I'm stuck at the following integral
$I(x,t) = \int_{-\infty}^{\infty}\{F(k)exp(it\psi(k)) \}dk$ with $\psi(k) = (k-k_0)(\frac{x}{t}) - (\beta(k)-\beta_0)$ where $\beta_0=\beta(k_0)$ and $F(k)= \frac{1}{k^2-k_0^2}$. $\beta(k)^2 = gktanh(kh)$, where $\beta(k)$ is taken as the positive root when $k \in \mathbb{R}$. Here $g,h$ are constants. The constraint on parameters is $t>0 $ & $x>0$.
In addition to this, take $\psi'(k_0) <0$ for the given $\frac{x}{t}$.
For $\lim_{t\rightarrow\infty}$ and finite $\frac{x}{t}$, an asymptotic expansion with the dominant first term is required. The stationary phase method is not directly applicable as $F(k_0)$ is singular.
I've tried to use the theorem 15 in sec 3.1.5 in lecture notes by Avramidi.
In here, i tried to turn the denominator into two partial fractions viz. $F(k)= \frac{1}{k^2-k_0^2} = \frac{1}{2k_0}\{ \frac{1}{k-k_0} - \frac{1}{k+k_0}\}$.
Thus $f(x)\equiv 1$. However, according to the hypothesis of the theorem $f(x)$ is compact. Though there is no limit specified on the extent of $supp \ f$ by the theorem. So if we know that the expansion is dominated by $x=0$, i think the theorem must be applicable to this as well. However, i'm not sure with this reasoning.
Does anyone have a better idea how to evaluate this integral?