Is it always true that $$\delta(\omega) = \frac{1}{2\pi}\int_{-\infty}^{+\infty} e^{i \omega t} dt , $$ regardless of your Fourier convention?

For example, if I choose to use the Fourier convention where $$F(\omega) = \int_{-\infty}^{+\infty} f(t) e^{i \omega t} dt , $$ wouldn't $$ \delta(\omega)= \int_{-\infty}^{+\infty} e^{i \omega t} dt , $$ in this case since the delta function and the uniform function (1 everywhere) are conjugate pairs? Or am I thinking about this incorrectly?


2 Answers 2


The equation you wrote is a mathematical statement and as such it should not (and it doesn't) depend on any convention. Its proper meaning is intended in the sense of distribution. The distribution on the left hand side is well know. It becomes meaningful after applying it to test functions. The proper definition of the distribution on the right hand side is a bit more involved.

In any case to find the constant $C$ in

$$ \delta(\omega) C= \int_{-\infty}^{+\infty} e^{i \omega t} dt \ , $$

you can apply the above to a gaussian and integrate over $\omega$:

\begin{align} C \int_{-\infty}^{+\infty} d\omega e^{-\omega^2} \delta(\omega) &= C\\ & =\int_{-\infty}^{+\infty} d\omega \int_{-\infty}^{+\infty}dt e^{-\omega^2+it\omega} \\ &= \int_{-\infty}^{+\infty}dt \sqrt{\pi} e^{-t^2/4}\\ &= 2\pi. \end{align}

You may note that the above manipulations are common in physics but are incorrect mathematically.

First, there is no function $\delta(\omega)$ which is zero almost everywhere but integrates to one. The first equation above is the definition of the $\delta$ distribution. Its definition is that, for a test function $\phi$,

$$ \langle \delta, \phi \rangle = \phi(0)$$

In a similar vein, interchange of integrals above is not justified. But the manipulations are correct once a proper definition of the distribution of the right hand side is given.

This state of affairs is quite common. After all, distributions where first invented by physicists who would do the wildest (and more mathematically incorrect) manipulations with them. After some time Schwartz developed a proper mathematically rigorous theory of distributions. That's part of the beauty of being a physicist.


The identity $$ \delta(\omega) = \frac{1}{2\pi} \int dt\, e^{i\omega t} $$ does not depend on the Fourier convention (but of course requires regularization of the integral).

When you use a different conventions, what changes is whether the factor $\frac{1}{2\pi}$ is seen as the function you Fourier transform, or part of the normalization factor (or split between them).


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