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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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    $L_2(\mathcal{R})$ 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.2012-08-31

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Hint:

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...

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    @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!2012-08-31