Proving Continuous Fourier Transform Formulas Given a continuous non-periodic function, its Fourier transform is defined as:

$$f(x) = \int_{-\infty}^\infty c(k) e^{ikx} dk, \ \ \ \ \ \ \ \ \ \ \ \ \ c(k) = \frac{1}{2\pi} \int_{-\infty}^\infty f(x) e^{-ikx} dx$$

The problem is proving this is true by evaluating the integral when $c(k)$ is plugged into the equation for $f(x)$.

This ends up with a long integral:

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{-ikx'} dx' \right) e^{ikx} dk$$

I'm not sure really how to proceed from here. I moved the $e^{ikx}$ into the inner integral, which I figured was fine since it's constant relative to $x'$.

$$f(x) = \int_{-\infty}^\infty \left( \frac{1}{2\pi} \int_{-\infty}^\infty f(x') e^{ik(x-x')} dx' \right) dk$$

I tried to kill at least one of the integrals by seeing if something evaluated to a Dirac Delta but I can't seem to get that result. I also tried integrating by parts, but that led me nowhere.
Let $f\in L^1(\mathbb{R}^d)$ such that $\widehat{f}\in L^1(\mathbb{R}^d)$ and:

$$
\forall x\in\mathbb{R}^d, \forall \epsilon\in\mathbb{R}, \phi(x)=\pi^{-\frac{d}{2}}\exp(-|x|^2), \phi_\epsilon(x)=\frac{1}{\epsilon^d}\phi\left(\frac{1}{\epsilon}x\right)\\
$$
Lemma:$$
\mathcal{F}(\exp(-a|\cdot|^2))(\xi)=\left(\frac{\pi}{a}\right)^{\frac{d}{2}}\exp\left(-\frac{|\xi|^2}{4a}\right)
$$
Then:
$$
\begin{equation}
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi=\lim\limits_{\epsilon\to0}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi,x\rangle)\widehat{\phi_\epsilon}(x)\widehat{f}(\xi)\mathrm{d}\xi
\end{equation}
$$

By Fubini theorem:
$$
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\int\limits_{\mathbb{R}^d\times\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{\phi_\epsilon}(\xi)f(y)\mathrm{d}\xi\mathrm{d}y
$$
So:
$$
(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, y-x)\widehat{\phi_\epsilon}(\xi)\mathrm{d}\xi=(2\pi)^{-d}\left(\frac{4\pi}{\epsilon^2}\right)^{\frac{d}{2}}\exp\left(-\frac{|x-y|^2}{\epsilon^2}\right)=\phi_\epsilon(x-y)
$$
And:
$$
\forall \epsilon&gt;0, (\phi_\epsilon\star f)(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle x, \xi\rangle)\widehat{\phi_\epsilon}(\xi)\widehat{f}(\xi)\mathrm{d}\xi
$$
Let $(\epsilon_n)_{n\in\mathbb{N}}$ be a sequence of reals, $\epsilon&gt;0$, that converge to 0. Then $\lim\limits_{n\infty}\|\phi_{\epsilon_n}\star f-f\|_{L^1(\mathbb{R}^d)}=0$, there is a subsequence such that for allmost all $x\in\mathbb{R}^d$,
$$
f(x)=(2\pi)^{-d}\int\limits_{\mathbb{R}^d}\exp(i\langle\xi, x\rangle)\widehat{f}(\xi)\mathrm{d}\xi
$$
From your work:
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}x' \mathrm{d}k$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x')e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{ik(x-x')} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')\int_{-\infty}^{\infty} e^{-ik(x'-x)} \mathrm{d}k\mathrm{d}x'$$
$$f(x)=\frac{1}{2\pi}\int_{-\infty}^{\infty}f(x')2\pi \delta(x'-x)\mathrm{d}x'$$
$$f(x)=\int_{-\infty}^{\infty}f(x')\delta(x'-x)\mathrm{d}x'$$