The Fourier series transform of a function $f(x)$ is given by $$ \mathcal{F} [f(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x) e^{-i k x} dx $$
Apparently, we also have $$ \mathcal{F} [f(ax)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(ax) e^{-i k x} dx $$ However, I am unsure of why this would not be expressed as $$ \mathcal{F} [f(ax)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(ax) e^{-i k a x} dx $$
Can anyone please explain to me why this is?
On
Think of the transformation as an operator that takes in a function $f$, and spit out another function $\mathcal F[f]$.
Now you have another function $g(x) = f \circ l(x)$, where $l(x) = ax $
But no matter how complicated your $g(x)$, the operator still treat it as an input and map it to another function the same way.
On
Read your course again.
If $a > 0$ then (change of variable $y = ax$) $$\mathcal{F} [f(ax)](k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(ax) e^{-i k x} dx = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(y) e^{-i k y/a} \frac{dy}{a} = \frac{1}{a} \mathcal{F} [f(x)](k/a) $$
If $a < 0$ then $$\mathcal{F} [f(ax)](k) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(ax) e^{-i k x} dx = \frac{1}{\sqrt{2 \pi}} \int_{\infty}^{-\infty} f(y) e^{-i k y/a} \frac{dy}{a} = \frac{1}{|a|} \mathcal{F} [f(x)](k/a) $$
And this last formula $\mathcal{F} [f(ax)](k)=\frac{1}{|a|} \mathcal{F} [f(x)](k/a)$ is true whenever $a \in \mathbb{R}^*$. See also the big table.
Just call $g(x) = f(ax)$, then you have
$$ \mathcal{F} [g(x)] = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty g(x) e^{-i k x} dx = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(a x) e^{-i k x} dx = \mathcal{F} [f(ax)] $$