The proof is given for the model quadratic phase $\varphi(y) = |y|^2$. The general case follows from this model by a change of variables (Morse lemma) that reduces the phase to its second-order Taylor expansion near the critical point. The strategy is to regularise the oscillatory [integral](/page/Integral) using a Gaussian damping factor, evaluate the result using the known [Fourier transform](/page/Fourier%20Transform) of a Gaussian, and expand the phase correction to identify the leading term. **Step 1 (Gaussian regularisation).** Write \begin{align*} I(\lambda) = \lim_{\varepsilon \to 0} \int_{\mathbb{R}^n} e^{(i\lambda - \varepsilon)|y|^2} a(y)\, d\mathcal{L}^n(y). \end{align*} For $\operatorname{Re}(z) > 0$, the Fourier transform identity $\mathcal{F}(e^{-z|y|^2})(\xi) = (2z)^{-n/2} e^{-|\xi|^2/(4z)}$ and Parseval's formula give \begin{align*} \int_{\mathbb{R}^n} e^{-z|y|^2} a(y)\, d\mathcal{L}^n(y) = \int_{\mathbb{R}^n} (2z)^{-n/2} e^{-|\xi|^2/(4z)}\, \check{a}(\xi)\, d\mathcal{L}^n(\xi), \end{align*} where $\check{a} = \mathcal{F}^{-1}(a)$. Setting $z = \varepsilon - i\lambda$ and taking $\varepsilon \to 0$: \begin{align*} I(\lambda) = \frac{1}{(2i\lambda)^{n/2}} \int_{\mathbb{R}^n} e^{|\xi|^2/(4i\lambda)}\, \check{a}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} **Step 2 (Expansion of the phase correction).** The integrand contains the factor $e^{|\xi|^2/(4i\lambda)}$, which converges pointwise to $1$ as $\lambda \to \infty$. We split: \begin{align*} \int_{\mathbb{R}^n} e^{|\xi|^2/(4i\lambda)}\, \check{a}(\xi)\, d\mathcal{L}^n(\xi) = \int_{\mathbb{R}^n} \check{a}(\xi)\, d\mathcal{L}^n(\xi) + \int_{\mathbb{R}^n} \left(e^{|\xi|^2/(4i\lambda)} - 1\right) \check{a}(\xi)\, d\mathcal{L}^n(\xi). \end{align*} [claim:Remainder Estimate] $\left|\int_{\mathbb{R}^n} (e^{|\xi|^2/(4i\lambda)} - 1)\, \check{a}(\xi)\, d\mathcal{L}^n(\xi)\right| \le C_a\, \lambda^{-1}$. [/claim] [proof] Using $|e^{i\theta} - 1| \le |\theta|$ for $\theta \in \mathbb{R}$, we bound the integrand by $\frac{|\xi|^2}{4\lambda} |\check{a}(\xi)|$. Since $a \in C_c^\infty(\mathbb{R}^n)$, the inverse Fourier transform $\check{a} \in \mathcal{S}(\mathbb{R}^n)$, so $|\xi|^2 |\check{a}(\xi)| \in L^1(\mathbb{R}^n)$. Therefore \begin{align*} \left|\int (e^{|\xi|^2/(4i\lambda)} - 1)\, \check{a}(\xi)\, d\mathcal{L}^n(\xi)\right| \le \frac{1}{4\lambda} \int_{\mathbb{R}^n} |\xi|^2 |\check{a}(\xi)|\, d\mathcal{L}^n(\xi) = \frac{C_a}{\lambda}. \end{align*} [/proof] **Step 3 (Leading term).** The [Fourier inversion theorem](/theorems/633) gives $\int \check{a}(\xi)\, d\mathcal{L}^n(\xi) = (2\pi)^{n/2} a(0)$. Combining Steps 1–2: \begin{align*} I(\lambda) = \frac{(2\pi)^{n/2}}{(2i\lambda)^{n/2}}\, a(0) + O(\lambda^{-n/2-1}) = \frac{C_n}{\lambda^{n/2}}\, e^{i\pi n/4}\, a(0) + O(\lambda^{-n/2-1}), \end{align*} where $C_n = (\pi/2)^{n/2} \cdot 2^{n/2} \cdot (2\pi)^{-n/2} \cdot e^{i\pi n/4}$. For general $\varphi$ with $\nabla\varphi(y_0) = 0$ and $\nabla^2\varphi(y_0)$ non-degenerate, the Morse lemma provides a local $C^\infty$ diffeomorphism $y \mapsto w$ near $y_0$ such that $\varphi(y) = \varphi(y_0) + Q(w)$, where $Q$ is a non-degenerate quadratic form with $|\det \nabla^2 Q| = |\det \nabla^2\varphi(y_0)|$. Applying the quadratic result in the new coordinates and accounting for the Jacobian yields the stated formula.