Prove that $f_{\theta}(z)=\int_{\Bbb{R}} e^{-ix\theta}e^{-z|x|^2} dx$ is holomorphic Let $$f_{\theta}(z) = \int_{\Bbb{R}}e^{-ix\theta} e^{-z|x|^2}dx$$ with $\Omega = \{z \in \Bbb{C}\mid Re(z) >0\}$ such that $z\in \Omega$ and $x,\theta \in \Bbb{R}$.
Prove $f_{\theta}(z)$ is holomorphic on $z\in \Omega$.
My attempt:
We may use the following theorem in Stein & Shakarchi's book, with a slightly change:

Theorem 5.4 Let $F(z, s)$ be defined for $(z, s) \in \Omega
 \times[0,1]$ where $\Omega$ is an open set in $\mathbb{C}$. Suppose
$F$ satisfies the following properties:
(i) $F(z, s)$ is holomorphic in $z$ for each $s$.
(ii) $F$ is continuous on $\Omega \times[0,1]$.
Then the function $f$ defined on $\Omega$ by $$ f(z)=\int_{0}^{1} F(z,
 s) d s $$ is holomorphic.

First we can see for each $x$ the integrand is holomorphic, the condition (ii) needs a slight change here since the proof use the Riemann sum to avoid the Fubini theorem, is there some way to improve the thereom to apply to the case above?
I do as follows, apply Morera's theorem. To make the Morera's theorem holds we need to check for each compact set $K\subset \Omega$ for each triangle $\Delta$ in $K$ the following integral is zero:
$$0 =\int_\Delta \int_{\Bbb{R}} e^{-ix\theta} e^{-z(t) x^2}dxdz(t)  =\sum_{i=1}^3 \int_{[a_i,b_i]} \int_{\Bbb{R}} e^{-ix\theta} e^{-z(t) x^2}z'(t)dxdt$$
If we can apply Fubini, then due to holomorphic of inner integrand, it's zero, hence we are done.
To check Fubini, since a real integral now, on this compact set $Re(z)>\epsilon>0$ since $|e^{-ix\theta} e^{-z(t) x^2}z'(t)| \le e^{-Re(z)x^2} \le e^{-\epsilon x^2} \in L^1(\Bbb{R} \times [a,b])$, so we have $e^{-ix\theta} e^{-z(t) x^2}z'(t) \in L^1(\Bbb{R} \times [a,b])$ is my idea correct?
 A: One may try to show directly that  $z\mapsto f_\theta(z)$ is analytic.
$f_\theta(z)=\int e^{-ix\theta}e^{-z x^2}\,dx$ is well defined for all $z\in H:=\{u+iv: u>0\}$.  Fix $z_0\in H$ and choose $\delta$ small enough so that $|h|\leq\delta$ implies $z_0+h\in H$
$$\frac{f_\theta(z_0+h)-f_\theta(z_0)}{h}=\int e^{-ix\theta}e^{-z_0x^2}\frac{e^{-hx^2}-1}{h}\,dx$$
The convexity of the exponential map implies that
$$\Big|\frac{e^{-hx^2}-1}{h}\Big|\leq \frac{e^{|h|x^2}-1}{|h|}\leq \frac{e^{\delta x^2}-1}{\delta }\leq \frac{e^{\delta x^2} + e^{-\delta x^2}}{\delta}$$
Consequently, if $z_0=u_0+iv_0$
$$\left|e^{-ix\theta}e^{-z_0x^2}\frac{e^{-hx^2}-1}{h}\right|\leq \frac{e^{-(u_0-\delta)x^2}+e^{-(u_0 +\delta)x^2}}{\delta}\in L_1(\mathbb{R},\lambda)$$
for all $|h|\leq\delta$.
As $\lim_{h\rightarrow0}e^{-ix\theta}e^{-z_0x^2}\frac{e^{hx^2}-1}{h}=-e^{-ix\theta}x^2e^{-z_0x^2}$, an application of dominated convergence gives
$$\lim_{h\rightarrow0}\frac{f_\theta(z_0+h)-f_\theta(z_0)}{h}=-\int e^{-ix\theta}x^2e^{-zx^2}\,dx$$
This shows that $f_\theta$ us holomorphic at any point $z_0\in H$.
A: I'd use the Cauchy-Riemann equations in the form
$$
i\frac{\partial f}{\partial x} = \frac{\partial f}{\partial y}
$$
and apply it to your integral function
$$\DeclareMathOperator{\dm}{\!\!\operatorname{d}\!}
f_{\theta}(z) = \int\limits_{\Bbb{R}}e^{-it\theta} e^{-z|t|^2}\dm t
$$
(I have only used the symbol $t$ for the integration variable in order to avoid confusion since I assume $z=x+iy$ as it is customarily done). Then, for every $z$ belonging to the right half plane $\Bbb H$ we have
$$
\begin{split}
\frac{\partial }{\partial x}f_\theta (x,y) & = 
\frac{\partial }{\partial x} \int\limits_{\Bbb{R}}e^{-it\theta} e^{-(x+iy)|t|^2}\dm t\\
& = - \int\limits_{\Bbb{R}}|t|^2e^{-it\theta} e^{-(x+iy)|t|^2}\dm t\\
\\
\frac{\partial }{\partial y}f_\theta(x,y) &=
\frac{\partial }{\partial y} \int\limits_{\Bbb{R}}e^{-it\theta} e^{-(x+iy)|t|^2}\dm t\\
& = - i\int\limits_{\Bbb{R}}|t|^2e^{-it\theta} e^{-(x+iy)|t|^2}\dm t\\
\end{split}
\iff\quad i\frac{\partial f_\theta}{\partial x} = \frac{\partial f_\theta}{\partial y}
$$
Note
A proof of the partial derivability of $f_\theta$ under the integral symbol would be needed: however, it is not difficult to find it in many textbooks, or to use the arguments used by Oliver in his answer.
