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

Emanuel Keith
2022-06-27
Answered

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

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massetereqe

Answered 2022-06-28
Author has **21** answers

$I={\int}_{0}^{\mathrm{\infty}}\mathrm{cos}\left(x(\frac{{t}^{3}}{3}-t)\right)dt$

Knowing the properties of the Airy functions, the answer is easy :

Let $\{\begin{array}{l}{T}^{3}=x{t}^{3}\\ zT=xt\end{array}\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}z={x}^{2/3}$

$I={x}^{-1/3}{\int}_{0}^{\mathrm{\infty}}\mathrm{cos}(\frac{{T}^{3}}{3}-zT)dT={x}^{-1/3}\pi \text{Ai}(-z)$

Ai is the Airy function :

$I=\pi {x}^{-1/3}\text{Ai}(-{x}^{2/3})$

Equivalent at infinity :

$\phantom{\rule{1em}{0ex}}\text{Ai}(-z)\sim \frac{1}{\sqrt{\pi}\phantom{\rule{mediummathspace}{0ex}}{z}^{1/4}}\mathrm{sin}(\frac{2}{3}{z}^{3/2}+\frac{\pi}{4})+O\left(\frac{1}{{z}^{3/2}}\right)$

$I={\int}_{0}^{\mathrm{\infty}}\mathrm{cos}\left(x(\frac{{t}^{3}}{3}-t)\right)dt\phantom{\rule{1em}{0ex}}\sim \phantom{\rule{1em}{0ex}}\sqrt{\frac{\pi}{x}}\mathrm{sin}(\frac{2}{3}x+\frac{\pi}{4})+O\left(\frac{1}{x}\right)$

Knowing the properties of the Airy functions, the answer is easy :

Let $\{\begin{array}{l}{T}^{3}=x{t}^{3}\\ zT=xt\end{array}\phantom{\rule{1em}{0ex}}\to \phantom{\rule{1em}{0ex}}z={x}^{2/3}$

$I={x}^{-1/3}{\int}_{0}^{\mathrm{\infty}}\mathrm{cos}(\frac{{T}^{3}}{3}-zT)dT={x}^{-1/3}\pi \text{Ai}(-z)$

Ai is the Airy function :

$I=\pi {x}^{-1/3}\text{Ai}(-{x}^{2/3})$

Equivalent at infinity :

$\phantom{\rule{1em}{0ex}}\text{Ai}(-z)\sim \frac{1}{\sqrt{\pi}\phantom{\rule{mediummathspace}{0ex}}{z}^{1/4}}\mathrm{sin}(\frac{2}{3}{z}^{3/2}+\frac{\pi}{4})+O\left(\frac{1}{{z}^{3/2}}\right)$

$I={\int}_{0}^{\mathrm{\infty}}\mathrm{cos}\left(x(\frac{{t}^{3}}{3}-t)\right)dt\phantom{\rule{1em}{0ex}}\sim \phantom{\rule{1em}{0ex}}\sqrt{\frac{\pi}{x}}\mathrm{sin}(\frac{2}{3}x+\frac{\pi}{4})+O\left(\frac{1}{x}\right)$

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