Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

I will be able to say that

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

Why?

share|improve this question
3  
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.r.to $w$. –  Mhenni Benghorbal Jun 25 '13 at 13:29
add comment

2 Answers 2

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 $x\mapsto xf(x)$).

share|improve this answer
    
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
add comment

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?

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.