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I am able to derive the following equation by substituting the definition of a Fourier transform into it’s inverse.

$$2\pi\delta(x-x’) = \int_{-\infty}^{\infty} e^{ik(x-x’)} dk$$

How do you prove that the Dirac Delta is equal to an integral of the exponential function? How do you prove the above equation is true?

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Let $$h_a(x)= \int_{-a}^a e^{i k x} dk = \frac{2 \sin(a x)}{x}= a \, H'(ax), \\ H(x) = \int_{-\infty}^x \frac{2 \sin(y)}{y}dy, \qquad H(-\infty) = 0, H(+\infty) = C$$

where for some reason $C = 2\pi$

If $\phi,\phi’$ are $L^1$ then

$$\lim_{a \to \infty}\int_{-\infty}^\infty h_a(x) \phi(x) dx = -\lim_{a \to \infty}\int_{-\infty}^\infty H(ax) \phi'(x) dx\\ = -\int_{-\infty}^\infty H(+\infty x) \phi'(x) dx = -\int_0^\infty C \phi'(x) dx= 2\pi \phi(0)$$

ie. in the sense of distributions $$\int_{-\infty}^\infty e^{ik x}dk \overset{def}=\lim_{a \to\infty} h_a = 2\pi \delta$$

Note how this proves the Fourier inversion theorem.

We can give a meaning to $\int_{-\infty}^{\infty} e^{ikx} \, dk$ by introducing a damping factor $e^{-\frac12\epsilon k^2}$ inside the integral and at the end let $\epsilon \to 0$:

$$

\int_{-\infty}^{\infty} e^{-\frac12\epsilon k^2} e^{ikx} \, dk

= \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} e^{-\frac12 x^2/\epsilon} \, dk \\

= e^{-\frac12 x^2/\epsilon} \int_{-\infty}^{\infty} e^{-\frac12\epsilon (k-ix/\epsilon)^2} \, dk

= \sqrt{\frac{2\pi}{\epsilon}} \, e^{-\frac12 x^2/\epsilon}

\to 2\pi \, \delta(x)

$$

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