Let $x\in \Bbb R^n, y\in\Bbb R$, and $z\in\Bbb C$. For a fixed $\lambda\in\Bbb R^n$ and a smooth, compactly supported function $\varphi\in\mathcal D(\Bbb R^{n+1})$, I want to show that $$ f(z):=\int_{\Bbb R^n} e^{-i\lambda\cdot x} \left( \int_{\Bbb R} e^{-izy} \varphi(x,y) dy \right) dx $$ is an analytic function.
The hint says that using Fubini's theorem and Morera's theorem is the way to go. My knowledge of complex analysis is very rusty so I'd really appreciate any help or advice.
By Morera's theorem, we need to verify
$$\int_{\partial\Delta} f(z)\,dz = 0\tag{1}$$
for all triangles $\Delta \subset \mathbb{C}$. Using the definition of $f$ and Fubini's theorem, we have
$$\int_{\partial\Delta} f(z)\,dz = \int_{\mathbb{R}^n} e^{- i \lambda\cdot x} \int_{\mathbb{R}} \varphi(x,y) \int_{\partial \Delta} e^{-izy}\,dz\,dy\,dx.$$
Since for every fixed $y$ the function $z \mapsto e^{-izy}$ is holomorphic, the innermost integral vanishes for all $y$, and thus $(1)$ follows.
The application of Fubini's theorem is justified because $(x,y,z) \mapsto e^{-i\lambda x} e^{-izy}\varphi(x,y)$ is continuous, hence measurable, and the integral of the modulus is finite.