Differentiation under integral sign holomorphic function. While reading this set of notes  I found the following result, 

Let  $f\in\mathbb L^2_{\nu}(\mathbb R)$ (the space of square integrable functions with respect to the standard gaussian measure) be orthogonal to the span of
  all monomials $(1,x,x^2,\cdots)$ ,  and define the complex valued
  function $g$ by:
\begin{equation} g(z)=\int_{\mathbb R} e^{i x z} f(x)\nu(d
x)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb R} e^{i x z} f(x) e^{-x^2/2} dx.
 \end{equation}
Hence we have that all derivatives of the holomorphic $g$ vanishes at
  $z=0$ (because of the orthogonality with monomials).

In order for the previous assertion to hold I should take derivatives under the integral sign. Doing some research on google I've found the following result:

Let $h:\mathbb R\to \mathbb C$be locally integrable with respect to
  Lebesgue measure, let $$H:=\big\{z\in\mathbb C:\int_{\mathbb R}
 |e^{izx}h(x)|dx<\infty\big\},$$ $$\hat{h}(z):=\int_{\mathbb R} e^{izx}h(x)dx, z\in
H,$$
Assume the interior of $H$, is not empty. Then $\hat{h}$ is
  holomorphic in the interior of $H$ and may be differentiated under the
  integral.

So I wanted to apply this to my function $g$ above in order to show that interchanging the derivative with the integral is legit.
I barely know anything about complex analysis so be patient with me and please let me know if I am doing something wrong.
First notice that  $f(x)e^{-x^2/2}$ is (globally) integrable on $\mathbb R$ with respect to the Lebesgue measure, and let
$$H:=\big\{z\in\mathbb C:\int_{\mathbb R} |e^{izx}f(x) e^{-x^2/2}|dx<\infty\big\}=\big\{(a+ib) :\int_{\mathbb R} |e^{i(a+ib)x}f(x) e^{-x^2/2}|dx<\infty\big\}$$
(since the notes say "the holomorphic function $g$, without specifying any subset of the complex, I assume they intend to say "holomorphic all over the complex plane", hence $H=\mathbb C$ right?)
One can see that 
$$\int_{\mathbb R} |e^{i(a+ib)x}f(x) e^{-x^2/2}|dx= \int_{\mathbb R} |e^{iax}|\cdot |e^{-bx}f(x) e^{-x^2/2}|dx\leq \int_{\mathbb R} |f(x)|e^{-bx} e^{-x^2/2}dx=e^{b^2/2}\int_{\mathbb R} |f(x)| e^{-(x+b)^2/2} dx<?$$
Hence as I understand $a$ could take any value in $\mathbb R$, but in general terms I cannot tell whether there exists some $b\neq 0$ such that the last integral is finite. 
Hence here's my question, with the information regarding $f$ we have, is it legit to differentiate under the integral sign?
Thanks in advance!
 A: $$H:=\big\{z\in\mathbb C:\int_{\mathbb R} |e^{izx}f(x) e^{-x^2/2}|dx<\infty\big\}=\big\{a,b\in\mathbb R :\int_{\mathbb R} |e^{i(a+ib)x}f(x) e^{-x^2/2}|dx<\infty\big\}$$
One can see that the following holds because of the boundedness of the complex exponential (implying that $a$ could be any real number).
$$\int_{\mathbb R} |e^{i(a+ib)x}f(x) e^{-x^2/2}|dx= \int_{\mathbb R} |e^{iax}|\cdot |e^{-bx}f(x) e^{-x^2/2}|dx\leq \int |f(x)|e^{-bx} e^{-x^2/2}dx =\sqrt{2\pi}  \int_{\mathbb R} |f(x)|e^{-bx} \nu(dx) $$
Finally a straightforward application of Cauchy-Schwartz inequality shows that 
$$\int_{\mathbb R}|f(x)|e^{-bx} \nu(dx) \leq \left(\int_{\mathbb R} f(x)^2 \nu(dx)\right)^{1/2}\cdot\left(\int_{\mathbb R} e^{-2bx}\nu(dx)\right)^{1/2}$$
Clearly the first integral on the right hand side is finite by assumption, while for the second integral; letting $2b=c$ we have
$$\int_{\mathbb R} e^{-(x^2+2xc)/2}dx=e^{c^2/2}\int_{\mathbb R} e^{-(x^2+2xc+c^2)/2}dx=e^{c^2/2}\int_{\mathbb R}  e^{-(x-(-c))^2/2}dx=e^{c^2/2}\sqrt{2\pi}=e^{2b^2}\sqrt{2\pi}<\infty $$
Hence we can see that the integral $\int_{\mathbb R} |e^{i(a+ib)x}f(x) e^{-x^2/2}|dx$ is finite for any real values of $a$ and $b$, i.e. $H=\mathbb C$.
