Showing that a limit exists and showing $f$ is not integrable. Show that $\lim\limits_{n \to \infty} \int_{1}^{n} f$ exists while $f$ is not integrable over $[1,\infty)$.
Define $f(x)=\frac{\sin(x)}{x}$ for $1 \leq x <\infty$.
 A: $\def\limn{\lim_{n\to\infty}}$We have by partial integration
\begin{align*}
  \limn\int_1^n \frac{\sin x}x \,dx &= \limn \left(\left.-\frac{\cos x}x\right|_1^n - \int_1^n \frac{\cos x}{x^2}\, dx\right)\\
   &= \cos 1 - \int_1^\infty \frac{\cos x}{x^2}\, dx
\end{align*}
where the latter integral converges as one can see by comparision with $\int_1^\infty x^{-2}\, dx$.
For the non-integrability, note that one has
$$\int_{n\pi}^{(n+1)\pi} \frac{|\sin x|}x \, dx\ge \frac 1{(n+1)\pi}\int_0^{\pi}|\sin x|\, dx = \frac 2{(n+1)\pi} $$
and hence
$$ \int_1^\infty \frac{|\sin x|}x\, dx \ge \sum_{n=1}^\infty \frac 2{(n+1)\pi} = \infty. $$
A: Recalling the Sine integral
$$ Si(x)=\int_{0}^{x} \frac{\sin(x)}{x}dx ,$$
then we can write our integral as 
$$  \lim_{n \to \infty} Si(n) = \frac{\pi}{2} . $$
See the link below to see that the integral converges. Off course, there are several techniques to evaluate the above integral. On the other hand, it is a well known fact that the integral
$$ \int_{0}^{\infty} \left|\frac{\sin(x)}{x}\right| dx $$ does not converge. Then by the theorem "A function $f$ is Lebesgue integrable if and only if $|f|$ is.", one can show directly that $\frac{\sin(x)}{x}$ is not Lebesgue integrable.   
