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We can see that the improper integral $$\int_{0}^{\infty}\sin\left(\,{x^{2}}\,\right) {\rm d}x $$ converges as the transformation $u = x^{2}$ helps us to use the Dirichlet's test on $$ \int_{0}^{\infty}\frac{\sin\left(\,{u}\,\right)} {2\,\sqrt{\,{u}\,}\,}\,{\rm d}u. $$ However, I need to know the absolute convergence of the same, whether it is conditionally convergence or not.

I feel that the area of contribution is eventually being demolished for $\left\vert\,{\sin\left(\,{x^{2}}\,\right)}\,\right\vert$ as similar to $\sin\left(\,{x^{2}}\,\right)$. But I couldn't certain the absolute convergence.

Thanks in advance.

Felix Marin
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Messi Lio
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    No way $\int_0^{\infty}|\sin(x^2)|,dx$ could converge. It has a saw tooth shaped minorant of average height $1/2$ (each tooth narrower and spikier than the preceding one, but they all have maximum height $1$). – Jyrki Lahtonen Nov 28 '22 at 04:49
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    To do the absolute/conditional convergence part, same change of variable also works, and note that $|\sin (u)| \geqslant \sin(u)^2 = (1 - \cos (2u))/2$, and the integral split into two parts, one of which converges and the other one diverges. – xbh Nov 28 '22 at 04:52
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    I actually think that xbh's suggestion is better. It may be a bit painful to rigorously prove the kind of convexity required to get that piecewise linear lower bound. True, some other lower bounds would still work. – Jyrki Lahtonen Nov 28 '22 at 04:57
  • @xbh the direct comprison ($f(u) \geq g(u)$) referred here lacks non-negativity condition, right? – Messi Lio Nov 28 '22 at 05:03

2 Answers2

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Let $f(x) = |\sin x^2 | $, $a_n = \sqrt{n \pi+{1 \over 4} \pi}, b_n = \sqrt{n \pi+{3 \over 4} \pi}$. Suppose $n \ge 2$ is even, note that $f(x) \ge {1 \over \sqrt{2}} 1_{[a_n,b_n]}(x)$ for $x \in [a_n,b_n]$.

Hence $\int_{a_n}^{b_n} f(x) dx \ge {1 \over \sqrt{2}} (b_n-a_n) \ge {1\over \sqrt{2}}{b_n^2-a_n^2 \over b_n+a_n} \ge K {1 \over \sqrt{n+1}}$. Since the latter is not summable, it follows that $f$ is not integrable.

copper.hat
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This is a typical textbook-example level question.

As I said in the comment, change the variable $x = \sqrt u$. Note the following $$ |\sin (u)| \geqslant |\sin(u)|^2 = \sin (u)^2 $$ where $\geqslant$ holds because $|\sin (u)| \in [0,1]$, and $=$ follows the operation rules of$ |\cdot|$. These are non-negative, so nothing to worry here. Proceed: break the integral as $$ \int_0^\infty \frac {|\sin (u)|} {2\sqrt u} \,\mathrm du = \left(\int_0^{1} + \int_1^\infty\right) \cdots, $$ then the $[0,1]$ part converges since $|\sin (u)| = \sin (u) \sim u [u \to 0^+]$, and the whole function under $\int$ satisfies $|\sin (u)|/2 \sqrt u \sim \sqrt u/2 [u \to 0^+]$.

For the $[1, +\infty)$ part, apply the trig equality $$ \sin (u)^2 = \frac 12 (1 - \cos (2u)), $$ and note that $$ \int_1^\infty \frac 1{4\sqrt u} \mathrm du $$ diverges [via direct computation], while $$ \int_1^\infty \frac {\cos (2u)} {4\sqrt u}\,\mathrm du $$ converges by the Dirichlet test: $1/4 \sqrt u \searrow 0 [u \to +\infty]$, and $\int_1^A \cos (2u)\, \mathrm d u$ is bounded for $A \in (1, +\infty)$. All in all, $$ \int_1^\infty \frac {1 -\cos (2u)}{4\sqrt u}\, \mathrm du $$ diverges as a sum of a convergent and a divergent improper integrals. Comparison test [valid, since $|\sin u| \geqslant (1 - \cos (2u))/2 \geqslant 0$ holds for all real $u$] concludes that $$ \int_1^\infty \frac {|\sin u|}{2\sqrt u} \, \mathrm du $$ diverges.

Finally, $$ \int_0^\infty \frac {|\sin u|}{2 \sqrt u} \,\mathrm du $$ diverges as a sum of two improper integrals, one of which converges while the other diverges. Therefore $$ \boldsymbol{ \int_0^\infty \sin (x^2) \,\mathrm dx } $$ converges conditionally.

xbh
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