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I need to find either those two integrals converges or not :

$$\int_0^\infty \sin ( \sin (x) )dx$$ $$\int_0^\infty \frac{\sin ( \sin (x) )} xdx$$

I don't want a proof that computes the integral ! (if it is possible in anyway, I don't know if it is even possible).


There are some suggestions. I'll show you i've done :

My attemps :

  • For the first one, I told that we know : If the infinite sum of a sequence converges, then the sequence converges to zero. Thus, it is the same for the integral. Here, since $\sin ( \sin (x) )$ does not have any limit at infinity, the integral can't be defined properly.

What do you think of my attemp? If it is okay, do you have any other idea to solve my problem?

  • Then, a second proof, following the suggestion :

$$ k \in \mathbb Z, \, x \in [ - \frac \pi 2 + 2 k \pi, \frac \pi 2 + 2 k \pi] : \, \ | \frac {x- 2k \pi }{2} | \leq | sin(x) | \leq | x - 2k \pi | $$ But then I don't know what to do... I thought that maybe we can use the squeeze theorem but I don't know how from there...

enter image description here

  • Finally for the second integral, I have no clue at all... I was suggested to compare the integral for $x$ and for $x+\pi$ and actually :

$$\int_0^\infty \frac{\sin ( \sin (x + \pi) )} { x + \pi } dx = \int_0^\infty -\frac{\sin ( \sin (x) )} { x + \pi } dx $$

thank you for reading me :)

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    $\begingroup$ The first doesn't converge, the second does. $\endgroup$ – peterh Jun 16 '18 at 14:48
  • $\begingroup$ that's also what I think, but why? $\endgroup$ – Marine Galantin Jun 16 '18 at 14:48
  • $\begingroup$ The first behaves like a sin, thus the area below the line oscillates like a cos. It has no limit. The second behaves like an $\frac{1}{x}-\frac{1}{x+1}$, thus it converges. These are only my intuitive impressions and they are not correct, axiomatical reasonings, this is why I give a comment and not an answer. Hopefully others will give you full answers, too. $\endgroup$ – peterh Jun 16 '18 at 14:52
  • $\begingroup$ I see that it behaves like a sin but why ? I was not sure why... You have that $-1 < sin x < 1 $ but then, what about $ sin(sin(x)) $? $\endgroup$ – Marine Galantin Jun 16 '18 at 15:02
  • $\begingroup$ Sin is zero in zero, and it becomes 1 roughly at 1.5 (half pi). Around zero, it is nearly linear. Thus, sin sin oscillates between sin 1 and sin -1. $\endgroup$ – peterh Jun 16 '18 at 15:04
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Let $F(t)=\int_0^t\sin(\sin(x))$. You have $\sin(\sin(x))\geq 0$ for all $0\leq x\leq\pi$ hence $a:=F(\pi)>0$. Since $x\mapsto\sin(\sin(x))$ is odd and $2\pi$-periodic we have $$\int_0^{2\pi}\sin(\sin(x))dx=\int_{-\pi}^\pi\sin(\sin(x))dx=0$$ and $$F(n\pi)=\int_0^{n\pi}\sin(\sin(x))dx= \begin{cases} a&2\nmid n\\ 0&2\mid n \end{cases}$$ hence cannot converge for $n\to\infty$.

Since $F(2n\pi)=0$ for $n\in\Bbb N$ and since $F$ is bounded, integration by parts yelds for $n>0$ $$\left|\int_{2n\pi}^{2(n+1)\pi}\frac{\sin(\sin(x))}xdx\right| =\left|\int_{2n\pi}^{2(n+1)\pi}\frac{F(x)}{x^2}\right| \leq\frac{\sup|F|}{2\pi n^2}$$ Consequently, $$\int_0^\infty\frac{\sin(\sin(x))}xdx=\int_0^{2\pi}\frac{\sin(\sin(x))}xdx+\sum_{n=1}^\infty\int_{2n\pi}^{2(n+1)\pi}\frac{\sin(\sin(x))}xdx$$ converges because $\sin (\sin (x))\sim x $ for $x\to 0$.

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  • $\begingroup$ In somehow, you just proved that the sequence $\left(\int_0^{2n\pi}\frac{\sin(\sin(x))}{x}\,\mathrm d x\right)_{n\in\mathbb N}$ converges, not that $\int_0^\infty \frac{\sin(\sin(x))}{x}\,\mathrm d x$ converges. $\endgroup$ – Surb Apr 17 at 10:24
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$\sin\sin x$ is a smooth and periodic function with mean zero, hence $\int_{0}^{M}\sin\sin(x)\,dx$ is bounded but it is not convergent as $M\to +\infty$. On the other hand Kronecker's lemma (or just integration by parts) ensures that $\lim_{M\to +\infty}\int_{0}^{M}\sin\sin(x)\frac{dx}{x}$ is convergent. By expanding $\sin\sin(x)$ as a Fourier sine series we have

$$ \sin\sin(x) = 2 J_1(1)\sin(x) + 2 J_3(1)\sin(3x) + 2J_5(1) \sin(5x)+\ldots $$ with $J_n$ being a Bessel function of the first kind. We have $0\leq J_n(1)\leq \frac{1}{2^n n!}$ and $$\lim_{M\to +\infty}\int_{0}^{M}\frac{\sin(mx)}{x}\,dx = \frac{\pi}{2}$$ for any $m\in\mathbb{N}^+$, hence the identity $$ \int_{0}^{+\infty}\frac{\sin\sin(x)}{x}\,dx = \pi\sum_{m\geq 0}J_{2m+1}(1) $$ provides an efficient numerical computation of the LHS, which is $\approx 1.44470914981$.

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