Is there an asymptotic bound for this oscillatory integral? I have an oscillatory integral:
$$ \int u(x,y) e^{i\lambda f(x,y)} dx $$
with $f(x,y)\in \mathbb{C}^{\infty}$ a complex-valued function in a neighborhood of $(0,0)$ satisfying:
$$ \text{Im} f \geq 0  \quad \text{Im} f(0,0) =0 \quad f'_x(0,0) = 0 \quad \text{det}f'_{xx}(0,0) \neq 0$$
I also have $u(0,0)=0$. However, $u$ is not differentiable at $(0,0)$. This does not allow me to apply stationary phase theorems from Hormander (The analysis of linear partial differential operators, v.1). Is there an asymptotic ($\lambda \rightarrow +\infty$) bound on this integral?
Edit
I also checked that $u$ has the form of inner product on Hilbert space:
$$ u(x,y) = \langle g(y), P_{(0,\infty)}(\text{sgn}(x) A) h(y) \rangle $$
with $P_{(0,\infty)}$ spectral projector onto positive eigenvalues, $\text{sgn}(x)$ sign function, $A$ linear self-adjoint operator. The spectrum of $\text{sgn}(x) A$ is discrete.  
 A: Let me assume that $u$ is depending only on  the $y$ variable, that $f$ is smooth and depends also only on $y$ and is such that
$$
\Im f\ge 0,\quad f(0)=0, df(0)=0,\quad \det f''(0)\not=0.
$$
Then there exists a neighborhood $V$ of $0$ such that for $u\in L^\infty_{\text{comp}}(V)\cap C^1 (V\backslash\{0\})$, $u'\in L^1$,
$$
I(\lambda, u)=\int_{\mathbb R} e^{i\lambda f(x)} u(x) dx=O(\lambda^{-1/2}).
$$
To prove this, we can use a van der Corput method. Let $\chi$
be a smooth compactly supported function equal to $1$ on $[-1/2, 1/2]$, supported in $[-1,1]$.
We have 
$
I(\lambda, \chi(\cdot \lambda^{1/2})u)=O(\lambda^{-1/2})
$
since the integration takes place on $x\in [-\lambda^{-1/2}, \lambda^{-1/2}]$ and $u e^{if}$ is bounded. Moreover, checking
$$
I(\lambda, (1-\chi(\cdot \lambda^{1/2})) u)=
\int_{\mathbb R} e^{i\lambda f(x)} (1-\chi(x\lambda^{1/2}))u(x) dx,
$$
we use the identity (note that $f'(x)\not=0$ on $V\backslash\{0\}$)
$$
e^{i\lambda f(x)}=\frac{1}{i\lambda f'(x)}\frac{d}{dx}\bigl(e^{i\lambda f(x)}\bigr),
$$
and we integrate by parts. We need to check
$$
\lambda^{-1}\frac{d}{dx}\left(\frac{u(x) \bigl(1-\chi(\lambda^{1/2}x)\bigr)}{f'(x)}\right).
$$
Since $\vert (1-\chi(\lambda^{1/2}x))u'/f'\vert$ is bounded above by $\vert u'(x)\vert \frac{1}{\vert x\vert}$ and supported on $\vert x\vert\lambda^{1/2}\ge 1/2$, $\vert x\vert\lesssim 1$, we need to check
$$
\lambda^{-1}\int_{\lambda^{-1/2}\le \vert x\vert\lesssim 1}\frac{\vert u'(x) \vert}{\vert x\vert} dx\le \lambda^{-1/2}\int\vert u'(x) \vert dx,
$$
which is $O(\lambda^{-1/2})$ thanks to the assumption $u'\in L^1$.
Another term comes from $\lambda^{-1}(1-\chi)uf''/(f')^2$, 
and amounts to bound from above, since $u$ is bounded 
$$
\lambda^{-1}\int_{\vert x\vert\ge \frac12\lambda^{-1/2}} \frac{dx}{\vert x\vert^2}
\lesssim \lambda^{-1+\frac12},
$$ 
which is $O(\lambda^{-1/2})$.
We are left with
$$
\left\vert\lambda^{-1}\int \frac{u(x)}{f'(x)}\chi'(\lambda^{1/2}x)\lambda^{1/2} dx\right\vert
\le C\lambda^{-1/2}
\int_{\frac12 \lambda^{-1/2}\le \vert x\vert \le \lambda^{-1/2}}\frac{\vert u(x)\vert}{\vert x\vert} dx\lesssim\int_{\frac12 \lambda^{-1/2}\le \vert x\vert \le \lambda^{-1/2}} \vert{u(x)}\vert dx=O(\lambda^{-1/2}),
$$
since $u$ is bounded.
