^{2}}\frac{1}{1+z^{2}} dz \bigg| \rightarrow 0.$$
Finally combining $(**)$, $(***)$ we get the following
$$\pi = \lim_{R \rightarrow \infty} \oint_{\gamma_{R}} \frac{1}{1+z^{2}}dz = \lim_{R \rightarrow \infty} \oint_{\gamma_{R}^{1}}\frac{1}{1+z^{2}}dz + \lim_{R \rightarrow \infty }\oint_{\gamma_{R}^{2}}\frac{1}{1+z^{4}}dz = \int_{-\infty}^{\infty} \frac{1}{1 + t^{4}}dt + 0.$$