According to the German Wikipedia, the $\delta$ distribution can be approximated by sequences of integrable functions $\delta_k$ satisfying

  1. $\delta_k(x) \geq 0$
  2. $\int_{\mathbb R} \delta_k(x) \mathrm dx = 1$
  3. $\lim_{k \to \infty} \int_{\mathbb R \setminus B_\varepsilon(0)} \delta_k(x) \mathrm dx = 0$ for all $\varepsilon > 0$,

called Dirac sequences.

One example given is the Fresnel representation $$ \delta_\varepsilon(x) = \frac{1}{\sqrt{i \pi \varepsilon}} \exp\left(\frac{i x^2}{\varepsilon}\right) , $$ where $\lim_{\varepsilon \to 0} \delta_\varepsilon = \delta$.

My questions are

  • These are complex functions, so 1. cannot be satisfied. How does it still qualify as a Dirac sequence?

  • To check 2. I tried to calculate the Integral $$ I := \int_{\mathbb R} e^{i x^2} \mathrm dx $$ and found $$ I^2 = \int_{\mathbb R} \int_{\mathbb R} e^{i (x^2 + y^2)} \mathrm dx \mathrm dy = 2 \pi \int_0^\infty \rho e^{i \rho^2} \mathrm d\rho = 2 \pi \cdot \left. \frac{1}{2 i} e^{i \rho^2} \right|_0^\infty = i \pi - \lim_{R \to \infty} i e^{i R^2} , $$ which does not converge. However, my intuition as well as wolfram|alpha tell me that $I$ does converge. Where did I go wrong?

  • $\begingroup$ Rewrite $i\pi-ie^{iR^2}$ as $\displaystyle\lim_{a\to0^{+}}\left(i\pi-ie^{(i-a)R^2}\right)$ $\endgroup$
    – Diffusion
    Commented Nov 25, 2017 at 20:32
  • $\begingroup$ @Messney, how does this help me? The limits cannot be switched. $\endgroup$
    – schtandard
    Commented Nov 26, 2017 at 15:44
  • $\begingroup$ How do you know that the limits cannot be switched? Try it, and compare your answer with Wolfram's. $\endgroup$
    – Diffusion
    Commented Nov 26, 2017 at 21:25
  • $\begingroup$ @Messney, the fact that you found some arbitrary way to make it look like the right result does not make your method correct. (You can even ask wolfram|alpha about that). My error is not in evaluating the final integral. It has to be in transforming the coordinates, but I cannot see why. $\endgroup$
    – schtandard
    Commented Nov 26, 2017 at 22:14

1 Answer 1


The integral \begin{align} I&=\frac{1}{\sqrt{i\pi\varepsilon}}\int_{-\infty}^\infty e^{ix^2/\varepsilon}dx\\ &=\frac{2}{\sqrt{i\pi}}\int_0^\infty e^{iy^2}dy \end{align} can be calculated using several method. With $e^{iy^2}=\cos y^2+i\sin y^2$, see for example the numerous answers here or here \begin{align} I&=\frac{2}{\sqrt{i\pi}}\sqrt{\frac{\pi}{8}}\left( 1+i \right)\\ &=1 \end{align}


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