Limit involving the Sine integral function $$
\mbox{Prove that}\qquad
\lim_{x \to \infty}\left[\vphantom{\large A}%
x\,\mathrm{si}\left(x\right)+ \cos\left(x\right)\right]
= 0
$$
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.
 A: Hint: Use integration by parts to show that
$$-si(x)=\frac{\cos(x)}{x}-\int_x^{+\infty}\frac{\cos(t)}{t^2}dt=\frac{\cos(x)}{x}+\frac{\sin(x)}{x^2}-2\int_x^{+\infty}\frac{\sin(t)}{t^3}dt$$
A: Define $$Si(x) = \int^x_0 \frac{sin t}{t} dt$$
By the relation $Si(x) = \frac{\pi}{2} + si(x) $
And the asymptotic series expansion of $si(x)$:
$$si(x) = -\frac{\cos x}{x} (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\frac{\sin x}{x} (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7)) $$
Therefore
$$xsi(x)+\cos x = -\cos x (1- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\sin x (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7))+\cos x \\
=   -\cos x (- \frac{2!}{x^2} +\frac{4!}{x^4} -O(1/x^6))-\sin x (\frac{1}{x}- \frac{3!}{x^3} +\frac{5!}{x^5} -O(1/x^7))$$
When x goes up to infinity, $\cos x$ falter between $[-1,1]$ but $x^n$ rise to infinity  so $1/x^n$ goes to 0 where $n \geq 0$ is a natural number. 
A: 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 $(3)$,
$$ \lim_{x\to +\infty}f(x) = 1-\int_{1}^{+\infty}\frac{dt}{t^2} = \color{red}{0}.\tag{5}$$
