# Fourier transform of an exponentially attenuated function

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 $\beta$?

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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

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...
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
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