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I had a curiosity question rise up in the middle of the night regarding the behavior of the Dirac Delta. Because it's not a function per-se, I am not sure how a concept like "integration" symmetry might affect it. Particularly, it's well accepted that:

$$ \delta(\omega) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i{\omega}t}dt $$

However, I was wondering if (by some type of "symmetry" argument, it might be reasonable to discuss that:

$$ \delta(\omega) = \frac{2}{2\pi}\int_{0}^{\infty}e^{i{\omega}t}dt= \frac{1}{\pi}\int_{0}^{\infty}e^{i{\omega}t}dt $$

Does this seem reasonable? Thanks for any takers!

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  • $\begingroup$ But $\frac{1}{\pi} \times 2 \neq \frac{1}{2\pi}$. $\endgroup$ – Don Larynx Nov 8 '13 at 5:58
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It is true that (in the sense of tempered distributions) $\delta_0$ is the Fourier transform of a constant function. Note that the Fourier transform sends compact supported functions (and distributions) into real-analytic functions. This is illustrated by $\delta_0$ having compact support, and a constant function being real-analytic.

However, your integral from $0$ to $\infty$ represents the Fourier transform of $1_{[0,\infty)}$, which is not continuous, let alone real-analytic. Hence, this transform cannot be $\delta_0$; it will not be compactly supported, and will have a heavy (non-integrable) tail at infinity.

But something similar to what you wrote holds. A constant function is even, and the Fourier transform of an even function is the cosine transform of its restriction to $[0,\infty)$. Hence, you can write $$\delta(\omega) = \frac{1}{\pi}\int_{0}^{\infty}\cos \omega t\,dt$$

Which can be interpreted as follows: for every test function $\phi$, $$\phi(0) = \frac{1}{ \pi} \lim_{T\to\infty} \int_{\mathbb R} \phi(\omega)\,d\omega \int_{0}^{T} \cos \omega t \,dt $$ or, to put it simpler, $$\phi(0) = \frac{1}{ \pi} \lim_{T\to\infty} \int_{\mathbb R} \phi(\omega) \frac{\sin \omega T}{\omega}\,d\omega $$

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