It is fairly easy to demonstrate that the Dirac delta function is the Fourier transform of the plane wave function, and hence that:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]\delta(x)=∫_{-∞}^{∞}e^{ikx}dk[/itex] (eg Tannoudji et al 'Quantum Physics' Vol 1 p101 A-39)

Hence it should be the case that [itex]∫_{-∞}^{∞}e^{ik}dk = \delta(1) = 0[/itex]

However the integral on the LHS does not even seem to exist, let alone equal zero. We can write:

[itex]∫_{-∞}^{∞}e^{ik}dk = ∫_{-∞}^{∞}cos(k)+i sin(k) dk[/itex]

[itex]= ∫_{-∞}^{∞}cos(k)dk+i ∫_{-∞}^{∞}sin(k)dk[/itex]

and neither of these integrals exist.

Yet, based on the Fourier argument, they must both exist and equal zero.

I have probably done something silly to get myself into this contradiction. Can somebody please help me by showing what it is?

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# Dirac Delta function as a Fourier transform

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