Prove that $\int_{0}^{\infty}\frac{\sin(x)}{x}dx=\frac{\pi}{2}$ using contour integral 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
 A: Let $$I = \int_0^\infty \frac{\sin(x)}{x} dx$$

The trick is to work with $\frac{e^{i z}}{z}$ rather than $\frac{\sin(z)}{z}$ when doing a contour integral.

In the case of $\frac{\sin(z)}{z}$ the singularity is removable. With $\frac{e^{i z}}{z}$ the singularity at 0 is essential. Consider the following contour:

We will let $R \to \infty$.
By Cauchy's theorem
$$
\begin{align}
0 =& \int_\gamma \frac{e^{i z}}{z} dz \\
 =& \int_{A \oplus B \oplus C \oplus D} \frac{e^{i z}}{z} dz \\
 =& \, 2 i \cdot I - i \pi + 0 \\
\end{align}
$$
thus
$$I = \tfrac{\pi}{2}.$$

$$
\begin{align}
& \int_{B \oplus D} \frac{e^{i z}}{z} dz \\
=& \int_{1/R}^{R} \frac{e^{i z}}{z} dz + \int_{-R}^{-1/R} \frac{e^{i z}}{z} dz \\
=& \int_{1/R}^{R} \frac{e^{i z}}{z} dz + \int_{R}^{1/R} -\frac{e^{-i z}}{-z} dz \\ =& \int_{1/R}^{R} \frac{e^{i z}}{z} dz - \int_{1/R}^{R} \frac{e^{-i z}}{z} dz \\
=& \int_{1/R}^{R} \frac{e^{i z} - e^{-i z}}{z} dz \\
=& \, 2i \cdot I \\
\end{align}
$$

For paths $A$ and $C$ note that $\gamma'(t) = i \pi \gamma(t)$.
$$
\begin{align}
& \int_{A} \frac{e^{i z}}{z} dz \\
=& - \int_0^1 \frac{e^{i (1/R) e^{i \pi t}}}{\gamma(t)} i \pi \gamma(t) dt \\
=& - i \pi \int_0^1 e^{i (1/R) e^{i \pi t}} dt \\
\to& -i \pi
\end{align}
$$

For the final part we will use estimation to show that
$$
\begin{align}
& \int_{C} \frac{e^{i z}}{z} dz \\
=& \int_0^1 \frac{e^{i R e^{i \pi t}}}{\gamma(t)} i \pi \gamma(t) dt \\
=& \, i \pi \int_0^1 e^{i R e^{i \pi t}} dt \\
\to& \, 0
\end{align}
$$
because
$$
\begin{align}
& \left| \int_0^1 e^{i R e^{i \pi t}} dt \right| \\
\le& \int_0^1 \left| e^{i R e^{i \pi t}} \right| dt \\
\le& \int_0^1 \frac{\left| e^{i R \cos(\pi t)} \right|}{\left| e^{R \sin(\pi t)} \right|} dt \\
\le& \int_0^1 \frac{1}{\left| e^{R \sin(\pi t)} \right|} dt \\
\le& 2 \int_0^{1/2} \frac{1}{\left| e^{R \sin(\pi t)} \right|} dt \\
\le& 2 \int_0^{1/2} \frac{1}{\left| e^{R \pi t} \right|} dt \\
\to& \, 0
\end{align}
$$
A: By the definition of the Laplace transform, we have
$$F(s) = \mathcal{L}\left\{\frac{\sin(t)}{t}\right\}= ∫_0^{+∞} \frac{\sin(t)}{t} e^{-st}dt = \tan^{-1} \left(\frac{1}{s}\right).$$
It follows that
$$F(0)= ∫_0^{+∞} \frac{\sin(t)}{t}dt = \lim_{s\to 0} \tan^{-1} \left(\frac{1}{s}\right) = \lim_{u\to∞} \tan^{-1} (u) = \frac{π}{2}.$$
No contour integral required and we get things done in 2 minutes! Let's go fishing!!!
