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The limit of $\sin\frac{1}{x^2}$ when $x\to0$ does not exist but this integral can be calculated, the result is $\sqrt{2\pi}$, can anyone explain?

Gary
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    Observe that the integral is symmetric. So it is $2\int_0^\infty \sin(1/x^2)dx$. Now change the integration variable from $x$ to $t$ by $t=1/x^2$, and use https://math.stackexchange.com/q/3806166 – Gary Oct 25 '21 at 02:10
  • If you're having trouble to translate from your language to English, try this. It's not perfect, but it works. – jjagmath Oct 25 '21 at 02:11
  • @jjagmath https://www.deepl.com/translator is probably better. – Gary Oct 25 '21 at 02:12
  • This can be related to a special Integral called as "Fresnel integral". It has applications in quantum physics. – RAHUL Oct 25 '21 at 09:44

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