We have
$$ \int\limits_0^{\pi} \frac{ \sin x }{x} \mathrm{d} x $$
The natural thing to do is to realize that $\sin x < x $ on this interval. Thus,
$$ \frac{ \sin x }{x} < 1 $$
and thus the comparison test gives convergence. Is there another way to show convergence?
The integral is not improper as written. Observe also that if $f(x)=g(x)$ for all values of $x\in [a,b]$ except for a single point $y\in[a,b]$, then $\int_a^b f(x)\,dx=\int_a^b g(x)\,dx$. If we define $$f(x)=\begin{cases}\frac{\sin(x)}{x} & x\neq 0\\ 1 & x=0,\end{cases}$$ then $f$ is continuous on $[0,\pi]$. Therefore, the fundamental theorem of calculus guarantees that $\int_0^\pi f(x)\,dx$ converges. Since $f(x)=\sin(x)/x$ except at a single point ($x=0$), $$\int_0^\pi f(x)\,dx=\int_0^\pi\frac{\sin(x)}{x}\,dx.$$
For all $x$ $$\sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$$ $$\frac{\sin x}x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n}$$ $$\int\frac{\sin x}x dx= \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} x^{2n+1}$$ $$\int_0^\pi\frac{\sin x}x dx= \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} \pi^{2n+1}$$ which is an alternating series.Let $$u_n=\frac{\pi^{2n+1}}{(2n+1)(2n+1)!} $$ $$\frac{u_{n+1}}{u_n}=\frac{\pi ^2 (2 n+1)}{2 (n+1) (2 n+3)^2}$$ which, for large $n$ behaves like $ \frac{\pi^2}{4 n^2}$ and then convergence.
For illustration purposes, let us compute the partial sums $$S_p=\sum^{p}_{n=0} \frac{(-1)^n}{(2n+1)(2n+1)!} \pi^{2n+1}$$ We should get $$\left( \begin{array}{cc} p & S_p \\ 0 & 3.1415926535897932385 \\ 1 & 1.4190217269064698954 \\ 2 & 1.9290545348819389842 \\ 3 & 1.8434453164075401160 \\ 4 & 1.8525726371421099192 \\ 5 & 1.8519025979652267964 \\ 6 & 1.8519384674118243051 \\ 7 & 1.8519370063882611164 \\ 8 & 1.8519370531650493604 \\ 9 & 1.8519370519572373079 \\ 10 & 1.8519370519829166634 \\ 11 & 1.8519370519824593386 \\ 12 & 1.8519370519824662594 \\ 13 & 1.8519370519824661694 \\ 14 & 1.8519370519824661704 \end{array} \right)$$
I cannot resist to once more use the approximation proposed more than 1,400 years ago (see here) $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ which would give $$\int\frac{\sin x}x dx=-2 \left(\log \left(4 x^2-4 \pi x+5 \pi ^2\right)+\tan ^{-1}\left(\frac{1}{2}-\frac{x}{\pi }\right)\right) $$ $$\int_0^\pi\frac{\sin x}x dx=2 \tan ^{-1}\left(\frac{4}{3}\right)\approx 1.85459$$
Using the approximation discussed here $$\sin(x)=\sum_{i=1}^3 a_i\big((\pi-x)x\big)^i\qquad (0\leq x\leq\pi)$$ we should arrive to $$\int_0^\pi\frac{\sin x}x dx=\frac{12972960}{\pi ^7}-\frac{1474704}{\pi ^5}+\frac{16296}{\pi ^3}\approx 1.85196$$
