Incorrect proof of Fourier translation theorem

Question

I'm trying to follow the proof of the Fourier shift theorem, which states: $$ F[f(x-a)]=e^{-iwa}F[f(x)] $$

There's one other similar question here. However, the only answer boils down to "what you're trying to do makes no sense". I'm asking again to make the question more explicit.

What is wrong with the following logic?

$$ \begin{align*} \text{The Fourier transform is }F[f(x-a)] & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x-a)e^{-iwx}dx \end{align*} $$ Now my inclination is to define $\lambda=x-a$,so we can substitute $dx=d\lambda$ and $x=\lambda+a$. But this results in a restatement of the Fourier transform with $\lambda$ instead of x, and with a different and incorrect exponent. $$ \begin{align*} F[f(\lambda)] & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw(\lambda+a)}d\lambda\\ & =\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}e^{-iwa}d\lambda\\ & =e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}d\lambda\\ & =\boxed{e^{-iwa}F[f(\lambda)]} \tag{This does not follow!}\\ F[f(x-a)] & =e^{-iwa}F[f(x-a)] \end{align*} $$

This result is inconsistent and contradicts the theorem. Clearly I have a fundamental misunderstanding here somewhere... but what?

Answer 1

Your proof is almost correct, just in the last step observe: \begin{align} e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}d\lambda=e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-iwx}dx, \end{align} And now one can continue your derivation as \begin{align} & F[f(x-a)]=e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}d\lambda= e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-iw x}dx= e^{-iwa}F[f(x)]. \end{align}

Answer 2

Apparently I forgot what integrals do. The following step is so elementary that the proofs I reviewed didn't bother to state it:

$$\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}d\lambda = \int_{-\infty}^{\infty}f(x)e^{-iwx}dx\\$$

So taking over my "incorrect" proof at the integral wrt $\lambda$: $$ \begin{align*} & =e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(\lambda)e^{-iw\lambda}d\lambda\\ & =e^{-iwa}\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x)e^{-iwx}dx\\ & =e^{-iwa}F[f(x)] \end{align*} $$