Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


$$h(u)= \int_{-\infty}^{\infty} g(x) \exp(iux) \, \text{d}x$$

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?

share|cite|improve this question
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

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.