$\sin x^2$ 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.

  • 3
    $\begingroup$ This is a problem from Rudin's Principles of Mathematical Analysis (#6.13 in my/the latest edition) where he has you fill out his outline. $\endgroup$
    – Tyler
    Feb 3, 2012 at 0:39
  • $\begingroup$ Do you at least have a basis for this statement? A.k.a. do you know if it's true before you attempted to prove it? What evidence do you have? If you can come up with an informal proof or some sort of educated conjecture then that might help you get started. I'm not saying it's not true, but you have to convince yourself that it's true before formally proving it. $\endgroup$
    – chharvey
    Feb 3, 2012 at 4:26
  • $\begingroup$ This is called Fresnel integral. $\endgroup$
    – user9464
    Dec 9, 2012 at 16:31
  • $\begingroup$ by the alternated series test, $\int_a^\infty \sin(x f(x)) dx$ converges whenever $f(x)$ is non-decreasing and $\to +\infty$. $\endgroup$
    – reuns
    May 18, 2016 at 17:43
  • $\begingroup$ @Jozef: you can also use the Abel-Dirichlet's criterion. Appently, according to a comment to this similar question, you could use Riemann-Lebesgue lemma. $\endgroup$
    – Watson
    Jul 25, 2016 at 10:14

7 Answers 7


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.


$x\mapsto \sin(x^2)$ 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 $$\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot t^{-3/2}\frac{-1}2dt,$$ and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2=\frac{\cos 1}2$ 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.

  • 8
    $\begingroup$ See what magic one can do when one does calculus carefully?One doesn't need fancy schmancy function spaces and measures all the time to do real math-sometimes all you have to do is know the basics and think it through carefully. : ) $\endgroup$ Feb 3, 2012 at 3:36
  • $\begingroup$ $\int_0^c \frac{\sin t}{\sqrt t} \mathrm{d}t= - \int_1^c \frac{\mathrm{d} \cos t}{\sqrt t} = - \frac{\cos t}{\sqrt t}\big|_1^c + \int_1^c \cos t (-1/2t^{-3/2}) \mathrm{d}t = -\frac{\cos c}{\sqrt c} + \cos 1 - 1/2 \int_1^c \cos t t^{-3/2} \mathrm{d}t$ $\endgroup$
    – RHS
    May 1, 2013 at 5:57
  • $\begingroup$ "and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite)" - where is the $\cos t$? $\endgroup$ Jun 7, 2015 at 14:17
  • 1
    $\begingroup$ @Elimination It prove that $\int_1^\infty|\cos t| t^{-3/2}dt$ converges. $\endgroup$ Jun 7, 2015 at 14:54

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

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

  • 1
    $\begingroup$ Page not found! $\endgroup$
    – Say OL
    Oct 10, 2021 at 14:57

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.


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


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, $$\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$$


$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{2}}\,\dd x} = {1 \over 4}\int_{0}^{\infty}x^{\color{red}{3/4} - 1}\,\,\, {\sin\pars{\root{x}} \over \root{x}}\,\dd x \end{align} Note that $\ds{{\sin\pars{\root{x}} \over \root{x}} = \sum_{k = 0}^{\infty}\,\,\color{red}{\Gamma\pars{1 + k} \over \Gamma\pars{2 + 2k}} \,{\pars{-x}^{k} \over k!}}$.

With Ramanujan-MT: \begin{align} &\bbox[5px,#ffd]{\int_{0}^{\infty}\sin\pars{x^{2}}\,\dd x} = {1 \over 4}\Gamma\pars{\color{red}{3 \over 4}}\, {\Gamma\pars{1 - \color{red}{3/4}} \over \Gamma\pars{2 - 2\bracks{\color{red}{3/4}}}} = {1 \over 4\,\Gamma\pars{1/2}}\,{\pi \over \sin\pars{\pi/4}} \\[5mm] = &\ {1 \over 4\root{\pi}}\,{\pi \over 1/\root{2}} = \bbx{{\root{2} \over 4}\,\root{\pi}} \approx 0.6267 \\ \end{align}


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