How would one prove the existence of the following indefinite integral

Question

Find out whether or not the following integral exists

$$\int_{0}^{\infty}\frac{\sin(x)}{x(x+1)}dx$$

My idea was to split the integral into two

$$\int_{0}^{\infty}\frac{\sin(x)}{x(x+1)}dx = \int_{0}^{1}\frac{\sin(x)}{x(x+1)}dx + \int_{1}^{\infty}\frac{\sin(x)}{x(x+1)}dx$$

I'm pretty sure both integrals exist but I can't seem to find a way to prove this. Can someone please explain how one would argue.

Answer 1

Actually, it's much simpler to note that$$\bigl(\forall x\in[1,+\infty)\bigr):\left|\frac{\sin x}{x(x+1)}\right|\leqslant\frac1{x^2}.$$Note that there's no problem at $0$, since$$\lim_{x\to0}\frac{\sin x}{x(x+1)}=1.$$

Answer 2

Split in two

• near $0$: $\;\sin x\sim_0 x$, so $$\frac{\sin x}{x(x+1)}\sim_0 \frac 1{x+1},$$ and as the latter is integrable on $[0,1]$ and they're positive functions, the former is too.
• near $\infty$ : $$\biggl|\frac{\sin x}{x(x+1)}\biggr|=O\biggl(\frac 1{x^2}\biggr),$$

which is integrable on $[1,+\infty)$.