Let g(x) be the function from R to R defined by $g(x)= 1$ if $x=0$, $\frac{\sin x}x$ otherwise.
Define the function $g_n (x)= g(x)$ if $-n < x < n$ and $x=0$ otherwise.
Show that for every n, $g_n$ is integrable on $\mathbb R$. And that $g_n$ converges to $g$.
$ |g_n(x) - g(x)| \neq 0 \Rightarrow |x|\ge n $ so $$ \sup_x |g_n(x) - g(x)| = \sup_{|x|\ge n} \frac{|\sin x|}{|x|} \le \frac 1n $$ then $g_n\to g$ and the convergence is uniform.
As far as the integrability is concerned, as $g_n$ is bounded and non zero on a bounded interval, it is Lebesgue integrable.