Problem understanding the Stationary-Phase-Approximation example of Wikipedia? I am trying to understand the Stationary-Phase Approximation from the Wikipedia example showed in here, but there is something I don´t understand how is made work. Following the article's example:
$$f(x,t) = \frac{1}{2\pi} \int\limits_{\mathbb{R}} F(w)\,e^{i[k(w)x-wt]}\,dw \,\,\,\,\texttt{(Eq. 1)}$$
Then, the phase term $\phi = k(w)x-wt$ is stationary when:
$$\frac{d}{dw}\left( k(w)x-wt \right)=0 \,\,\Rightarrow w_0\,\,\,\,\texttt{(Eq. 2)}$$
Or equivalently:
$$ \frac{dk(w)}{dw} = \frac{t}{x}\,\,\,\,\texttt{(Eq. 3)}$$
Then, through Taylor series approximations and other manipulations the Stationary-Phase Approximation is given by:
$$ f(x,t) \approx \frac{|F(w_0)|}{2\pi}\sqrt{\frac{2\pi}{x|k''(w_0)|}} \cos\left( k(w_0)x-w_0t \pm\frac{\pi}{4} \right) \,\,\,\,\texttt{(Eq. 4)}$$
I believe that the following term of Eq. 4 is interpreted as:
$$k''(w_0) \cong \frac{d^2}{dw^2}\left(k(w) \right)\Big|_{w=w_0}$$
But since the right-side of Eq. 3 is independent of $w$, it means that:
$$ \frac{d^2}{dw^2}\left(k(w) \right) = \frac{d}{dw}\left( \frac{dk(w)}{dw} \right) = \frac{d}{dw}\left( \frac{t}{x} \right) = 0 \,\,\,\forall \,w \Rightarrow k''(w_0) = 0$$
Which will means that I have a division by zero happening in Eq. 4.
So, How is possible that Eq. 4 is not undetermined?? What I am understanding wrongly??

We are considering a point $w_0$ where we have the equality $\frac{dk}{dw}\Big|_{w=w_0} = \frac{t}{x}$. There is no claim that the identity holds for all $w$. For example, let $k=w^2$. Then $\frac{dk}{dw} = 2w$ and we are considering a point where $w = \frac{t}{2x}$. But clearly $\frac{d^2 k}{dw^2} =2 \neq 0$.
