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How do you prove that $$ \delta(x-x') = \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} e^{-ik_x(x-x')} \, \mathrm{d} k_x $$

Attempt, take fourier transform of delt function and using the sifting property $$ \frac{1}{\sqrt{2\pi}} \int\limits_{-\infty}^{\infty} \delta(x-x') e^{-ik_xx} \, \mathrm{d} k_x = \frac{1}{\sqrt{2\pi}}e^{-ik_xx'} $$

Now take inverse fourier transform on both sides yields $$ \frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \delta(x-x') e^{-ik_xx} \, \mathrm{d} k_x e^{ik_xx} \mathrm{d}x = \int\limits_{-\infty}^{\infty}\frac{1}{2\pi}e^{-ik_xx'} e^{ik_xx}\mathrm{d}x $$

I feel like I'm one step away but I can't figure out how to pull the delta out from the left hand side

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Be careful with the integration variable. The integral on the left-hand side of the second equation should have a $dx$ instead of a $dk_x$:

$$\mathcal{F}\{\delta(x-x')\}=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\delta(x-x')e^{-ik_x x}dx=\frac{1}{\sqrt{2\pi}}e^{-ik_xx'}\tag{1}$$

After this step you're already done because from (1) you have

$$\delta(x-x')=\mathcal{F}^{-1}\left\{\frac{1}{\sqrt{2\pi}}e^{-ik_xx'}\right\}= \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{ik_x(x-x')}dk_x\tag{2}$$

Note that (2) is identical to your first equation because you can change the sign of the exponent by substituting $k_x=-\tilde{k}_x$.

So you don't actually need your last equation for the proof, but the problem there is that you use the variable $x$ for two different purposes: as integration variable and as independent variable. The correct integral would be:

$$\frac{1}{2\pi} \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \delta(y-x') e^{-ik_xy} \, dy\, e^{ik_xx} dk_x$$

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