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?

  • $\begingroup$ Intuitively, if there are no crazy cancellations which I am missing at the moment, this integral does not converge because the numerator is something of order 1, so something which oscillates back and forth between -1 and 1 and does not help with convergence. So you are left with the integral of $\frac{1}{x}$ which does not converge. $\endgroup$
    – Luke
    Aug 2, 2017 at 12:16
  • $\begingroup$ I think it does converge. It's not easy to prove that the integral of $\sin \left( \cos \left( x \right) +\sin\left( x\sqrt {3} \right) \right)$ is bounded, but it should be possible: this is a uniformly almost continuous function, the integral is bounded if $0$ is not its spectrum. But by the addition theorem, the function is a sum of products of purely periodic functions with incommensurable periods, so that should be the case. It's non-trivial to make that rigorous, though. $\endgroup$
    – user436658
    Aug 2, 2017 at 12:39
  • $\begingroup$ Oops, I meant "uniformly almost periodic function", sorry! $\endgroup$
    – user436658
    Aug 2, 2017 at 12:57

2 Answers 2


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.


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.

  • 1
    $\begingroup$ I'm afraid with this technique, you could prove the divergence of the integral over $\sin x/x,$ too. $\endgroup$
    – user436658
    Aug 2, 2017 at 12:49
  • $\begingroup$ @ProfessorVector Thank you for pointing out that. I was mistaken. $\endgroup$
    – cdamle
    Aug 2, 2017 at 12:54
  • $\begingroup$ I agree with @ProfessorVector $\endgroup$ Aug 2, 2017 at 12:54
  • $\begingroup$ I think we can solve it like this: if we take $f(x)=\int_{1}^{x}{sin(cos(t)+sin(t\sqrt{3}))dt}$ for any x>1. $f'(x)=sin(cos(x)+sin(x\sqrt{3})=0$ therefore $cos(x)+sin(x\sqrt{3})=0$ $(because -2<cos(x)+sin(x\sqrt{3})<2)$. $cos(x)+sin(x\sqrt{3})=0$ means $cos(x)=cos(\frac{\pi}{2}+x\sqrt{3})$ therefore $x=\frac{1}{\sqrt{3}-1}(\frac{\pi}{2}+2k\pi) \ or \ x=\frac{1}{\sqrt{3}+1}(\frac{\pi}{2}+2n\pi).$ then we study the sign of $f''(x)$ to know if it a max or a min $\endgroup$ Aug 2, 2017 at 14:24
  • $\begingroup$ @ron macguire The arguments of the maxima and minima are certainly easy to find, but there are infinitely many of them, and we still need $f(x)$ at those arguments to decide if it's bounded or not. $\endgroup$
    – user436658
    Aug 2, 2017 at 20:14

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