Showing $\hat{\tilde{f}}=\tilde{\hat{f}}$ where $\hat{f}$ is the Fourier transform, and $\tilde{f}(x) = f(-x)$ I'm trying to prove that $\hat{\tilde{f}}=\tilde{\hat{f}}$ for any integrable function $f$, where $\hat{f}$ denotes the Fourier transform of $f$ and $\tilde{f}$ denotes the mapping $f(x)\to f(-x)$, though I'm currently stuck at $$\hat{\tilde{f}}(x)=\int_{\mathbb{R}^n}f(-t)e^{-2\pi it\cdot x}dt\tag*{(1)}$$ and $$\tilde{\hat{f}}(x)=\int_{\mathbb{R}^n}f(t)e^{2\pi it\cdot x}dt\tag*{(2)}.$$ Surely the substitution $u=-t$ in $(1)$ gives $$\hat{\tilde{f}}(x)=\int_{\mathbb{R}^n}-f(u)e^{2\pi iu\cdot x}du,$$ which in general is not equal to $(2)$? Many thanks in advance.

You need to be careful when doing the change of variables. The differentials change by the absolute value of the determinant of the Jacobian of the transformation, which in this case is $|(-1)^n| = 1$. 
You can see your error if you write your integrals as iterated integrals. Write $\hat{\tilde{f}}(x)$ as $\int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(-t)e^{-2\pi it\cdot x}dt_1\dots dt_n$. Then the variable substitution $u = -t$ corresponds to the $n$ variable substitutions $u_k = - t_k$. For each $k$ we see that $du_k = -dt_k$, but also, the limits of integration swap. That is,
\begin{align*}
\hat{\tilde{f}}(x) &= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(-t)e^{-2\pi it\cdot x}dt_1\dots dt_n\\ 
&= \int_{\infty}^{-\infty}\dots\int_{\infty}^{-\infty}f(u)e^{2\pi iu\cdot x}(-1)^ndu_1\dots du_n\\ 
&= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}(-1)^nf(u)e^{2\pi iu\cdot x}(-1)^ndu_1\dots du_n\\ 
&= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(u)e^{2\pi iu\cdot x}|(-1)^n|du_1\dots du_n\\
&= \int_{-\infty}^{\infty}\dots\int_{-\infty}^{\infty}f(u)e^{2\pi iu\cdot x}du_1\dots du_n\\
&= \int_{\mathbb{R}^n}f(u)e^{2\pi iu\cdot x}du.
\end{align*}
Replacing $u$ by $t$, you then obtain $\hat{\tilde{f}}(x) = \tilde{\hat{f}}(x)$.
