Dirac delta function is said to be Fourier transformation of 1,

$$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\ e^{i bx} . 1\ =\ \delta(b) $$

This is usually shown by considering the definition of the dirac-delta and looking for its Fourier transform. If one directly evaluate the LHS under the limit going from -L to L then we get,

$$ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} dx\ e^{i bx} . 1\ =\ \frac{2\sin bL}{b} $$

Now here is the question how do I show,

$$ \lim_{L\rightarrow \infty} \frac{2\sin bL}{b}\ =\ \delta(b) $$


You don't. Not really. The integral does not exist as integral, you have to apply it as a distribution to a suitable test function.

What the formula then really means is the Fourier inversion theorem, i.e., that the inverse Fourier transform of the Fourier transform of a suitably smooth and bounded function returns the same function. This is a very deep result in Fourier analysis, you can not hope to solve it with a simple limit.

If you want an intuitive way to understand it, approximate the constant 1 by a (slowly) exponentially decaying function, i.e., consider for small $0<α$

$$\int_{-\infty}^{\infty} e^{-α|b|}\cdot e^{ibx}\,db= \frac{2α}{α^2+x^2}$$

which gives a proper approximation of the delta distribution, positive function values, constant integral over $\mathbb R$ with value $2\pi$ and pointwise convergence to zero except in $x=0$.

This approximation is usually also used in the proof of the Fourier inversion theorem.

Another familiar approximation uses the Gauss bell function, again for small positive $α$

$$\int_{-\infty}^{\infty} e^{-\tfracα2 b^2}\cdot e^{ibx}\,db =\int_{-\infty}^{\infty} e^{-\tfracα2 (b-i\tfrac{x}{α})^2}\cdot e^{-\tfrac{x^2}{2α}}\,db=\sqrt{\tfrac{2\pi}{α}}e^{-\tfrac{x^2}{2α}}$$

gives the same properties of an delta approximating family, more exactly, it converges as family of tempered distributions to $2\pi\delta_0$.

For a nice presentation of the necessary details see Carl Offner: "A little harmonic analysis"


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