Leading order asymptotic behaviour of the integral $\int^1_0 \cos(xt^3)\tan(t)dt$ I'm trying to get the leading order asymptotic behaviour of the integral:
$$\int^1_0 \cos(xt^3)\tan(t)dt$$
I'm trying to use the Generalised Fourier Integrals and the Stationary Phase Method, but I can't understand how to start this.
THIS IS WHAT I HAVE TRIED:
We want to get:
$$\Re \int^1_0e^{ixt^3}\tan(t)dt$$
$\phi(t)=t^3$ has a stationary point at $t=0$ so the main contribution is around that point and due to the small angle approximation $\tan(t) \approx t$.
$$\Re\int^1_0e^{ixt^3}tdt$$
I tried some substitutions but I can't find anything useful after that.
Any ideas,
Many thanks!
 A: I'll show a quick way to also include the oscillatory nature of the asymptotic expression in addition to the solution already presented by @Svyatoslav. Having
$$
I = \int_0^1 \exp(ixt^3)\tan(t)dt
$$
the minimum of $p(t)=t^3$ occurs at the lower limit $a=0$. Using the notation here.
Taylor expanding around $a=0$ gives that $\tan(t)=t+\frac{t^3}{3}+O(t^5)$. From the link, we identify the parameters $P=1,\mu=3,Q=1,\lambda=2$. Then, to leading term when only considering the point $t=0$, the asymptotic expansion is given by
$$
I(x) \sim \exp(\lambda \pi i/(2\mu))\frac{Q}{\mu}\Gamma\left(\frac{\lambda}{\mu} \right) \frac{\exp(ixp(a))}{(Px)^{\lambda/\mu}} $$
with real part
$$\Re(I) \sim \Gamma\left(\frac{2}{3} \right) \frac{1}{6x^{2/3}}$$
In the case of the upper limit being finite (here $b=1$), an additional term is introduced. Also only considering the leading term (se the mentioned link for the general expression), the full expression is
$$
I(x) \sim \exp(\lambda \pi i/(2\mu))\frac{Q}{\mu}\Gamma\left(\frac{\lambda}{\mu} \right) \frac{\exp(ixp(a))}{(Px)^{\lambda/\mu}} - \exp(ixp(1))P_0(1)\left(\frac{i}{x} \right)
$$
with $P_0(t) = \frac{\tan(t)}{3t^2}\Rightarrow P_0(1) = \frac{\tan(1)}{3}$. Now, taking the real part of this gives that
$$
\Re(I) \sim \Gamma\left(\frac{2}{3} \right) \frac{1}{6x^{2/3}} + \frac{\tan(1)}{3} \frac{\sin(x)}{x}
$$
This is seen in the figure below, where the legend one term denotes the asymptotic expansion stemming from $a=0$ and two term denotes when also considering the effect of $b=1$.

A: Using the series expansion of $\tan(t)$, we have
$$\sum_{n=0}^\infty  \frac{(-1)^n 4^{n+1} \left(4^{n+1}-1\right) B_{2 n+2}}{\Gamma (2n+3)}\int_0^1 \cos(xt^3)\,t^{2n+1}\,dx$$ The antiderivative exists and
$$\int_0^1 \cos(xt^3)\,t^{2n+1}\,dx=\frac{\,
   _1F_2\left(\frac{n}{3}+\frac{1}{3};\frac{1}{2},\frac{n}{3}+\frac
  {4}{3};-\frac{x^2}{4}\right)}{2 (n+1)}$$ which is asymptotically
$$\sqrt{\pi } \frac{2^{\frac{2 n-1}{3}} \Gamma \left(\frac{n+1}{3}\right)}{3 \Gamma
   \left(\frac{3-2n}{6}\right)}x^{-\frac{2}{3}   (n+1)} $$ So, for large values of $x$
$$\frac{\Gamma \left(\frac{2}{3}\right)}{6 x^{2/3}}\left(1+\frac{2 \sqrt[3]{2} \sqrt{\pi }}{3 x^{2/3} \Gamma
   \left(-\frac{1}{6}\right)}\right)+O\left(\frac{1}{x^2}\right)$$
