While reading the book "Modern Electrodynamics" by Andrew Zangwill, on page 13 I've encountered a Dirac function integral representation. The proof the book provides is the following:

\begin{align*} \int_{-\infty}^\infty \text{d}k\, e^{ikx} &= \int_0^\infty \text{d}k\, e^{ikx} + \int_0^\infty \text{d}k\, e^{-ikx} \\ &= \lim_{\epsilon \to 0} \left[ \int_0^\infty \text{d}k\, e^{ik(x+i\epsilon)} + \int_0^\infty \text{d}k\, e^{-ik(x-i\epsilon)} \right]. \end{align*} The convergence factors make the integrands zero at the upper limit, so $$ \int_{-\infty}^\infty \text{d}k\, e^{ikx} = \lim_{\epsilon \to 0} \left[ \frac{i}{x + i \epsilon} - \frac{i}{x - i \epsilon} \right] = \lim_{\epsilon \to 0} \frac{2 \epsilon}{x^2 + \epsilon^2} = 2 \pi \delta(x). $$ (Original image here.)

I don’t see how the convergence makes the first term inside the limit vanish at the upper limit. I can understand the rest from there as the the $k$-independent factor appears while integrating and the final conclussion is from one of the informal dirac delta definitions.


1 Answer 1


Looking at the first one:

$$\int_0^\infty dk e^{ik(x+i \epsilon)} = \int_0^\infty dk e^{ikx} e^{-k \epsilon}.$$

Now $e^{ikx}$ is bounded in modulus while $e^{-k \epsilon}$ (for fixed $\epsilon$) decays exponentially in modulus. So the integral converges by comparison to $\int_0^\infty dk e^{-k \epsilon}$.

The second one is basically the same.

Note that the first step is quite informal (one cannot justify it by the application of some convergence theorem which can be formulated without distribution theory.) I think you might get more insight by looking at a formal definition of the Fourier transform on tempered distributions. The formal definition is quite simple, and the desired statement (that the Fourier transform of the Dirac delta is a constant function) follows quite easily from it.

  • $\begingroup$ I missed looking at the modolus. Can you recomend me some literature on the subject? Had very little exposition to distributions formally as a physicist. $\endgroup$ Commented Feb 12, 2015 at 23:20
  • $\begingroup$ Consider volume 1 of Hormander's The analysis of partial differential equations. The notes for my current course are from there (I can check as to whether I am allowed to simply share them). $\endgroup$
    – Ian
    Commented Feb 12, 2015 at 23:38
  • $\begingroup$ Would greatly apreaciate if so. Looking foward to that $\endgroup$ Commented Feb 12, 2015 at 23:56

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