Solving the following exercise I have some doubts. I want to show that
\begin{equation\*}
 \int\_{0}^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}
\end{equation\*}

I tried the following: Let $\gamma$ be the upper semicircle
\begin{equation\*}
 \int\_{\gamma} \frac{\sin(z)}{z} dz = 0
 \end{equation\*}
how do we have to
\begin{equation\*}
 f(x) = \frac{\sin(x)}{x} = \frac{\sin(-x)}{-x} = f(-x)
 \end{equation\*}
Note that
\begin{eqnarray\*}
 0 = \int\_{\gamma} \frac{\sin(z)}{z} dz
 & = & \int\_{-R}^{R} \frac{\sin(z)}{z} dz + \int\_{C\_{R}} \frac{\sin(z)}{z} dz\\
 & = & \int\_{-R}^{R} \frac{\sin(x)}{x} dx + \int\_{C\_{R}} \frac{\sin(z)}{z} dz\\
 & = & 2\int\_{0}^{R} \frac{\sin(x)}{x} dx + \int\_{C\_{R}} \frac{\sin(z)}{z} dz\\
 \end{eqnarray\*}
then
\begin{eqnarray\*}
 -2 \int\_{0}^{R} \frac{\sin(x)}{x}dx 
 & = & \int\_{C\_{R}} \frac{\sin(z)}{z}dz\\
 & = & \int\_{0}^{\pi} \frac{\sin(Re^{i\theta})}{Re^{i\theta}} Rie^{i\theta} d\theta\\
 & = & i\int\_{0}^{\pi}\sin(Re^{i\theta})d\theta\\
 & = & -\int\_{u(0)}^{u(\pi)}\sin(u)udu\\
 & = & -\left(u\cos(u)|\_{u(0)}^{u(\pi)}-\int\_{u(0)}^{u(\pi)}-\cos(u)du \right)\\
 & = & -\left(Re^{i\theta}\cos(Re^{i\theta})|\_{0}^{\pi}-\sin(Re^{i\theta})|\_{0}^{\pi} \right)\to -\pi\text{ when }R\to\infty\text{ ?}\\
 \end{eqnarray\*}
I do not think this last step is correct, however I see it necessary, I do not know if my reasoning is wrong or I should approach the problem from another point of view.

Is my reasoning correct? I have my doubts about being able to take the closed curve, because at $ z = 0 $ there is a singularity and it is located on the border of the region enclosed by my curve

PD: This requires the use of contour integral, although another method is also welcome