Degenerate endpoint in Laplace's method

Suppose a real phase has its unique minimum at the left endpoint and $\varphi (t)-\varphi (0)\sim c{t}^m$, with $c>0$ and integer $m\ge 1$. For a continuous amplitude $a$ with $a(0)\ne 0$, localization followed by $s=(\nu c{)}^{1/m}t$ givesThis uses the Gamma integral. It assumes sufficient decay away from the minimum to make the remaining integral negligible. A cubic endpoint has width ${\nu }^{-1/3}$; an ordinary quadratic endpoint has width ${\nu }^{-1/2}$.

For a Laplace integral with local phase $\varphi (t;\delta )=\delta t+b{t}^3/3+\cdots$, $b>0$, an interior minimum approaches the endpoint as $\delta \uparrow 0$. The distinguished limit is $t=O({\nu }^{-1/3})$, $\delta =O({\nu }^{-2/3})$. With $s=(\nu b{)}^{1/3}t$, its leading local integral isThe cubic Laplace transition integral joins the ordinary endpoint estimate for positive $\delta$, the cubic endpoint estimate at zero, and the interior Laplace's method estimate for negative $\delta$. Separate fixed-parameter formulas fail to be uniform when the minimum is within its own width of the endpoint.