How to determine the leading-order asymptotic behaviour of this integral? $\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt$
as x → ∞
I am stuck on how to apply the stationary phase method when $\tan(t)$ vanishes at the stationary point. Should I expand tan in its Taylor series? Is the stationary point $0\,$? Thanks in advance
 A: Just to show something different from @joriki's answer to [this question], working first the antiderivative
$$I=\int\cos(xt^{3})\tan(t)\, dt$$ let $x t^3=u^3$ to make
$$I=\frac 1{x^{1/3}}\int\cos \left(u^3\right) \tan \left(\frac{u}{x^{1/3}}\right)\,du$$ Since $x\to \infty$, using $\tan(\epsilon)\sim \epsilon$,we have
$$I\sim \frac 1{x^{2/3}}\int\ u\, cos \left(u^3\right)\,du$$ The integral can be computed using the gamma function or the exponential integral function to make
$$\int\ u\, cos \left(u^3\right)\,du=-\frac{1}{6} u^2 \left(E_{\frac{1}{3}}\left(-i
   u^3\right)+E_{\frac{1}{3}}\left(i u^3\right)\right)$$ Back to $t$
$$I \sim -\frac{1}{6} t^2 \left(E_{\frac{1}{3}}\left(-i t^3
   x\right)+E_{\frac{1}{3}}\left(i t^3 x\right)\right)$$ Using the bounds
$$J=\int_0^1\cos(xt^{3})\tan(t)\, dt\sim\frac{1}{6} \left(\frac{\Gamma
   \left(\frac{2}{3}\right)}{x^{2/3}}-E_{\frac{1}{3}}(-i x)-E_{\frac{1}{3}}(i x)\right)$$
Limited to the first order
$$E_{\frac{1}{3}}(-i x)+E_{\frac{1}{3}}(i x)\sim -\frac {e^{-i x }} x$$
So the leading order of $J$ is
$$\frac{\Gamma \left(\frac{2}{3}\right)}{6\, x^{2/3}}$$
A: Just to provide a bit different approach to the evaluation of the leading asymptotic term.
$$I(x)=\int^{1}_{0} \cos(xt^{3})\tan(t)\, dt=\frac{1}{3}\int^{1}_{0} \frac{\cos(xs)\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds=\Re\frac{1}{3}\int^{1}_{0} \frac{e^{ixs}\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds$$
Next, we change the variable $s\to{is}=se^{\frac{\pi{i}}{2}}$. Integral along a closed contour (from small $r$ to $1$ of axis $X$; along a quarter-circle (angle $\phi$ - from $0$ to $\frac{\pi}{2}$); from $i$ to $i{r}$ and along the quarter-cercle of small radius $r$ clockwise to the starting point) is zero - there is no singularity inside the contour. Integral along a small quarter-circle (around $s=0$) $\,\to0$ at $r\to0$ - due to the integrand asymptotics near zero.
Integral along the quarter-circle of radius $1$ counter clockwise is limited by
$$\int_{R=1}\frac{e^{ixs}\tan(s^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds\leqslant{C}\int_0^{\frac{\pi}{2}}e^{-sin(\phi){x}}dx\leqslant{C}\frac{\pi}{2x}\,,$$ where C is some constant.
Representing $\tan((is)^{\frac{1}{3}})$ as a series and expanding integration to $\infty$  we get the leading asymptotic term:
$$I(x)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}}\int^{1}_{0} \frac{e^{-xs}\tan((is)^{\frac{1}{3}})}{s^{\frac{2}{3}}}\, ds+O\biggl(\frac{1}{x}\biggr)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}}\int^{\infty}_{0} e^{-xs}\frac{{e}^{\frac{\pi{i}}{6}}s^{\frac{1}{3}}}{s^{\frac{2}{3}}}\, ds+O\biggl(\frac{1}{x}\biggr)$$
$$I(x)\sim\frac{1}{3}\Re \,{e}^{\frac{\pi{i}}{6}} \biggl(\frac{{e}^{\frac{\pi{i}}{6}}\Gamma(\frac{2}{3})}{x^{\frac{2}{3}}}\biggr)+O\biggl(\frac{1}{x}\biggr)\sim\frac{\cos(\frac{\pi}{3})\Gamma(\frac{2}{3})}{3x^{\frac{2}{3}}}+O\biggl(\frac{1}{x}\biggr)$$
$$I(x)\sim\frac{\Gamma(\frac{2}{3})}{6x^{\frac{2}{3}}}+O\biggl(\frac{1}{x}\biggr)$$
