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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?

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I have tried to improve your post a bit, check if it is still what you are trying to say. Note the difference between the value $g(x)$ and the function $g$. –  Jonas Teuwen Oct 10 '11 at 19:00

1 Answer 1

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.

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