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Consider a function $f\colon \mathbb{R}\to \mathbb{R}$ whose Fourier transform is given by $$\mathcal{F}[f(x)]=\hat{f}(k)=\int_{-\infty}^{\infty}f(x)e^{-ikx} \, dx.$$

I'm asked to prove that the inverse transform is given by \begin{align} f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty} \hat{f}(k)e^{ikx} \, dk,\end{align} using the Dirac delta function.

I've started by inserting the definition of $\hat{f}(k)$ into the expression for $f(x),$ which yields \begin{align}\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(z)e^{-ikz} \, dz \, e^{ikx} \, dk &=\frac{1}{2\pi}\int_{-\infty}^{\infty} f(z)\int_{-\infty}^{\infty}e^{ik(x-z)} \, dk \, dz \\ & = \int_{-\infty}^{\infty} f(z)\frac{1}{2\pi}\int_{-\infty}^{\infty}\mathcal{F}[\delta(x-z)]e^{ikx} \, dk \, dz,\end{align} after swpping the order of integration and using the fact that the Fourier transform of the Dirac delta function is $1$.

Am I along the right lines or is there a better way?

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    $\begingroup$ Why can you swap the order of integration? $\endgroup$ – Jonas Lenz Apr 27 '18 at 22:07
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    $\begingroup$ This is formally OK, but of course there is a number of mathematical subtleties that you are sweeping under the rug (see the other comment). However, if your goal is a formal-only proof (whatever that means), then you are on the right track. Reduce the last expression to $\int f(z)\delta(x-z)\, dz$. $\endgroup$ – Giuseppe Negro Sep 21 '18 at 9:21

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