TITLE: Limit involving the Sine integral function
QUESTION [9 upvotes]: $$
\mbox{Prove that}\qquad
\lim_{x \to \infty}\left[\vphantom{\large A}%
x\,\mathrm{si}\left(x\right)+ \cos\left(x\right)\right]
= 0
$$
where we define 
$$\mathrm{si}\left(x\right) =
- \int^{\infty}_{x}\frac{\sin\left(t\right)}{t}\,\mathrm{d}t
$$
I have no clue how to start. I have verified the result using wolframalpha.

REPLY [3 votes]: If we set, using standard notations,
$$ f(x) = -\frac{\pi x}{2}+\cos(x)+x\,\text{Si}(x) = x\,\text{si}(x)+\cos(x) \tag{1}$$
we have:
$$ f'(x) = -\frac{\pi}{2}+\text{Si}(x) = -\int_{x}^{+\infty}\frac{\sin t}{t}\,dt \tag{2}$$
and since $f(0)=1$,
$$ f(x) = 1-\int_{0}^{x}\int_{u}^{+\infty}\frac{\sin t}{t}\,dt\,du =1-\int_{1}^{+\infty}\frac{1-\cos(t x)}{t^2}\,dt\tag{3}$$
Now 
$$ \lim_{x\to +\infty}\int_{1}^{+\infty}\frac{\cos(tx)}{t^2}\,dt = 0\tag{4}$$
by the Riemann-Lebesgue lemma, hence, by $(3)$,
$$ \lim_{x\to +\infty}f(x) = 1-\int_{1}^{+\infty}\frac{dt}{t^2} = \color{red}{0}.\tag{5}$$