In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $\text{Proposition (1)}$ is sound ?
$\text{Proposition (1)}$
$$\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$
$\text{Proof}$
Assume that $R>1$ define $\gamma_{R}$ such that,
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R$$
$$\gamma_{R}^{2}(t) = Re^{it} \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $ 
Consider our choice $f$ and that,
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz.$$
Clearly it's obvious that
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3}dz = \sum_{\psi = 1,2} \oint_{\gamma_{R}^{\psi}}\frac{e^{iz}}{z^{2}+3}dz.$$
It's trivial that,
$$\oint_{\gamma_{R}^{1}}e^{iz}/({z^{2}+3})dx \rightarrow \int_{0}^{\infty} \frac{e^{ix}}{x^{2}+3}\operatorname{dx}$$
It's natural to claim that,
$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \Bigg| \rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{iz}}{z^{2}+3}|\leq \pi R \cdot \frac{1}{R^{2} – 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty$$
Thus,
$$ \operatorname{Re}\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \operatorname{Re} \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}} = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}.$$
However we need to consider that
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz = 2 \pi i \sum_{j=1,2} \operatorname{Ind_{\gamma}} \cdot \operatorname{Res_{f}(P_{j})}$$
It's easy to observe that,
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz = \big(2\pi i \cdot \frac{ie^{\sqrt{3}}}{2 \sqrt{3}}) = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$
Best Answer
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$\text{Proof}$
To proceed one must consider that,
$t:=\sqrt{3}x$
Assume that $R>1$ define $\gamma_{R}$ such that,
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R$$
$$\gamma_{R}^{2}(t) = Re^{it} \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
Consider our choice $f$ and that,
$$\frac{1}{\sqrt{3}}\,\oint_{\gamma_{R}}\,\frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z $$
Clearly it's simple that $$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \sum_{\psi = 1,2} \frac{1}{\sqrt{3}} \oint_{\gamma_{R}^{\psi}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz.$$
Now it's imperative to claim that
$$\lim_{R\to\infty}\,\oint_{\gamma_R^1}\,\frac{\exp(\sqrt{3}z)}{z^2+3}\,\text{d}z= \frac{1}{\sqrt{3}} \,\int_{0}^{\infty}\, \, \frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$
Now at this leg of our journey important to conjecture that,
$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz \Bigg| \rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{(\sqrt{3}z)}}{z^2+3}dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{(\sqrt{3}z)}}{z^2+3}|\leq \pi R \cdot \frac{1}{R^{2} - 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty.$$
Thus, taking
$$\operatorname{Re} \bigg( \frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\text{d}t \bigg) \, = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}$$
However we need to consider that
After a trivial calculation we have that,
$$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}} \frac{e^{(\sqrt{3}z)}}{z^2+1}\, dz = 2 \pi i + \operatorname{Res}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$