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I am interested in the numerical calculation of $$ F(\eta)=\frac{2}{π}\int_0^{+\infty}\sin(t\eta-t^3)\frac{dt}{t}\quad\text{for $\eta\ge 0$}. $$ I believe that the function $F$ is bounded, but I do not know its upper bound. It is easy to get $F(0)=-1/3$ and various other values for $\eta \in [0,40]$ but getting above the value $40$ for $\eta$ seems to raise computational difficulties for Mathematica. It is also interesting to see that $F$ takes values strictly above 1 with the seemingly largest value $1.5$ for $\eta =3$.

As a motivation, one may note that with $g(t)= e^{it^3}\text{pv}\frac{1}{πt}$, the operator of convolution with $g$ is bounded on $L^2(\mathbb R)$ whenever its Fourier transform is bounded. We have here $F=\hat g$, up to some normalization, and a precise bound in $L^\infty$ for $F$ would provide the $\mathcal B(L^2)$ operator-norm of that convolution.

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Nemo's representation of $F(\eta)$ in terms of a hypergeometric function can be evaluated without difficulty for large $\eta$: $$F(\eta)=\frac{\sqrt{3} {\eta}^2 \Gamma \left(\frac{2}{3}\right) \; _1F_2\left(\frac{2}{3};\frac{4}{3},\frac{5}{3};-\frac{{\eta}^3}{27}\right)}{6\pi }-\frac{12 {\eta} \; _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{{\eta}^3}{27}\right)}{6\Gamma \left(-\frac{1}{3}\right)}-\frac{1}{3}$$

$$F(\eta)\rightarrow 1+\sqrt[4]{3} \sqrt{2/\pi}\frac{1}{\eta^{3/4}} \left[\sin \left(\frac{2 \eta^{3/2}}{3 \sqrt{3}}\right)-\cos \left(\frac{2 \eta^{3/2}}{3 \sqrt{3}}\right)\right],\;\;\eta\gg 1.$$

see plot, blue is $F(\eta)$, gold is the large-$\eta$ approximation (nearly indistinguishable for $\eta>5$). So $F(\eta)$ oscillates around 1, with an amplitude that decays as $\eta^{-3/4}$. The maximum 1.5487 is reached at $\eta= 3.37213$.

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  • $\begingroup$ Great! Thank you very much. $\endgroup$ – Bazin Feb 11 at 9:14
  • $\begingroup$ In your asymptotic expression, it seems that the term $\sqrt{2π}$ should be replaced by $\sqrt{2/π}$. $\endgroup$ – Bazin Feb 11 at 12:40
  • $\begingroup$ Sorry to bother you again with that matter. I would be very grateful if you could double-check your constant $\sqrt{2\pi}$, which I believe should be replaced by $\sqrt{2/\pi}$. Thanks. $\endgroup$ – Bazin Feb 13 at 10:31
  • $\begingroup$ you are right, it was my typing error, apologies for the confusion. $\endgroup$ – Carlo Beenakker Feb 13 at 10:42
  • $\begingroup$ Thanks a lot for caring to check this; of course I renew my thanks for your very helpful and thorough answer to my question. $\endgroup$ – Bazin Feb 13 at 16:05

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