# Fourier transform

#### Poirot

##### Banned
What is the fourier transform of $f'(ax)$, where a>0 is a constant? Firstly, I reasoned that (lets say $F[f]$ is the fourier transform of f) $F[f('x)]=\frac{1}{a}F[f](\frac{k}{a})$ by scaling theorem, then using the derivative rule we get $F[f'(ax)]=\frac{ik}{a}F[f(x)](\frac{k}{a})$. But when I did it manually (substitution and integration by parts), I got an extra factor of $\frac{1}{a}$. So which is correct and why the discrepency?

#### CaptainBlack

##### Well-known member
What is the fourier transform of $f'(ax)$, where a>0 is a constant? Firstly, I reasoned that (lets say $F[f]$ is the fourier transform of f) $F[f('x)]=\frac{1}{a}F[f](\frac{k}{a})$ by scaling theorem, then using the derivative rule we get $F[f'(ax)]=\frac{ik}{a}F[f(x)](\frac{k}{a})$. But when I did it manually (substitution and integration by parts), I got an extra factor of $\frac{1}{a}$. So which is correct and why the discrepency?
Let: $$g(x)=f'(x)$$ and $$h(x)=g(a x)$$, and $$H(\omega)$$, $$G(\omega)$$ and $$F(\omega)$$ be the FT of $$h(x)$$, $$g(x)$$ and $$h(x)$$ respectively.

then:

$H(\omega)=\frac{1}{a}G\left(\frac{\omega}{a}\right)$

and:

$G(\omega) = i \omega F(\omega)$

Now substitute the second into the first to get:

$H(\omega)=\frac{1}{a} \frac{ i \omega}{a}F\left(\frac{\omega}{a}\right) = \; \frac{i \omega}{a^2} F\left( \frac{\omega}{a}\right)$

CB

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