Improper Integral I can't solve $ (1+\sin x)/x^2$ I've tried to figure out if this improper integral convergse or diverges:
$$\int_1^\infty\frac{\sin(x) +1}{x^2}.$$
I want to use the "direct comparison" (sorry but I'm italian and in my 
 book it's called "Confronto Diretto"). My integrand function $f(x)$ is always  equal/bigger than zero, because  $\sin(x)\in [-1,1]$, so I can apply the method. 
I have to find a function always bigger/equal  than mine like:
$$g(x)=\frac{2}{x^2}$$
$G(x)$ converges and therefore also $F(x)$ converges. Is it right?
Thanks for the help
 A: Yes, it is correct. For $x\geq 1$,
$$0\leq\frac{\sin(x) +1}{x^2}\leq \frac{2}{x^2}.$$
which implies that $F(t)=\int_1^t\frac{\sin(x) +1}{x^2}dx$ is an increasing in $[2,+\infty)$ with a finite upper bound
$$F(t)\leq 2\int_1^{+\infty}\frac{1}{x^2}dx=2.$$
Therefore the limit $\lim_{t\to+ \infty}F(t)=\int_1^{+\infty}\frac{\sin(x) +1}{x^2}dx$ exists and it is finite.
P.S. We may say that we are using the direct comparison test for integrals.
A: As a minor addendum, I will outline an approach for a numerical evaluation.
We have $\int_{1}^{+\infty}\frac{dx}{x^2}=1$ and by a useful property of the Laplace transform 
$$\begin{eqnarray*} \int_{1}^{+\infty}\frac{\sin(x)}{x^2}\,dx &\stackrel{=}{=}& \int_{0}^{+\infty}\frac{\sin(1)\cos(x)+\cos(1)\sin(x)}{x^2}\,dx\\ &\stackrel{\mathcal{L}}{=}&\int_{0}^{+\infty}\frac{\sin(1)s^2+\cos(1)s}{s^2+1}e^{-s}\,ds\\&=&\sin(1)-\int_{0}^{+\infty}\frac{\sin(1)-\cos(1)s}{s^2+1}e^{-s}\,ds \end{eqnarray*}$$
where the last integral can be manipulated by integration by parts / estimated through the Cauchy-Schwarz inequality. For instance
$$\left|\int_{0}^{+\infty}\frac{\sin(1)-\cos(1)s}{s^2+1}e^{-s}\,ds\right|\leq\sqrt{\frac{1}{2}\int_{0}^{+\infty}\frac{\left(\sin(1)-\cos(1)s\right)^2}{(s^2+1)^2}\,ds}=\sqrt{\frac{\pi-2\sin 2}{8}}. $$
