I am currenty struggling with the integral $$\int_{-\infty}^{\infty}f(t)\exp(ix(t^3-t))dt$$ where $f(t)$ is smooth and $f\rightarrow 0$ as $t\rightarrow \pm\infty$. I want to obtain the leading asymptotic beahviour as $x\rightarrow \infty$.

I would not have a problem if the boundaries of the integral are finite, as stated here http://www.math.unl.edu/~scohn1/8423/intasym4.pdf (Formula (2))

$g(t)=t^3-t$ and $g'(t)=0$ at $\pm\sqrt{\frac{1}{3}}$.

What can I do to use the formula stated in the link above?

  • $\begingroup$ Split the integral up like $\int_{-\infty}^{-1} + \int_{-1}^{1} + \int_{1}^{\infty}$ then make the change of variables $t^3-t = s$ in the first and last integrals. Integrate by parts to show that they're each $o(1/\sqrt{x})$. The remaining integral, $\int_{-1}^{1} f \cdot \exp(\cdots)\,dt$, is of the form described in the pdf. I think this should get you where you want to go. $\endgroup$ Mar 2, 2014 at 7:40
  • $\begingroup$ Thanks for your comment. Using the substituion $t^3-t=s$ the expression $t=...$ gets really ugly, is this supposed to be or is there a workaround somehow? For the middle integral do I have to change anything, because there are two points where $g'(t)=0$ $\endgroup$
    – Alkibiades
    Mar 2, 2014 at 15:25
  • $\begingroup$ It does get ugly, yeah. You shouldn't need to work with the exact expression for $t$, though. I'll write an answer describing the process for $\int_1^\infty$. The other integral should be similar. $\endgroup$ Mar 2, 2014 at 15:38

2 Answers 2


If we split the interval of integration up into four parts,

$$ \int_{-\infty}^{\infty} = \int_{-\infty}^{-1} + \int_{-1}^0 + \int_0^1 + \int_1^\infty, $$

the inner two integrals are of the type considered in the PDF you linked, so it just remains to show that the first and last integrals are asymptotically smaller than them as $x \to \infty$ (and hence that they do not contribute to the leading-order asymptotic).

I'll just consider the last integral,

$$ I(x) = \int_1^\infty f(t) \exp\left[ix(t^3-t)\right]\,dt, $$

since the process for the first, $\int_{-\infty}^{-1}$, should be similar.

The substitution $s = t^3-t$ defines an increasing, concave bijection $t(s) : [0,\infty) \to [1,\infty)$. For large $s$ we have

$$ t \sim s^{1/3} $$

and for small $s$ we have

$$ t = 1 + \frac{s}{2} + O(s^2). $$

We'll then write

$$ f(t)\,dt = f(t(s))t'(s)\,ds, $$

so that

$$ I(x) = \int_0^\infty f(t(s))t'(s)e^{ixs}\,ds. $$

Note that, since $t \sim s^{1/3}$ for large $s$, we have

$$ t'(s) \sim \frac{1}{3}s^{-2/3} $$

for large $s$. Integrating by parts thus yields

$$ \begin{align} I(x) &= \frac{1}{ix}\left[f(t(s))t'(s)e^{ixs}\right]_0^\infty - \frac{1}{ix} \int_0^\infty \frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr]e^{ixs}\,ds \\ &= -\frac{f(1)}{2ix} - \frac{1}{ix} \int_0^\infty \frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr]e^{ixs}\,ds, \tag{1} \end{align} $$

since $t(0) = 1$ and $t'(0) = \frac{1}{2}$. Now

$$ \frac{d}{ds} \Bigl[f(t(s))t'(s)\Bigr] = f'(t(s))t'(s)^2 + f(t(s))t''(s), $$

and for large $s$ we have

$$ f'(t(s))t'(s)^2 \sim f'(t(s)) \left( \frac{1}{3} s^{-2/3} \right)^2 \tag{2} $$


$$ f(t(s))t''(s) \sim f(t(s)) \left( -\frac{2}{9} s^{-5/3} \right). \tag{3} $$

Since $f(t(s)) \to 0$ as $s \to \infty$ the expression in $(3)$ is integrable, and if the expression in $(2)$ is integrable as well (for instance if $f'(r)$ is bounded) then the integral in $(1)$ exists and is bounded. Thus

$$ I(x) = O\left(\frac{1}{x}\right). $$

This is smaller than the estimates you would get for the integrals over finite intervals, which would be something on the order of $1/\sqrt{x}$. So, you can throw the tails of the integral out if all you're interested in is its leading-order approximation.

