Emanuel Keith

2022-06-27

Find the leading behavior of ${\int }_{0}^{\mathrm{\infty }}\mathrm{cos}\left(x\left(\frac{{t}^{3}}{3}-t\right)\right)dt$

massetereqe

$I={\int }_{0}^{\mathrm{\infty }}\mathrm{cos}\left(x\left(\frac{{t}^{3}}{3}-t\right)\right)dt$
Knowing the properties of the Airy functions, the answer is easy :
Let $\left\{\begin{array}{l}{T}^{3}=x{t}^{3}\\ zT=xt\end{array}\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}z={x}^{2/3}$
$I={x}^{-1/3}{\int }_{0}^{\mathrm{\infty }}\mathrm{cos}\left(\frac{{T}^{3}}{3}-zT\right)dT={x}^{-1/3}\pi \text{Ai}\left(-z\right)$
Ai is the Airy function :
$I=\pi {x}^{-1/3}\text{Ai}\left(-{x}^{2/3}\right)$
Equivalent at infinity :
$\phantom{\rule{1em}{0ex}}\text{Ai}\left(-z\right)\sim \frac{1}{\sqrt{\pi }\phantom{\rule{mediummathspace}{0ex}}{z}^{1/4}}\mathrm{sin}\left(\frac{2}{3}{z}^{3/2}+\frac{\pi }{4}\right)+O\left(\frac{1}{{z}^{3/2}}\right)$
$I={\int }_{0}^{\mathrm{\infty }}\mathrm{cos}\left(x\left(\frac{{t}^{3}}{3}-t\right)\right)dt\phantom{\rule{1em}{0ex}}\sim \phantom{\rule{1em}{0ex}}\sqrt{\frac{\pi }{x}}\mathrm{sin}\left(\frac{2}{3}x+\frac{\pi }{4}\right)+O\left(\frac{1}{x}\right)$

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