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Consider the Fourier pairs: $$\psi(x,t) \stackrel{\mathrm{FT}}{\longleftrightarrow} \Psi(k,t)$$ $$\text{If } \quad \quad\Psi(k,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \psi(x,t) e^{-ikx} \, dx \quad \quad \dots(i)$$ $$\text{then, can we derive: } \quad \quad \psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \Psi(k,t) e^{ikx} \, dk \quad \quad \dots(ii) \quad ?$$ I have used comparison method to proof $eq.(ii) $
i.e., $$\text{consider:} \quad \quad x(t) \stackrel{\mathrm{FT}}{\longleftrightarrow} X(\omega)$$ $$\text{then, } \quad \quad X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-i\omega t} \, dt \quad \quad \dots(iii)$$ $$\text{and, } \quad \quad x(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty} X(\omega) e^{i\omega t} \, d\omega \quad \quad \dots(iv)$$

Now, comparing $eq(iii)$ with $eq(i)$ , we find:
$w \to k$
$t \to x$
$x(t) \to \frac{\psi(x,t)}{\sqrt{2\pi}}$
$X(w) \to \Psi(k,t)$ , putting these values in $eq(iv)$, we get:
$$\frac{\psi(x,t)}{\sqrt{2\pi}}=\frac{1}{2\pi}\int_{-\infty}^{\infty} \Psi(k,t) e^{ikx} \, dk $$ $$\implies \psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \Psi(k,t) e^{ikx} \, dk$$

But can we derive $eq(ii)$ from $eq(i)$? {without using any comparison}

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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 $(iii)$:

$$\begin{align}\int_{-\infty}^{\infty}X(j\omega)e^{j\omega t}d\omega&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}x(\tau)e^{-j\omega\tau}d\tau e^{j\omega t}d\omega\\&=\int_{-\infty}^{\infty}x(\tau)\underbrace{\int_{-\infty}^{\infty}e^{j\omega (t-\tau)}d\omega}_{2\pi\delta(t-\tau)} d\tau\\&=2\pi x(t)\end{align}\tag{3}$$

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