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If we define the fourier transform of f as

$$\hat{f}(\omega) = \frac{1}{\sqrt{2}}\int_{-\infty}^\infty f(x) e^{-i\omega x} dx$$

then if f is differentiable, and the integrals for $\hat{f}$ and $\hat{f'}$ both converge then

$$\hat{f'}(w) = \frac{1}{i\omega}\hat{f}(\omega)$$

My first attempt was trying to use the fact that since both $\hat{f}$ and $\hat{f'}$ converge, then the improper integral in their respective definitions can be rewritten using the residue theorem, but this is as far as I've gotten and I am stuck. Someone suggested I try using integration by parts but I am unsure where to use this.

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  • $\begingroup$ The residue theorem doesn't apply.. Integrate by parts the given integral, differentiating $f$. And the constant is $\frac1{\sqrt{2\pi}}$ $\endgroup$
    – reuns
    Dec 4 '18 at 8:22
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\begin{eqnarray} \int_{-\infty}^{+\infty} \frac{{\rm d}f(x)}{{\rm d}x} e^{-i\omega x}~{\rm d}x &=& \int_{-\infty}^{+\infty}\left[\frac{{\rm d}}{{\rm d}x}(f(x)e^{-i\omega x}) - f(x)\frac{{\rm d}e^{-i\omega x}}{{\rm d}x} \right]{\rm d}x \\ &=& \left.f(x)e^{-i\omega x}\right|_{-\infty}^{+\infty} + i\omega \int_{-\infty}^{+\infty}f(x)e^{-i\omega x}~{\rm d}x \end{eqnarray}

The first term vanishes if the function $f$ goes to 0 at infinity. From here

$$ \hat{f'}(\omega) = (i\omega) \hat{f}(\omega) $$

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