# Fourier Transform of a Derivative [duplicate]

I'm trying to prove that:

$$F\,\{f'(x)\} = -i\omega F(\omega) \qquad (1)$$

where $\, F(\omega) = F\,\{f(x)\}$

This is my procedure so far:

$$F\,\{f'(x)\} = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} f'(t)e^{i\omega t} dt$$

Integrating by parts I obtained:

$$=\frac{1}{\sqrt{2\pi}} \big[ \space f(t) e^{i\omega t} \space \big|_{-\infty}^{\infty} - \int_{-\infty}^{\infty} i\omega f(t)e^{i\omega t} dt \space \big]$$

Now, in order to (1) to be true I need to get:

$$f(t)e^{i\omega t} \space \big|_{-\infty}^{\infty} =0 \qquad (2)$$

I developed it and got the following:

$$f(\infty)e^{i\omega \infty} - f(-\infty)e^{-i\omega \infty } = f(\infty)e^{i\omega \infty} - 0= f(t)e^{i\omega \infty}$$

I undarstand that the second term is $0$, since f(t) must have a value to accomplish Dirichlet conditions (and $e^{-\infty} = 0$), but I don't see how $f(\infty)e^{i\omega \infty}$ is $0$.

## marked as duplicate by Nosrati, Community♦Apr 17 '17 at 18:22

and $e^{-\infty} = 0$ but I don't see how $f(\infty)e^{i\omega \infty}$
It seems you are thinking in terms of exponentials with real exponents. But here we have imaginary exponents. The factors $e^{i\omega \infty}$ and $e^{-i\omega \infty}$ have the same behaviour, they are both bounded (further, their modulus equals $1$). Hence, it's enough that the function tends to zero.