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This question already has an answer here:

Does the improper integral $\int_{0}^{\infty}\sin(x^2)\;\mathrm dx$ converge?


So if it converges then $\lim_{b \to\infty}\int_{0}^{b}\sin(x^2)\;\mathrm dx$ exists and our integral converges to this limit. I have tried the substitution $y=x^2$ but this does not make the calculation easier.

Another way I know is to use the integral test but it also seems to be useless now.

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marked as duplicate by GEdgar, Michael Hoppe, marwalix, Daniel W. Farlow, Michael Grant Apr 23 '15 at 22:13

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It is a Fresnel integral. The limit exists since $$ I(b)=\int_{0}^{b}\sin t^2\,dt = \frac{1}{2}\int_{0}^{b^2}\frac{\sin x}{\sqrt{x}}\,dx$$ converges by Dirichlet's test (integral version), because $\sin x$ is a function with a bounded primitive and $\frac{1}{\sqrt{x}}$ is a monotonic function converging to zero as $x\to +\infty$. To compute it, we may use the Laplace transform, giving: $$ \mathcal{L}(\sin x)=\frac{1}{1+t^2},\qquad \mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)=\frac{1}{\sqrt{\pi t}}$$ from which: $$ \int_{0}^{+\infty}\sin t^2\,dt = \frac{1}{2\sqrt{\pi}}\int_{0}^{+\infty}\frac{dt}{(1+t^2)\sqrt{t}}=\frac{1}{\sqrt{\pi}}\int_{0}^{+\infty}\frac{du}{1+u^4}=\frac{1}{2}\sqrt{\frac{\pi}{2}}.$$

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  • $\begingroup$ Thank you! I understand your answer but I have not met the Dirichlet's test before. Do you think the question can be anwered without it (with the use of only more elemental techniques)? $\endgroup$ – moon1234 Apr 23 '15 at 19:21
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    $\begingroup$ @moon1234: you may use integration by parts, but it is essentially the same. "Dirichlet's" test is just a name for an ubiquitous technique, it is worth to learn it just to avoid to repeat the same steps over and over. $\endgroup$ – Jack D'Aurizio Apr 23 '15 at 19:28
  • $\begingroup$ @Jack D'Aurizio I'm trying to understand the step where you take $\frac{1}{2}\int_{0}^{\infty}\mathcal{L}(\sin x)\mathcal{L}^{-1}\left(\frac{1}{\sqrt{x}}\right)dt$. This is a new technique to me. Does it have a name so I can read about it somewhere? I understand $1=e^{-st}e^{st}$ but I don't understand why can you take two infinite integrals in the middle of the integrand. $\endgroup$ – user5389726598465 Jun 30 '17 at 2:34
  • $\begingroup$ @user135711: have a look at this paragraph - en.wikipedia.org/wiki/… $\endgroup$ – Jack D'Aurizio Jun 30 '17 at 3:33

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