Let
$$h(u)= \int_{-\infty}^{\infty} g(x) \exp(iux) \, \text{d}x$$
be the Fourier transform. Then let us suppose that for $ |x| \to \infty $ the function $g$ goes as
$$g(x) = \exp(-ax) \text{ for some positive $a$}$$
Does this mean that we can analytically continue the Fourier transform $h$ to the region of the complex plane where $-a < \mathop{Im}(z) < a$? Apart from $h(u)$ being defined for every real $u$?
If $g(x)$ tends to $0$ sufficiently fast does it mean I can define the Fourier transform $h(z)$ for every complex number $z$?
In fact if we put $u = iz$ for real $z$ then is
$$h(iz)= \int_{-\infty}^{\infty}g(x) \exp(-zx) \, \text{d}x$$
defined for every real $z$ in this case if $g$ is a Gaussian for example?
Best Answer
Yes, it is possible to continue $h$ analytically to $-a < \text{Im}(z) < a\ $ if $|g(x)| \le C\exp(-a|x|)\ $ for some $a>0$. The function is defined by putting $z$ instead of $x$ in the Fourier transform. If $g$ satisfy this estimate for any $a>0$ then $h$ is an entire function.