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I'm currently working on the problem:

If $\hat{f}(k)$ is the complex Fourier Transform of the function $f(x)$ and $a$ is a real constant with $a>0$, show that the complex Fourier Transform of $f(ax)$ is $\dfrac{1}{a} \hat{f}(\dfrac{k}{a})$.

I've shown this using a change of variables. However, now:

What is the corresponding result if $a<0$?

Why is this a seperate case from the above? Is it because $\hat{f}$ can only take positive inputs? If so, how would I go about finding the next result?

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The direction of integration over $\mathbb{R}$ after a change of variable depends on the sign of $a$. – WimC Apr 27 '13 at 18:05
up vote 1 down vote accepted

Suppose $a<0$. Then, \begin{align*} \hat{f}(ak)&=\int_{-\infty}^{\infty}f(ax)e^{-ik x}\, dx= \{ ax=s \} = \int_{\color{red}\infty}^{\color{red}-\color{red}\infty}\frac{f(s)}{a}e^{-ik \frac{s}{a}}\, ds= \\ &= -\int_{-\infty}^{\infty}\frac{f(s)}{a}e^{-i \frac{k}{a}s}\, ds=\int_{-\infty}^{\infty}\frac{f(s)}{|a|}e^{-i \frac{k}{a}s}\, ds=\frac{1}{|a|}\hat{f}\left( \frac{k}{a} \right) \end{align*} The trick is, as WimC already said, that since $a$ is negative, we must change the order of integration.

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