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According both to Wolfram Alpha and to sources like Wikipedia, the improper integral $\int_0^\infty \sin(t^2)dt$ converges, but this seems counterintuitive or even impossible. The sine function oscillates across the real line, as does $\sin(t^2)$, which should mean that the value of the integral $\int_0^x \sin(t^2)dt$ goes back and forth if $x$ increases gradually.

If we had an integrand whos magnitude went to zero, for example $\frac{1}{t}\sin(t)$, I could understand how the improper integral could converge, but that's not the case with $\sin(t^2)$, and unless someone convinces me to change my mind, I don't buy what Wolfram Alpha, Wikipedia and other sources are claiming.

Note that what I'm presenting here is a paradox. Simply a proof that the improper integral converges is not a satisfactory answer since it doesn't explain what's wrong with my argument that it can't converge. And therefore my question is not a duplicate of any question which has been asked earlier.

A A
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    Can you compute $$ \int_{\sqrt{2\pi k}}^{\sqrt{2 \pi (k+1)}} ; \sin(t^2) ,\mathrm{d}t \text{,} $$ so you can rewrite your integral as a sum of these finite (mostly cancelling) areas? – Eric Towers Oct 05 '24 at 16:43
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    $\int_0^{\infty} f(x) dx <\infty $ does not imply $f(x)\to 0$ as $x\to +\infty$. Perform the substitution $t^2=u$ to convince yourself that the integral converges – Sine of the Time Oct 05 '24 at 16:47
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    This post should fully answer your question. – user773458 Oct 05 '24 at 16:50
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    Is the value of that integral zero, Eric? If so, why doesn't a similar argument show that also $\int_0^\infty sin(t)$ converges? – A A Oct 05 '24 at 16:58
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    @AA Key difference is that the oscillation gets faster, i.e. the intervals with a given sign become smaller. As the integrand is bounded, this means that the integral is moving up/down by smaller and smaller amounts. – Severin Schraven Oct 05 '24 at 17:03
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    It isn't zero.The change of variables $t^2=x$ gives the integral as: $$\int_{2\pi k}^{2\pi(k+1)}\frac{\sin x}{2\sqrt x},dx.$$ I doubt this has a closed form, but I'd bet you can find a bound for this integral. – Thomas Andrews Oct 05 '24 at 17:04
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    It seems better to let $I_k$ to be the integral from $\pi k$ to $\pi(k+1),$ and then realize $I_k$ is an alternating sum decreasing (in absolute value) to $0.$ – Thomas Andrews Oct 05 '24 at 17:08
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    But, unlike $\sum_{k=0}^\infty a_k,$ where convergence means $a_k\to0,$ the same is not true for infinite integrals. All we can really say, knowing $\int_0^\infty f(x),dx$ converges, is that for any $\delta>0,$ $\lim_{x\to\infty}\int_x^{x+\delta}f(x)dx=0.$ – Thomas Andrews Oct 05 '24 at 17:12
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    "Unless someone convinces me to change my mind, I don't buy what Wolfram Alpha, Wikipedia and other sources are claiming." You'll find that people are more willing to help you if you don't frame your request as a demand to be supported in (wrongly) challenging established mathematical authorities. Anyway, suffice it to say there are functions which are not even bounded as $x\to\infty$ which are nevertheless integrable. – Charles Hudgins Oct 05 '24 at 17:26
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    This is a Fresnel integral and converges to $\sqrt{\frac\pi8} \approx $0.626657$ – Henry Oct 05 '24 at 17:27

2 Answers2

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It's not clear what precisely you take to be the paradox. Is it that an integrable function should have to decay to $0$, yet $\sin(t^2)$ does not?

In that case, thinking about another function may help you see that integrable functions can actually behave quite poorly. Consider $$ f(t) = \begin{cases} n & n\leq t < n + \frac{1}{n^3}, n \in \mathbb{N} \\ 0 & \text{otherwise} \end{cases} $$ Then $$ \int_0^\infty f(t) \,dt = \sum_{n=1}^\infty n\cdot\frac{1}{n^3} = \sum_{n = 1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $$ Note that $f$ not only doesn't decay to $0$, it is unbounded on $[M, \infty)$ for all $M > 0$.

Charles Hudgins
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    Convolve $f$ with your favorite mollifier to get a smooth function with the same behavior. – Charles Hudgins Oct 05 '24 at 18:24
  • Interesting construction! It's surprising and amusing to see the Basel problem put up in this context, but that this improper integral converges seems easier to accept since we have rapidly decaying contributions to the integral when we extend the integration interval. – A A Oct 05 '24 at 21:29
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    @AA don't read significance into the Basel problem showing up. One could just easily make intervals of width $1/3^k$ with height $2^k$ or whatever you like. The key is just picking a pair of sequences $a_n, b_n$ such that $a_n b_n$ is summable. – Charles Hudgins Oct 06 '24 at 04:11
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    And please note that it is just so with $\sin(t^2)$. Sufficiently rapidly decaying contributions to the integral mean that the function is integrable. – Charles Hudgins Oct 06 '24 at 04:13
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There is no problem near zero, so it suffices to prove that $\int_1^\infty\sin(x^2)\,dx$ converges. Put $t=x^2$. Then $dt=2\sqrt{t}dx$ and the integral becomes $$\frac{1}{2}\int_1^\infty \frac{\sin t}{\sqrt{t}}\,dt$$ And this integral converges by Dirichlet's theorem, as can easily be checked.

uniquesolution
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    One thing this answer illustrates, which the OP might find informative, is that the exact same integral can be represented by an integrand that does not decay to $0$ (as in the OP) and an integral that does decay to $0$ (as here). So clearly integrand-decays-to-$0$ is not an intrinsically important property of an integral. – Greg Martin Oct 05 '24 at 19:52
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    @GregMartin - Interesting comment, which raises the question whether every convergent integral can be represented as an integral of a function that decays to zero at infinity. – uniquesolution Oct 06 '24 at 08:30
  • Interesting question indeed. For the function in Charles Hudgins's answer, I believe setting $u=e^t$ doesn't quite work but setting $u=e^{e^t}$ does work. Now change the values in that example from $n$ to $n^2$ (and the widths of the intervals from $n^{-3}$ to $n^{-4}$) and I don't see a suitable change of variables.... – Greg Martin Oct 06 '24 at 16:40