How do I find the full asymptotic expansion of the following integral using steepest decent-? I came across this integral while working through the book on Mathematical Methods by Bender.
$$I=\int_0^1t^{\alpha}e^{ixt^3}dt$$.
to do this using Steepest descent, we have to deform the contour from $t= 0$ to $t= 1$ in the real part of t, in complex $t$ plane, into two contours of steepest descent joined by a contour with vanishing contribution- Now, $a(t) = it^3$, let $t = $u + i$v$, cubing $t$ we get- $a(t) = i(u^3- 3uv^2) - (3u^2v- v^3)$ for $t=0$, it's easy to see, $u=\sqrt3$, to get steepest descent contour, it's also not too hard to use laplace integral for the contour- $t= v( i + \sqrt(3))$, $0<=v<\infty$. But for $t= 1$, the contour itself is very messy and so is the integral along the contour through $t =1$. Any suggestions on how to proceed further? also, any idea on one would proceed, if integrated had higher powers of $t$ like- $t^5$? Thanks!