How does one prove that the Fresnel integrals $$ S(x) = \int_0^x \! \sin(t^2) \, \mathrm{d}t, \qquad C(x) = \int_0^x \! \cos(t^2) \, \mathrm{d}t $$ do not belong to $L^1(\mathbb{R})$?
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1Do you mean to show that $$\int^\infty_0\lvert \sin(t^2) \rvert dt ,,,,,, \text{ and } ,,, \int^\infty_0 \lvert \cos(t^2) \rvert dt$$ diverge or that $$\int_{-\infty}^\infty \lvert S(x) \rvert dx ,,,,, \text{ and } ,,,,, \int^\infty_{-\infty} \lvert C(x) \rvert dx$$ diverge? – User8128 Jul 25 '18 at 13:36
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The latter. I would like to show that $S, C \not\in L^1(\mathbb{R})$. – sleepingrabbit Jul 25 '18 at 13:40
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Since $\lim_{\lvert x \rvert \to \infty} \lvert S(x) \rvert = \sqrt{\pi/8}$, we can find $M \in \mathbb R$ such that for $\lvert x \rvert > M$, we have $$\lvert S(x) \rvert \ge \sqrt{\pi/16}.$$ But then $$\int^\infty_{-\infty} \lvert S(x) \rvert dx \ge \int^M_{-M} \lvert S(x) \rvert dx + \int^\infty_M \sqrt{\pi/16} \,\,dx + \int^{-M}_{-\infty} \sqrt{\pi/16}\,\, dx = \infty$$ and similarly for $C(x)$.
User8128
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