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I'm given the function:

$$\psi(x)=\begin{cases} A &: x_1<x<x_2\\ 0 &:\text{outside} \end{cases}$$

So, I'm using the Fourier transform and its inverse to show that I'll get the original function $\psi(x)$ back. $$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\psi(x)e^{-ikx}dx=\frac{1}{\sqrt{2\pi}}\int_{x_2}^{x_1}Ae^{-ikx}dx\Rightarrow \phi(k)=-\frac{A}{\sqrt{2\pi}}\frac{1}{ik}\left(e^{-ikx_2}-e^{-ikx_1}\right)$$

I'm struggling to apply the inverse Fourier transform to get back the original function: $$\psi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi(k)e^{ikx}dk=-\frac{A}{2\pi i}\int_{-\infty}^{\infty}\frac{1}{k}\left(e^{ik(x-x_2)}-e^{ik(x-x_1)}\right)dk$$ I'm not sure how to integrate this.

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  • $\begingroup$ You're right, I didn't notice. It is still difficult to integrate though; I have no idea of how to it. $\endgroup$ – davidllerenav Apr 4 at 2:07
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You can use the following trick:

$\int_{-\infty}^{\infty}\frac{1}{ik}\left(e^{ikb}-e^{ika}\right)dk =\int_{-\infty}^{\infty}\int_{a}^{b} e^{ikx} dx\, dk =\int_a^b\int_{-\infty}^{\infty} e^{ikx} dk\, dx =\int_a^b 2\pi\delta(x)\,dx=\begin{cases}2\pi, &\mathrm{if}\, 0\in(a,b)\\ 0, &\mathrm{if}\, 0\notin(a,b) \end{cases} $

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  • $\begingroup$ But how to show it without the detour into $\delta$-function formalism? $\endgroup$ – A rural reader Apr 4 at 2:54
  • $\begingroup$ Thank you! I wouldn't have noticed that trick. $\endgroup$ – davidllerenav Apr 4 at 19:00

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