Let $\delta_a$ be a dilation, such that $\delta_a f(x) = f(ax)$. Let $\hat{f}$ be the Fourier transform of $f$. Also, $E_n$ is a Euclidean space of dimension $n$.
Show that $\widehat{\delta_{-1} f} = \delta_{-1} \hat{f}$.
\begin{align*} \widehat{\delta_{-1} f}(x) & = \int_{E_n} \delta_{-1} f(t) e^{-2 i \pi t \cdot x} \mathrm{d}t \\ & = \int_{E_n} f(-t) e^{-2 i \pi t \cdot x} \mathrm{d}t \\ & = \int_{-\infty}^{\infty} f(-t) e^{-2 i \pi t \cdot x} \mathrm{d}t \\ & = \int_{\infty}^{-\infty} f(s) e^{2 i \pi s \cdot x} (-1)^{-n} \mathrm{d}s \\ & \quad \qquad \mbox{with substitution $s = -t$ } \\ & = \int_{\infty}^{-\infty} f(s) e^{2 i \pi s \cdot x} (-1)^{n} \mathrm{d}s \\ & = (-1)^{n+1} \int_{-\infty}^{\infty} f(s) e^{2 i \pi s \cdot x} \mathrm{d}s \\ & = (-1)^{n+1} \int_{-\infty}^{\infty} f(s) e^{-2 i \pi s \cdot (-x)} \mathrm{d}s \\ & = (-1)^{n+1} \hat{f}(-x) \\ & = (-1)^{n+1} \delta_{-1} \hat{f}(x). \end{align*}
I need to get rid of that pesky $(-1)^{n+1}$. I've investigated the even-ness and odd-ness of $\hat{f}$, but that doesn't help either. Any suggestions?
You cannot switch between $\int_{-\infty}^\infty$-integrals and $\int_X$-Integrals $(X:={\mathbb R}^n)$ without thinking.
Let $g(t):=f(-t)$ $(t\in X)$. Then $$\hat g(x)=\int_X f(-t)e^{-2\pi i\> t\cdot x}\>{\rm d}(t)\ .$$Parametrize $X$ by $t:=-t'$ $(t'\in X)$. The Jacobian has absolute value $1$, hence $$\hat g(x)=\int_X f(t')e^{2\pi i\> t'\cdot x}\>{\rm d}(t')=\hat f(-x)\ .$$