I got linked to this old question from a more recent one, and I hope that you don't mind me adding a somewhat bizarre way of doing calculating this integral, using Bessel functions.

I'm aware of that this way is not the shortest way of obtaining the result, and the facts I give on Bessel functions are standard, and can be found in (probably) any book on Bessel functions. Therefore, some details will be left to be checked by the interested reader.

I have never seen this way to calculate the integral $\int_0^{+\infty}\frac{\sin x}{x}\,dx$, but I claim no originality. If someone has seen it, please tell in a comment. Here it goes:

Let us define the $n$th Bessel function $J_n$ by the integral
$$
J_n(x)=\frac{1}{\pi}\int_0^\pi \cos(n\theta-x\sin \theta)\,d\theta.
$$
We will only work with the cases $n=0$ and $n=1$. The function $J_n$ solves the Bessel differential equation
$$
x^2y''(x)+xy'(x)+(x^2-n^2)y(x)=0,
$$
and moreover $D J_0(x)=-J_1(x)$ and $D(xJ_1(x))=xJ_0(x)$.

Our first statement is that
$$
\frac{\sin x}{x}=\int_0^1\frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy.\tag{1}
$$
Indeed, define $f$ as
$$
f(x)=\int_0^1\frac{xy J_0(yx)}{\sqrt{1-y^2}}\,dy
$$
Then
$$
f'(x)=\int_0^1 \frac{yJ_0(xy)-xy^2 J_1(xy)}{\sqrt{1-y^2}}\,dy
$$
and
$$
f''(x)=-\int_0^1 \frac{xy^3J_0(xy)+y^2J_1(xy)}{\sqrt{1-y^2}}\,dy,
$$
and so
$$
f''(x)+f(x)=\int_0^1 xy\sqrt{1-y^2}J_0(xy)-\frac{y^2}{\sqrt{1-y^2}}J_1(xy)\,dy=0.
$$
(Here we integrated by parts in the last step.) Moreover,
$$
f(0)=0,\quad\text{and}\quad f'(0)=\int_0^1 \frac{y}{\sqrt{1-y^2}}\,dy =1.
$$
Thus $f(x)=\sin x$ and the equality (1) follows. Thus, we can write
$$
\int_0^{+\infty} \frac{\sin x}{x}\, dx
= \int_0^{+\infty}\int_0^1 \frac{y J_0(yx)}{\sqrt{1-y^2}}\,dy\, dx.
$$
Next, we change the order of integration, and use the integral
$$
\int_0^{+\infty} J_0(u)\,du=1,\tag{2}
$$
to find that
$$
\int_0^{+\infty} \frac{\sin x}{x}\, dx = \int_0^1 \frac{1}{\sqrt{1-y^2}}\,dy=\arcsin 1-\arcsin 0=\frac{\pi}{2}.
$$
I remains to prove (2), which certainly follows by letting $x\to 0^+$ in
$$
\int_0^{+\infty} J_0(u)e^{-xu}\,du=\frac{1}{\sqrt{1+x^2}}.
$$
This is just the Laplace transform of $J_0$, and one can use the representation $J_0(u)=\frac{2}{\pi}\int_0^{\pi/2} \cos(u\cos\theta)\,d\theta$ to obtain it,
$$
\begin{aligned}
\int_0^{+\infty}J_0(u)e^{-xu}\,du &= \int_0^{+\infty} e^{-xu}\frac{2}{\pi}\int_0^{\pi/2}\cos(u\cos\theta)\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2} \int_0^{+\infty}e^{-xu}\cos(u\cos\theta)\,du\,d\theta\\
&=\frac{2}{\pi}\int_0^{\pi/2}\frac{x}{x^2+\cos^2\theta}\,d\theta\\
&=\frac{2}{\pi}\biggl[\frac{\arctan\Bigl(\frac{x\tan \theta}{\sqrt{1+x^2}}\Bigr)}{\sqrt{1+x^2}}\biggr]_0^{\pi/2}\\
&=\frac{1}{\sqrt{1+x^2}}.
\end{aligned}
$$