Proof integral converging : $\int \sin ( \sin (x) )dx$ 
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...



*

*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 :)
 A: 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 $x\to 0$.
A: $\sin\sin x$ 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 $\approx 1.44470914981$.
