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I'm trying to find the fourier transform of $t\cdot f''$. The rules I've found that relates to this seems to be that for a function $f(t)$ and it's Fourier transform $F(\omega)$ the following holds: $\frac{d^nf(t)}{dt^n} = (i\omega)^nF(\omega)$ and $(-it)^nf(t) = \frac{d^nF(\omega)}{d\omega^n}$.

Does this mean that the Fourier transform is $(i\cdot t)^2 \cdot \frac{dF(t)}{d\omega}$?

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Isn't the value of $\frac{dF(t)}{d\omega}$ zero? –  Dilip Sarwate Dec 8 '12 at 22:26
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up vote 0 down vote accepted

The Fourier transform is $$ i\frac{d}{d\omega}\bigl((i\,\omega)^2F(\omega)\bigr). $$

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Thanks for the answer, but since I'm slightly retarded, how do you reach that conclusion? –  Jonas Dec 8 '12 at 23:30
    
Let $g(t)=f''(t)$ and let $G(\omega)$ be the Fourier transform of $g$. Then, according to the rules in your post, the Fourier transform of $t\,g=i(-i\,t)g$ is $i\,\frac{d}{d\omega}G(\omega)$. Now, again according to the same rules, the Fourier transform of $G$ is $(i\,\omega)^2F(\omega)$. –  Julián Aguirre Dec 8 '12 at 23:39
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