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Let be $f\in L^1(\mathbb{R})$,

I will be able to say that

$$ \dfrac{\hat{df(w)}}{dx} = \int_{-\infty}^{\infty}\dfrac{df(x)}{dx}\exp(-2\pi j wx)dx $$?


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Are you trying to differentiate a function of $w$ with respect to $x$? That doesn't look very promising. Also, if you are asking whether the derivative of the Fourier transform is the Fourier transform of the derivative, that is wrong in general. – Lukas Geyer Oct 29 '12 at 4:17
I think you want to differentiate $w$. – Mhenni Benghorbal Jun 25 '13 at 13:29
up vote 2 down vote accepted

When $f$ decays enough at $\pm\infty$ and $\hat{f}$ given by $$\hat{f}(w)=\int_{-\infty}^{\infty}f(x)\exp(-ixw)\,dx,$$ then \begin{eqnarray}\frac{d}{dw}\hat{f}(w)&=&\int_{-\infty}^{\infty}f(x)\frac{d}{dw}\exp(-ixw)\,dx\\ &=&-i\int_{-\infty}^{\infty}xf(x)\exp(-ixw)\,dx = -i\widehat{xf}(w) \end{eqnarray} (where in the last expression we abused the notation by writing $xf$ for the function $x\mapsto xf(x)$).

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I see what you mean. BTW it took me a minute to realize that your notation made sense. – Ron Gordon Jun 25 '13 at 13:38

Your notation makes no sense - you cannot differentiate the FT with respect to $x$, as there is no dependence. (Well, OK, you get zero.) Here's how it works: consider

$$\int_{-\infty}^{\infty} dx \, f'(x) e^{-i 2 \pi w x}$$

(Sorry, I insist on using $i$ rather than $j$ - this is a math site after all).

Assume we can integrate by parts:

$$\left [ f(x) e^{-i 2 \pi w x}\right]_{-\infty}^{\infty} + i 2 \pi w \int_{-\infty}^{\infty} dx \, f(x) e^{-i 2 \pi w x}$$

In order for the above to make any sense,

$$\lim_{x \to \infty} f(x) = 0$$

Then we have that the FT of the derivative of $f$ is $i 2 \pi w \hat{f}(w)$. Differentiation in one domain becomes multiplication in another. Is this what you were after?

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I just noticed that this was posted in October 2012. Better late than never. Btw, combining our answers gives the duality differentiation and the multiplicative action of polynomials. – AD. Jun 25 '13 at 13:35

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