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I'm trying to stud the convergence of $$\int_0^\infty \frac{\sin(\sin(x))}{x}dx.$$

I really have no idea on how solve this integral. Unfortunately, I can't majorate absolutely $\frac{\sin(\sin(x))}{x}$ by a $L^1(0,\infty )$ function, so I'm thinking that it's doesn't converge, but I'm not able to prove it. Moreover, graphically it looks to converge slower than $x\longmapsto \frac{1}{x}$, so I imagine that it doesn't converge, but since $\frac{\sin(\sin(x))}{x}$ often change of sign, I could be surprised.

For the context, it's a question of an exam of 1st year bachelor in mathematic, so we can't use theorem as dominated or monotone convergence theorem. And the argument should also be more or less elementary (I hope).

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  • $\begingroup$ If $\int_0^M \sin(\sin(x))dx$ is uniformly bounded in $M$, then you're done. And I have a feeling a uniform bound should hold. $\endgroup$ – mathworker21 May 31 '18 at 7:40
  • $\begingroup$ Intuitively, in $[2k\pi,(2k+2)\pi]$ the numerator alternates symmetrically while the denominator remains quasi-constant hence the integral amounts to an alternating harmonic series. $\endgroup$ – Yves Daoust May 31 '18 at 8:08
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Hint. Show that $$F(x):=\int_{0}^x\sin(\sin(t))dt$$ is bounded in $[0,+\infty)$. Note that $f(t):=\sin(\sin(t))$ is a continuous periodic function of period $2\pi$ and $f(x+\pi)=-f(x)$.

Then, by integration by parts, $$\int_0^\infty \frac{\sin(\sin(x))}{x}dx=\left[\frac{F(x)}{x}\right]_0^{+\infty}+\int_0^\infty \frac{F(x)}{x^2}dx=\int_0^\infty \frac{F(x)}{x^2}dx$$ where $\lim_{x\to 0^+}\frac{F(x)}{x}=\lim_{x\to 0^+}\frac{\sin(\sin(x))}{1}=0$ and the integral on the right is absolutely integrable.

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    $\begingroup$ A possible bound is $\left|\int_{0}^x\sin(\sin(t))dt\right|\leq \int_{0}^{\pi}\sin(\sin(t))dt$, isn't it ? $\endgroup$ – Gabriel Romon May 31 '18 at 7:48
  • $\begingroup$ @GabrielRomon Yes, you are correct! $\endgroup$ – Robert Z May 31 '18 at 7:50
  • $\begingroup$ quite brilliant (and tricky). very nice (but I would never think about doing that). (+1) thanks a lot. $\endgroup$ – idm May 31 '18 at 8:34
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The function $f(x):=\sin(\sin x)$ satisfies $f(x+\pi)=-f(x)$, $\>f(\pi-x)=f(x)$, and

$${2\over\pi}\sin x\leq f(x)\leq \sin x\qquad\left(0\leq x\leq{\pi\over2}\right)\ .$$ Its graph therefore looks like the graph of an ordinary sine, and we have $$\int_a^{a+\pi} \bigl|f(x)\bigr|\>dx=2\int_0^{\pi/2}f(x)\>dx\geq{2\over\pi}\qquad\forall a\geq0\ .$$ It is therefore out of the question that the integral $$\int_0^\infty{|f(x)|\over x}\>dx$$ is finite as a Lebesgue integral. On the other hand the improper Riemann integral $$\lim_{b\to\infty}\int_0^b{f(x)\over x}\>dx$$ is convergent (like the alternating harmonic series) since the numbers $$a_n:=\int_{n\pi}^{(n+1)\pi}{|f(x)|\over x}\>dx=\int_0^\pi{f(x)\over n+x}\>dx$$ form a sequence which is monotonically decreasing to $0$.

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