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How can the improper integral $\int_0^\infty \sin(t^2)dt$ converge? [duplicate]

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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.


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