I'm stuck on one of the questions related to the method of stationary phase in Murray's book on Asymptotic analysis. The question is as follows;
If $h(t)$ has a single stationary point at $t_0$, where $a < t_0 < b$, and $h'(t_0) = h''(t_0) = 0$, $h'''(t_0) > 0$, show that $$\int_a^b e^{i \lambda h(t)} dt \approx \Gamma\left(\frac43\right) \left\lbrace \frac{48}{\lambda h'''(t_0)}\right\rbrace^{\frac13} e^{i \lambda h(t_0)} \cos(\frac{\pi}{6}) \text{ as } \lambda \to \infty.$$
I don't really think this is meant to be particularly hard, but I'm kind of confused by a few of the terms in that expression, especially the gamma function term and the cosine term.
Regardless, I started with the Taylor expansion of $h(t)$; $$h(t) \approx h(t_0) + \frac{1}{3!}h'''(t_0)(t-t_0)^3$$
Which then gives;
$$\int_a^b e^{i \lambda h(t)} dt \approx \int_{t_0 - \epsilon}^{t_0 + \epsilon} e^{i \lambda \left[h(t_0) + \frac16 h'''(t_0) (t-t_0)^3 \right]}dt$$
Expanding out the right hand side integral, we get (I think...); $$ e^{i \lambda h(t_0)} \int_{t_0 - \epsilon}^{t_0 + \epsilon} e^{i \lambda \left[\frac16 h'''(t_0) (t-t_0)^3\right]} dt$$
At this stage, I'm a bit confused with what to do. In class, we're generally left with a squared term in the integral, to which we make a substitution $\pm s^2 = \text{stuff inside the square brackets multiplied by } \lambda$, but in this case, I don't think that that will work. We've taken the substitution to be $s^2$, because then we get an integral that is easy to evaluate.
However, as I said, I'm a bit unsure as to how I should proceed. I've tried a couple of substitutions (such as $s^2$, $s^3$, etc...), but none of these really seem to give me what I want.
Any hints as to how to proceed would be very much appreciated :).