Circumference of a circle using integral representation of Dirac delta

Question

I am trying to obtain the circumference of a circle of radius $r$ in a rather complex way because I am practising with the integral representation of the Dirac delta. Let me refer to the circle as $C$ and let me write the expression to be computed as $$ \int_C dx dy = \int \int \delta(x^2+y^2 - r^2) || \nabla (x^2+y^2 - r^2) || dx dy \\ = 2r \int \int \delta(x^2+y^2 - r^2) dx dy .\tag{1} $$ I know that the last integral evaluates to $\pi$, which can be seen easily after changing to polar coordinates. However, I want to stick to Cartesian coordinates and use that $$ \delta(x^2 + y^2 - r^2) = \int \frac{d \omega}{2 \pi} {\rm{e}}^{i\omega (x^2 + y^2 - r^2)}.\tag{2} $$ Then the double integral can be rewritten as $$ \int \int \delta(x^2+y^2 - r^2) dx dy = \int \frac{d \omega}{2 \pi} \int \int {\rm{e}}^{i\omega (x^2 + y^2 - r^2)} dx dy.\tag{3} $$ And now we can perform the Gaussian integral over $x$: $$ \int_{-\infty}^{\infty} dx {\rm{e}}^{i\omega x^2} = \sqrt{\frac{\pi}{- i \omega}}.\tag{4} $$ And similarly over $y$. Therefore we arrive at $$ \int \int \delta(x^2+y^2 - r^2) dx dy = \frac{1}{2} \int {d \omega} \frac{{\rm{e}}^{-i\omega r^2}}{(-i \omega)}.\tag{5} $$ Then we can identify the integral as the inverse Fourier transform ($\mathcal{F}^{-1}()$) of the function $i/\omega$ evaluated at $r^2$ (and multiplied by the prefactor $\sqrt{2 \pi}$), i.e. $$ \int \int \delta(x^2+y^2 - r^2) dx dy = \sqrt{\frac{\pi}{2}} \left(\mathcal{F}^{-1}\left(\frac{i}{\omega} \right)\right)_{r^2} \\ = \frac{\pi}{2} \Theta(r^2) \\ = \frac{\pi}{2}.\tag{6} $$ Where $\Theta()$ refers to the Heaviside function. As you can see, I am off of the correct result by a factor of 1/2.

Answer 1

Hint: Eq. (4) is ill-defined if $\omega=0$. We can regularize eq. (4) as a distribution $$ \int_{\mathbb{R}} \!\mathrm{d}x~ e^{(i\omega-0^+) x^2} ~=~\sqrt{\frac{\pi}{0^+- i \omega}},\tag{4'} $$ so that eq. (3) becomes $$\begin{align}\iint_{\mathbb{R}^2}\!\mathrm{d}x~\mathrm{d}y~ \delta(x^2+y^2 - r^2) ~=~& \int_{\mathbb{R}} \!\frac{\mathrm{d}\omega}{2 \pi} \iint_{\mathbb{R}^2}\!\mathrm{d}x~\mathrm{d}y~ e^{i\omega (x^2 + y^2 - r^2)}\cr ~\stackrel{(4')}{=}~& \int_{\mathbb{R}} \!\frac{\mathrm{d}\omega}{2 \pi} \frac{\pi}{0^+- i \omega} e^{-i\omega r^2}\cr ~=~& \frac{i}{2}\int_{\mathbb{R}} \!\frac{\mathrm{d}\omega}{\omega+i0^+} e^{-i\omega r^2}\cr ~=~& \frac{i}{2} (-2\pi i) \Theta(r^2)\cr ~=~& \pi\Theta(r^2), \end{align} \tag{3'}$$ where we closed the $\omega$-contour in the lower (upper) $\omega$-plane if $r^2>0$ ($r^2<0$), respectively.