Existence of improper integral. I want to show that the improper integral $$\int_0^{\infty} \frac{\sin(x)}{x(1+x)}dx$$ exists. In order to do that, I split the integral into  $\int_0^1 \frac{\sin(x)}{x(1+x)}dx + \int_1^{\infty} \frac{\sin(x)}{x(1+x)}dx$, so that I can compute a value for $\int_0^1 \frac{\sin(x)}{x(1+x)}dx$, which proves the existence for that part.
For the other part, is it sufficient to show that $\int_1^{\infty} \frac{\sin(x)}{x(1+x)}dx < \int_1^{\infty} \frac{1}{x^2}dx$ knowing that $\int_1^{\infty} \frac{1}{x^2}dx$ exists, to say that $\int_1^{\infty} \frac{\sin(x)}{x(1+x)}dx$ exist and therefore the whole improper integral exists? Thanks for an answer in advance. 
 A: $\sin x$ may be negative in $[1,\infty)$, so $\int_1^\infty\frac{\sin x}{x(1+x)}\,dx < \int_1^\infty\frac1{x^2}\,dx$ is not quite correct. What you can say, by the squeeze theorem, is that
$$\left|\int_1^\infty\frac{\sin x}{x(1+x)}\,dx\right|\le\int_1^\infty\frac1{x(1+x)}\,dx\le\int_1^\infty\frac1{x^2}\,dx$$
and since the last integral converges, so does the leftmost integral without the absolute value signs.
A: Since $\mathcal{L}(\sin x)=\frac{1}{s^2+1}$ and $\mathcal{L}^{-1}\left(\frac{1}{x(x+1)}\right)=1-e^{-s}$ we have
$$ \int_{0}^{+\infty}\frac{\sin x}{x(x+1)}\,dx = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{ds}{(s^2+1)e^s} $$
where the last integral is clearly positive, but bounded by
$$\sqrt{\int_{0}^{+\infty}\frac{ds}{(s^2+1)^2}\int_{0}^{\infty}\frac{ds}{e^{2s}}}=\sqrt{\frac{\pi}{8}}$$
by the Cauchy-Schwarz inequality. $\frac{\pi}{2}-\sqrt{\frac{\pi}{8}}$ is a fairly sharp approximation of the actual value of the integral; a crude and much more elementary bound can be obtained by integrating $0\leq \frac{\sin x}{x(x+1)} \leq \frac{1}{x+1}$ over $(0,1]$ and $\frac{|\sin x|}{x(x+1)}\leq\frac{1}{x(x+1)}$ over $[1,+\infty)$.
