I am trying to prove the below identity where $f_c(x)=f(cx)$ such that c is a positive number.

$F_c(\alpha)=\frac 1 c F(\frac\alpha c)$

F above represents the Fourier transformed $f(x)$. I attempted this by representing $f(x)$ as a Fourier series such that we can represent its values in the frequency domain:

$f(x) = a_0 + \sum_{n=1}^\infty a_n \cos(n\pi x ) + \sum_{n=1}^\infty b_n \cos(n\pi x ) $

Because the coefficients are defined over your signal period, the coefficients do not change when writing f(cx). You can then prove that:

$f(cx) = a_0 + \sum_{n=1}^\infty a_n \cos(n\pi xc ) + \sum_{n=1}^\infty b_n \cos(n\pi xc ) $

This gives us an integral we can solve when plugging back into the Fourier transform:

$F_c(\alpha)=\int_{-\infty}^\infty \left(a_0 + \sum_{n=1}^\infty a_n \cos(n\pi xc ) + \sum_{n=1}^\infty b_n \cos(n\pi xc )e^{-i\alpha xc}\right)$

Euler's number can of course be represented as $\cos(\alpha xc)+i\sin(\alpha cx)$

For the sake of brevity, I'll apply this integral to only the $a_0$ term to represent my dilemma.

$F_c(\alpha)=a_0\int_{-\infty}^\infty (\cos(\alpha xc)+i\sin(\alpha cx))dx+...$

$F_c(\alpha)=\frac{a_0}{\alpha c}(\sin(\alpha xc)]^\infty _{-\infty}-\cos(\alpha xc)]^\infty _{-\infty} ) $

And here lies my dilemma. The above equation for f(x) is:

$F(\alpha)=\frac{a_0}{\alpha}(\sin(\alpha x)|^\infty _{-\infty}-\cos(\alpha x)|^\infty _{-\infty} ) $

Since $\sin(\alpha xc)|^\infty _{-\infty}\ne \sin(\frac{\alpha x}{c})|^\infty _{-\infty}$ I don't know how to prove the identity.