Derivation of Inverse Fourier transform from forward Fourier transform
I'll use the common DSP definition and notation of the Fourier transform and its inverse like given in Eqs $(iii)$ and $(iv)$ in the OP. Note that the argument $t$ in Eqs $(i)$ and $(ii)$ is irrelevant.
First of all, note that the Fourier transform of a delay $x(t)=\delta(t-\tau)$ is given by
$$\int_{-\infty}^{\infty}\delta(t-\tau)e^{-j\omega t}dt=e^{-j\omega \tau}\tag{1}$$
Consequently, we have to accept the following expression (in a distributional sense), following from the inverse Fourier transform of $(1)$:
$$\delta(t-\tau)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{j\omega (t-\tau)}d\omega\tag{2}$$
Now we can prove $(iv)$ from $(iii)$:
$$\begin{align}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega&amp;=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\tau)e^{-j\omega\tau}d\tau e^{j\omega t}d\omega\\&amp;=\int_{-\infty}^{\infty}x(\tau)\underbrace{\int_{-\infty}^{\infty}e^{j\omega (t-\tau)}d\omega}_{2\pi\delta(t-\tau)} d\tau\\&amp;=2\pi x(t)\end{align}\tag{3}$$