  • $\begingroup$ In general one should be careful when differentiating asymptotic equivalences; if $p \sim q$ it's not necessarily true that $p' \sim q'$. Here everything is nice enough that it works out. $\endgroup$ Mar 2, 2014 at 16:18
  • $\begingroup$ Thanks for your analysis of the integral, I have one last question concerning the boundaries of the original integral, would there be an obvious change in the behaviour when we replace the lower limit of the integral with $-\frac{1}{\sqrt{3}}$? $\endgroup$
    – Alkibiades
    Mar 2, 2014 at 16:59
  • $\begingroup$ Do you mean split it like $\int_{-\infty}^0 = \int_{-\infty}^{-1/\sqrt{3}} + \int_{-1/\sqrt{3}}^0$? The main contribution to the integral comes from a neighborhood of the stationary point, for instance a neighborhood like $(-1/\sqrt{3}-\epsilon, -1/\sqrt{3}+\epsilon)$. As such, if you split the integral up like that then the part $\int_{-\infty}^{-1/\sqrt{3}}$ would contain contributions from the stationary point as well and would not be negligible in the sense of the above answer. $\endgroup$ Mar 2, 2014 at 17:06
  • $\begingroup$ I chose to split the integral at the points $\pm 1$ and $0$ since they aren't stationary points and they're convenient when making the change of variables, but in theory you could split it at any points $a,b,c$ with $a < -1/\sqrt{3} < b < 1/\sqrt{3} < c$ and obtain the same result. $\endgroup$ Mar 2, 2014 at 17:09
  • $\begingroup$ I mean looking at $\int_{\frac{-1}{\sqrt{3}}}^{\infty}$ instead of $\int_{-\infty}^{\infty}$ $\endgroup$
    – Alkibiades
    Mar 2, 2014 at 17:09

I) OP's integrand reads

$$\begin{align} g(z)~:=~&f(z)e^{ixS(z)}, \qquad S(z)~:=~z^3-z, \cr S^{\prime}(z)~=~&3z^2-1,\qquad S^{\prime\prime}(z)~=~6z,\end{align} \tag{1}$$

with 2 critical points at

$$z_{\pm}~=~ \pm 3^{-\frac{1}{2}}\tag{2}$$

with angular steepest decent direction $\pm\frac{\pi}{4}$, respectively.

$\uparrow$ Fig. 1. The complex $z$-plane with OP's original integration contours $-{\cal C}_1$, and two other contours ${\cal C}_2\equiv {\cal C}_-$ and ${\cal C}_3\equiv{\cal C}_+$. The shaded regions denote exponentially decaying sectors. (Figure taken from Ref. [W].)

II) OP seems mostly interested in the stationary phase approximation for real smooth functions $f$ on the real line $\mathbb{R}$. In this answer we will just mention that if $f$ is also holomorphic in appropriate regions of the complex plane $\mathbb{C}$, then one may apply the method of steepest descent. OP's integral

$$ I ~=~ \int_{\mathbb{R}+i0^+} \mathrm{d}z~g(z)~=~I_- + I_+ \tag{3} $$

can in the latter case be deformed to a sum of 2 steepest descent contours $$\begin{align} I_{\pm}~:=~& \int_{{\cal C}_{\pm}} \mathrm{d}z~g(z) \cr ~\sim~& \sqrt{\frac{2\pi}{-ix S^{\prime\prime}(z_{\pm})}}g(z_{\pm}) \cr ~=~&e^{\pm\frac{i\pi}{4}} \sqrt{\frac{\pi}{x \sqrt{3}}} f(z_{\pm}) e^{ix(3^{-\frac{3}{2}}-3^{-\frac{1}{2}})} \qquad\text{for}\qquad x~\to~\infty \end{align} \tag{4}$$

through the 2 critical points $z_{\pm}= \pm 3^{-\frac{1}{2}}$, respectively.

Note that the final formula (4) for $I=I_- + I_+$ only refers to objects on the real line. Therefore we expect that a real analysis would lead to the same result for an appropriate class of real smooth functions $f$.


  • [W] E. Witten, Analytic Continuation Of Chern-Simons Theory, arXiv:1001.2933; p. 23-29, 48-49. A related 2015 KITP lecture by Witten, A New Look At The Path Integral Of Quantum Mechanics, can be found on YouTube.
  • $\begingroup$ In this problem, if assuming $x\to 0^+$ instead, can one use the method of Steepest Descent by rescaling $\tau=xt$? $\endgroup$ May 2, 2021 at 19:39

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