Find leading approximation of oscillatory integrals (by integrating by parts)

Question

I am thinking of integrals like the following

$$\int_{0}^{1} e^{-x}e^{it(x^3-12x)}dx$$

where the critical point of $x^3 - 12x$ is out of the given interval $[0,1]$. How to find the leading approximation of this integral? Thanks!

Answer 1

This doesn't answer the integration by parts part, but the leading approximation comes from the endpoints, you need to take a contour that goes from $0$ in the direction $-i$ and returns to $1$ in the direction $i$. Let $f(x) = i (x^3 - 12 x)$, then $$f(-i \xi) \sim -12 \xi, \quad f(1 - i \xi) \sim -11 i - 9 \xi, \quad \xi \to 0, \\ \int_0^1 e^{-x} e^{t f(x)} dx = -i \int_0^\infty e^{-12 t \xi} d\xi + i e^{-1} \int_0^\infty e^{-11 i t - 9 t \xi} d\xi + o(t^{-1}) = \\ \frac {4 \sin(11 t) + i (4 \cos(11 t) - 3 e)} {36 e t} + o(t^{-1}), \quad t \to \infty.$$