Method of steepest descent for $\cos$ integral

Question

I'm looking to do an asymptotic analysis for the following integral (similar to this but not quite the same) $$J(x)=\int_0^1 \cos(x (t^3/3+t))dt.$$ My idea is to write $$J(x)=\Re\int_\Gamma \exp(xi(z^3/3+z))dz$$ where $\Gamma=[0,1]$.

Now, we note that if $f(z)=i(z^3/3+z)$, then $f'(z)=0$ at $z=\pm i$ and $f''(i)=2i(i)=-2<0$. Next, we note that $$f(u+iv)=\frac{v (-3 u^2 + v^2 - 3)}{3} + i\frac{ u (u^2 - 3 v^2 + 3)}{3}$$ and hence we expect that the biggest contribution will come from the saddle point $x=i$ along a level curve of the contour curve $\Im f = c $, and so I'd like to deform the contour $\Gamma$ to somehow take this into account, but I don't know how to proceed.

Any hints are helpful, this is especially tricky (to me at least) since the bounds are finite.

Answer 1

I take $x\gg1$. We can relate your integral to $$ I(x) = \int_{-1}^1dz \,e^{xi(z^3/3+z)}. $$ As you point out, the function in the exponent $f(z)=i(z^3/3+z)$ has a stationary point at $z=i$. Thus, we want to deform the integration contour such that it has constant phase (imaginary part of $f(z)$) and passes through $z=i$.

The imaginary part of $f(z)$ looks like

The integrating along the red contour is equivalent to integrating along the original blue contour. The left and right parts of the red contour contribute the same and can be evaluated using Watson's lemma. The top contour passing through the stationary point can be treated using Laplace's method. In both cases the application is pretty straightforward as the contour has constant phases by construction.

Specifically, the first terms are $$ \begin{align*} \int_{\Gamma_\text{left}} dz \,e^{xi(z^3/3+z)} &= \frac{e^{4 i x/3}}{4 x}\left(2 i+\frac{1}{x}-\frac{i}{x^2}+\ldots\right),\\ \int_{\Gamma_\text{top}} dz \,e^{xi(z^3/3+z)} &=-\frac{\sqrt{\pi } e^{-2 x/3}}{4608 \sqrt{x}}\left(4608-\frac{480}{x}+\frac{385}{x^2}+\ldots\right). \end{align*} $$

The contribution from the top contour is exponentially small in $x\gg 1$ and can be dropped.

Answer 2

A general blueprint to solve this type of problems is explained in my Math.SE answer here. Applied to OP's example, one derives that

$$ \int_0^1 \! \mathrm{d}z~e^{ix P(z)}~=~J(0)-e^{ix\overbrace{P(1)}^{=4/3}}J(1), \qquad P(z)~=~\frac{z^3}{3}+z,\tag{1}$$

where

$$\begin{align} J(a)~:=~&\int_0^{\infty\exp\left(\frac{i\pi}{6}\right)} \! \mathrm{d}z~e^{ix S(z)}\cr ~\stackrel{iu=S(z)}{=}& ~i \int_0^{\infty} \! \mathrm{d}u~ \frac{\exp\left(-xu\right)}{S^{\prime}(\alpha_1(u))} \cr ~=~&\frac{i}{x} \int_0^{\infty} \! \mathrm{d}u~ \frac{\exp\left(-u\right)}{S^{\prime}(\alpha_1(u/x))} \cr ~=~&\frac{i}{x S^{\prime}(\alpha_1(0))}+O(x^{-2}) ,\end{align}\tag{2} $$

and where

$$\begin{align} S(z)~:=~&P(z+a)-P(a) ~\stackrel{(1)}{=}~\frac{z^3}{3}+z^2a +z(a^2+1),\cr S^{\prime}(z)~=~&z^2+2za +a^2+1,\cr S^{\prime\prime}(z)~=~&2(z+a),\end{align}\tag{3} $$

is a 3rd-order polynomial with 3 roots

$$ S(z)-iu~=~\frac{1}{3}\prod_{i=1}^3(z-\alpha_i(u)),\tag{4} $$

where

$$\alpha_1(u)~=~O(u^1).\tag{5}$$

So

$$\begin{align}J(a\!=\!0)~\stackrel{(2)}{=}~& \frac{i}{x S^{\prime}(0)}+O(x^{-2})~=~ \frac{i}{x}+O(x^{-2}),\cr J(a\!=\!1)~\stackrel{(2)}{=}~&\frac{i}{x S^{\prime}(0)}+O(x^{-2}) ~=~\frac{i}{2x}+O(x^{-2}).\end{align}\tag{6}$$

This leads to the OP's sought-for expansion

$${\rm Re}\int_0^1 \! \mathrm{d}z~e^{ix P(z)} ~\stackrel{(1)+(6)}{=}~\frac{1}{2x}\sin\frac{4x}{3}+O(x^{-2}).\tag{7}$$