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Can anyone point me to a derivation of $x^n f(x)$? I know that the answer is $(i)^n$ times the $n$-th derivative of the transform of $f(x)$, but I've searched for a derivation and can't find it.

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@ctype.h, every time you edit an old problem (and you've done dozens), you bump it onto the front page, where it takes up valuable real estate. Please limit yourself to, maybe, three of these a day. –  Gerry Myerson Nov 22 '12 at 6:04
    
@GerryMyerson Sorry, I was not aware that it would be problematic. I will scale down the editing. –  ctype.h Nov 22 '12 at 6:52
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1 Answer 1

up vote 2 down vote accepted

It suffices to prove the case with $n=1$ and then apply induction.

Consider applying $d/ds$ to both sides of the definition (below) and then multiplying both sides by $i$.

$$ F(s)=\mathcal{F}\{f(x)\}(s)=\int_{-\infty}^\infty f(x)e^{-isx}dx$$

(We must have sufficient regularity on $f$ for this derivation to be valid.)

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