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$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.

I'm trying to understand why and would like some help in working towards a formal proof.

  • 0
    @Jozef: you can also use the [Abel-Dirichlet's criterion](http://math.stackexchange.com/questions/413950/proving-abel-dirichlets-test-for-convergence-of-improper-integrals-using-integr). Appently, according to a comment to [this](http://math.stackexchange.com/questions/1153083) similar question, you could use Riemann-Lebesgue lemma.2016-07-25

6 Answers 6

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Just with elementary tools: see Here,How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

In fact we have: \begin{split} \int_0^\infty \sin(x^2) dx=\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx<\infty\end{split} See below

Employing the change of variables $2u =x^2$ after integration by parts we get \begin{split} \int_0^\infty \sin(x^2) dx&=&\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx\\& =&\frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx\\&=&\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx \\&= &\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\end{split}

Given that $ \sin 2x =(\sin^2x)'$ and $\lim_{x\to 0}\frac{\sin x}{x}=1$ However, $ \int^\infty_1\frac{\sin^2 x}{x^{3/2}}\,dx\le \int^\infty_1\frac{1}{x^{3/2}}\,dx<\infty$ since $|\sin x|\le |x|$ we have, $\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$

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The humps for $x\mapsto \sin(x^2)$ go up and down. Each has an area smaller than that of the last. The areas converge to 0 as you progress down the $x$-axis. By the alternating series test, this converges.

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$x\mapsto \sin(x^2)$ is integrable on $[0,1]$, so we have to show that $\lim_{A\to +\infty}\int_1^A\sin(x^2)dx$ exists. Make the substitution $t=x^2$, then $x=\sqrt t$ and $dx=\frac{dt}{2\sqrt t}$. We have $\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot t^{-3/2}\frac{-1}2dt,$ and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2=\frac{\cos 1}2$ and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite), we conclude that $\int_1^{+\infty}\sin(x^2)dx$ and so does $\int_0^{+\infty}\sin(x^2)dx$. This integral is computable thanks to the residues theorem.

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    @Elimination It prove that $\int_1^\infty|\cos t| t^{-3/2}dt$ converges.2015-06-07
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I solved this one integral as a particular case of the formula I provide here: http://www.mymathforum.com/viewtopic.php?f=15&t=26243 under the name Weiler.

$\int\limits_0^\infty {\sin \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} - \sin \frac{{{b^2}}}{a}} \right)$

$\int\limits_0^\infty {\cos \left( {a{x^2}} \right)\cos \left( {2bx} \right)dx} = \sqrt {\frac{\pi }{{8a}}} \left( {\cos \frac{{{b^2}}}{a} + \sin \frac{{{b^2}}}{a}} \right)$

So you have

$\int\limits_0^\infty {\sin \left( {{x^2}} \right)dx} = \sqrt {\frac{\pi }{8}} $

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This is also informative, and works when there is no aspect with closed form. Taking Davide's substitution, define $ A_n^+ = \int_{2 \pi n}^{2 \pi n + \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; , $ $ A_n^- = \int_{2 \pi n + \pi}^{2 \pi n + 2 \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; , $ and finally $ A_n = A_n^+ + A_n^- = \int_{2 \pi n }^{2 \pi n + 2 \pi} \; \frac{\sin t}{2 \sqrt t} \; dt \; , $

Next, I just used $\int_{m \pi}^{m \pi + \pi} \sin t dt = \pm 2,$ depending upon the integer $m,$ and took bounds based on the size of the denominators.

I suppose we need to start with $n \geq 1.$ With that, $ \frac{1}{\sqrt{2 \pi n + \pi}} \leq A_n^+ \leq \frac{1}{\sqrt{2 \pi n}}, $ $ \frac{-1}{\sqrt{2 \pi n + \pi}} \leq A_n^- \leq \frac{-1}{\sqrt{2 \pi n + 2 \pi}}, $ and $ 0 \leq A_n \leq \frac{1}{\sqrt{ 8 \pi} \; \; n^{3/2}}.$

What does this say about convergence? The integral is $ \sum_{n = 0}^\infty A_n. $ Convergence does not depend on the initial terms, so we may start at the more convenient $n=1.$ From the $3/2$ exponent in the estimate of $A_n,$ we see that the sum is a finite constant. We do see modest oscillation in the indefinite integral, however the $\sqrt n$ terms in the denominators of $A_n^+$ and $A_n^-$ tell us that eventually the indefinite integral stays within any desired distance of the infinite integral.

This idea, cancellation of alternating contributions, can be used with far worse integrands, $\sin (x^5 - x - 1)$ comes to mind.

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There are more involved methods to actually compute this integral. A real variables method is given in this answer and a contour integration method is given in this answer.

Here is a much simpler method to show only the convergence of the integral. $ \begin{align} \int_0^\infty\sin\left(x^2\right)\,\mathrm{d}x &=\int_0^\infty\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{1}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+2)\pi}\frac{\sin(x)}{2\sqrt{x}}\,\mathrm{d}x\tag{2}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\left(\frac1{\sqrt{x}}-\frac1{\sqrt{x+\pi}}\right)\mathrm{d}x\tag{3}\\ &=\sum_{k=0}^\infty\int_{2k\pi}^{(2k+1)\pi}\frac{\sin(x)}2\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\mathrm{d}x\tag{4}\\ &\le\int_0^\pi\frac{\sin(x)}2\left(1+\sum_{k=1}^\infty\frac1{4\sqrt{2\pi} k^{3/2}}\right)\mathrm{d}x\tag{5}\\ &=1+\frac{\zeta\!\left(\frac32\right)}{4\sqrt{2\pi}}\tag{6} \end{align} $ Explanation:
$(1)$: substitute $x\mapsto\sqrt{x}$
$(2)$: break the integral into $2\pi$ segments
$(3)$: $\sin(x+\pi)=-\sin(x)$
$(4)$: algebra
$(5)$: $\frac{\pi}{\sqrt{x}\sqrt{x+\pi}\left(\sqrt{x}+\sqrt{x+\pi}\right)}\le\min\left(1,\frac1{4\sqrt{2\pi} k^{3/2}}\right)$ for $x\ge2k\pi$
$(6)$: evaluate integral and sum