does the integral $\int _{1}^{\infty }\!{\frac {\sin \left( \cos \left( x \right) +\sin \left( x\sqrt {3} \right) \right) }{x}}{dx}$ converges? I want to know if this integral converges or not : $\int _{1}^{\infty }\!{\frac {\sin \left( \cos \left( x \right) +\sin\left( x\sqrt {3} \right)  \right) }{x}}{dx}.$ I tried to integrate by parts or to use dirichlet's test, but it seems impossible to prove that $\int _{1}^{x}\!\sin \left( \cos \left( t \right) +\sin \left( t\sqrt {3} \right)  \right) {dt}$ is bounded. Do you have any idea how to solve this problem?
 A: The function $\varphi(t) = \cos(t)+\sin(t\sqrt{3})$ is uniformly almost periodic and has mean value $0$. It's easy to see that $\sin\varphi(t)$ has the same properties. Unfortunately, that doesn't necessarily mean that $\int^x_1\sin\varphi(t)\,dt$ is bounded. If it is, it's a uniformly almost periodic function of $x$, too. A sufficient condition for that would be that $0$ is not a limit point of the spectrum of $\sin\varphi(t)$, but it is: we'll see below that the spectrum contains all $k+l\sqrt{3}$ with integer $k,l$ so that $k+l$ is odd, and so the spectrum is dense on the real line.
Still we can prove boundedness, but we'll have to investigate the approximation of $\sin\varphi(t)$ by trigonometric polynomials in some detail. It's natural to use the Taylor series $$\sin z=\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)!}z^{2n+1}$$ for that.
Lemma For $n\ge0$, we have the expansion $$\varphi^{2n+1}(t)=\sum_{(k,l)\in I_{n}} a_{kl}\,f_{kl}(t),\tag1$$ where $$I_{n}=\{(k,l): k,l\in\mathbb{Z}, l\ge0,|k|+l\le2n+1, k+l\, {\rm odd}\},$$ $$f_{kl}(t)=\sin(k+l\sqrt{3})t$$ for odd $l$, $$f_{kl}(t)=\cos(k+l\sqrt{3})t$$ for even$l$, and $$\sum_{(k,l)\in I_{n}} |a_{kl}|\le2^{2n+1}$$
Proof: This is obviously true for $n=0$. For an induction step, we'll need the elementary trigonometric identities $$\sin\alpha\,\sin\beta=\frac12[\cos(\alpha-\beta)-\cos(\alpha+\beta)],$$
$$\sin\alpha\,\cos\beta=\frac12[\sin(\alpha-\beta)+\sin(\alpha+\beta)],$$
and $$\cos\alpha\,\cos\beta=\frac12[\cos(\alpha-\beta)+\cos(\alpha+\beta)].$$ Using those, we obtain $$\varphi^2(t)=1+\frac12\cos2t+\sin(\sqrt{3}+1)t+\sin(\sqrt{3}-1)t-\frac12\cos2\sqrt{3}t,\tag2$$ first, where the sum of the absolute values of the coefficients is exactly $4$.
For the induction step, we have to multiply (1) by (2) and use the above identities. It's easy to see that the sum of the absolute values of the coefficients multiplies at most by $4$, and that $k+l$ stays odd. q.e.d.
Now, we have $$\left|\int^x_1f_{kl}(t)\,dt\right|\le\frac2{|k+l\sqrt{3}|}=\frac{2|k-l\sqrt{3}|}{|k^2-3l^2|}\le2|k+l|\sqrt{3}\le2\sqrt{3}\,(2n+1),$$ since $k,l$ can't be both $0$ and thus $|k^2-3l^2|\ge1.$ This gives $$\left|\int^x_1\varphi^{2n+1}(t)\,dt\right|\le2\sqrt{3}\,(2n+1)\sum_{(k,l)\in I_{n}} |a_{kl}|\le2\sqrt{3}\,(2n+1)\,2^{2n+1}.$$ From the Taylor series, we now obtain
$$\left|\int^x_1\sin\varphi(t)\,dt\right|\le\sum^\infty_{n=0}\frac{2\sqrt{3}\,(2n+1)\,2^{2n+1}}{(2n+1)!}=4\sqrt{3}\,\sum^\infty_{n=0}\frac{2^{2n}}{(2n)!}=4\sqrt{3}\cosh2.$$
From this, with the help of the Dirichlet criterion, we see that the integral $\int^\infty_1\frac{\sin\varphi(t)}t\,dt$ converges.
A: Here is my idea to solve this problem: 
Note that $\cos x$ has positive values on $I_n = (\frac{2n-1}{2} \pi, \frac{2n+1}{2} \pi)$ for every $n \in \mathbb{Z}$ and that $\sin (\sqrt{3} x)$ has positive values on $J_k = (\frac{2k}{\sqrt{3}} \pi, \frac{2k+1}{\sqrt{3}} \pi)$ for every $k \in \mathbb{Z}$. 
For each fixed $n \in \mathbb{Z}$, we can find $k \in \mathbb{Z}$ such that 
$$\left(\frac{2n-1}{2}\right) \pi < \frac{2k}{\sqrt{3}} \pi < \left(\frac{2n+1}{2} \right) \pi,$$
which implies that for each $I_n$, there exists $J_k$ such that $I_n \cap J_k \neq \emptyset$. 
Here, we can choose a small subinterval $L \subseteq I_n \cap J_k$ that makes $\cos x + \sin (\sqrt{3} x) > \alpha $ for some $0< \alpha < 1$. 
Thus $$\frac{\sin(\cos x + \sin(\sqrt{3} x) )}{x} > \frac{\beta}{x}$$
on some sub-subinterval $\tilde{L} \subseteq L$ for some $0<\beta<\alpha$ ($\sin x$ is continuous). 
Since there are infinitely many such disjoint subintervals $L \subseteq (1,\infty)$, so the integral diverges if the lengths of $L$ and $\tilde{L}$ and the number $\beta >0$ are chosen appropriately.
