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

Suppose I have a function $f(t) \in L^2(\mathcal{R})$ and it is specified by:

$$f(t) = \int_{-\infty}^{\infty} H(\omega) \exp(-\beta t \omega) \exp(i \omega t)\,d\omega$$

Suppose $H(\omega)\in L^2(\mathcal{R})$, and we know it is among the subset of $L^2(\mathcal{R})$ which is Fourier transformable,but we don't know anything more specific than that. Is it possible to find an analytical formula for $\hat{f}(\omega)$, the Fourier transform of $f(t)$, in terms of $H(\omega)$ and $\beta$?

share|cite|improve this question
Should I edit this question and assume that that $H$ is in the subset of functions in $L_2$ which are Fourier transformable? – ncRubert Aug 31 '12 at 21:01
We have $f(t)=\hat{H}(t+i\beta t)$, correct? So what is $\mathcal{F}\{ \hat{H}(\alpha t) \}$, assuming we can use the formula for real $\alpha$ also for complex $\alpha$, if the function is sufficiently nice. – Thomas Klimpel Aug 31 '12 at 21:25
$L_2(\mathcal{R})$ functions are not Fourier transformable in general. For example $f(x)=\frac{1}{1+ix}\in L_2(\mathcal{R})$, but it is not Fourier transformable. – Mhenni Benghorbal Aug 31 '12 at 21:31
up vote 1 down vote accepted


If $\tilde f(\omega)$ is the Fourier transform of $f(t)$, then $$ f(t) = \int_{-\infty}^{\infty} \tilde f(\omega) e^{i\omega t}\,d\omega;$$ (or something similar depending on your definition of the Fourier transform).

Comparing with your formula you can pretty easily figure out what $\tilde f(\omega)$ is...

share|cite|improve this answer
I think you need to read my question more closely. You're ignoring the $exp(-\beta \omega t)$ – ncRubert Aug 31 '12 at 21:03
I'm not ignoring nothing. I just wrote down the definition of the Fourier transform... – Fabian Aug 31 '12 at 21:05
Besides the exponential basis $exp(i \omega t)$ there is a function of both time and frequency under the integral sign. That is the complicated part. If I understand your response you want to write $\hat{f}(\omega) = H(\omega)exp(-\beta t \omega)$ which is not correct. – ncRubert Aug 31 '12 at 21:12
@ncRubert: it is very interesting that you know what I want to write. I am sure that I wrote what I wanted to write. I however believe that you did not think about my hint to hard. how about another hint: did you try to move the integration contour a bit in the direction of the imaginary axis, e.g., integrating along $\tilde \omega = \omega + i \beta$. Of course you need certain assumptions on $H(\omega)$ to show that the resulting integral is the same. But that is why my answer was a hint and not a solution! – Fabian Aug 31 '12 at 21:18

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.