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The delta function can be defined as:

$$ \delta(x) = \int_{-\infty}^{\infty} e^{-2\pi i k x} \, dk $$

Loosely speaking, I can understand this because unless $x=0$, the complex exponential oscillates rapidly over its domain and the integral vanishes on average. Meanwhile, when $x=0$ the integral diverges. Of course, there are more rigorous proofs.

My question is whether the following is an alternative expression for a delta function:

$$ \delta(x) \stackrel{?}{=} \int_{-\infty}^{\infty} e^{i \phi_{WN}(q)-i\phi_{WN}(x+q)} \, dq $$

where the function $\phi_{WN}(x)$ is a random white-noise process uniformly distributed between $0$ and $2\pi$. Intuitively, this seems like a candidate for such a definition because, as above, unless $x=0$, the argument of the complex exponential fluctuates randomly from $0$ to $2\pi$, and hence on average the integral should vanish. And when $x=0$, the integral diverges in the same way as for the conventional delta function definition. Has the delta function ever been defined in this manner?

The physical motivation for this question pertains to a problem involving wave scattering from a rough surface.

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  • $\begingroup$ The integral might oscillate but that doesn't mean it gives a zero answer, in fact this integral doesn't converge. $\endgroup$ Commented Sep 27, 2023 at 23:52
  • $\begingroup$ @Triatticus The OP's first (conventional) definition doesn't converge either, but we get good results from it anyway. Go figure. $\endgroup$ Commented Sep 27, 2023 at 23:57

1 Answer 1

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If $\phi(x)$ is truly white noise then $\phi(q)$ and $\phi(x+q)$ are iid. with $\int \mathbf{E}\left[e^{\mathfrak j\phi(q)}dq\right] = 0$. Therefore

$\mathbf{E}\left[\int e^{\mathfrak j\phi(q)}e^{\mathfrak j\phi(x+q)}dq\right] = \int \mathbf{E}\left[e^{\mathfrak j\phi(q)}e^{\mathfrak j\phi(x+q)}dq\right]\\ = \int \mathbf{E}\left[e^{\mathfrak j\phi(q)}\right]\mathbf{E}\left[e^{\mathfrak j\phi(x+q)}\right]dq = 0$

but $\mathbf{E}[\delta(x)] =\delta(x)$

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