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I am looking at the proof of the Fourier transform of a Gaussian, that is, for $z\in \mathbb C$ with $\Re z>0$, we have $\mathcal{F}(e^{-\frac{z|x|^2}{2}})(p)=\frac{1}{z^{d/2}}e^{-\frac{|p|^2}{2z}},$ where we interpret $z^{d/2}$ as $(\sqrt{z})^d$, and the branch cut of the square root is chosen along the negative real axis.

The proof is completed by showing that $\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}} e^{-zx^2/2}dx=\frac{1}{\sqrt z}$.

This is true for $z>0$ by applying the change of variables $y=\sqrt{z}x$.

For general $z\in \mathbb C$ it says this follows by analytic continuation.

To apply analytic continuation I need to know that the function $$z\mapsto \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{-zx^2/2}dx$$ is analytic in $\{z\in \mathbb{C}, \Re z>0\}$.

How can I show this? One method I thought of is Morera's theorem, but I don't know how to guarantee that Fubini's theorem apply here to exchange the integrals. I would greatly appreciate any help.

Gary
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nomadicmathematician
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    $\text{Re}(z)>0$ so the integral is exponentially decaying. This is more than enough to apply Fubini for Morera. You can also apply the holomorphic version of differentiation under the integral. See here for more details. There I give the application to the Gamma function, but the first part is still relevant here. – peek-a-boo Jan 27 '24 at 02:31
  • @peek-a-boo Thank you for a great theorem. Could you be more explicit about why the Fubini works because the integral is exponentially decaying? I don't know how to show that the double integral will be finite. – nomadicmathematician Jan 27 '24 at 02:43
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    Look at the theorem I quoted. Surely you can try to come up with a dominating function $g_a$ which works? In fact making a change of variables, you’ll see that the Gaussian integral is extremely closely related to the Gamma function (you don’t need to change variables… I’m just pointing out these two are very closely related). So 95% of what I wrote there (and the types of estimates and dominating functions I’m coming up with there) will work here as well. So I strongly suggest reading the details there more closely. – peek-a-boo Jan 27 '24 at 02:43
  • @peek-a-boo So I think for any closed curve $\gamma$, we can take $a=\min |z|$ in $\gamma$ and uniformly bound $e^{-zx^2/2}$ by $e^{-ax^2/2}$, which will be integrable in $\mathbb R$ and then is just a constant so integrable over $\gamma$ as well, which will bound the original function over the double integral. Is this correct? – nomadicmathematician Jan 27 '24 at 03:08
  • yup that’s it.. – peek-a-boo Jan 27 '24 at 03:24

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