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Section 11.6 The Exponential Representation of the Dirac Delta Function

As discussed in Section 11.4, the Dirac delta function can be written in the form

\begin{equation} \delta(x) = \frac{1}{2\pi}\int_{-\infty}^\infty e^{ikx}\, dk \text{.}\tag{11.6.1} \end{equation}

We outline here the derivation of this representation.

In order to evaluate the integral, we introduce a regularization factor, \(e^{-k\epsilon}\text{,}\) as follows:

\begin{align*} \int_{-\infty}^{\infty} e^{ikx}\, dk \amp = \int_{-\infty}^0 e^{ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ \amp = \int_0^\infty e^{-ikx}\, dk + \int_0^\infty e^{ikx}\, dk \nonumber\\ \amp = \int_0^\infty (e^{ikx} + e^{-ikx})\, dk \nonumber\\ \amp = \lim_{\epsilon\to0^+}\int_0^\infty (e^{ikx} + e^{-ikx}) \,e^{-k\epsilon}\, dk \nonumber\\ \amp = \lim_{\epsilon\to0^+}\int_0^\infty (e^{ik(x+i\epsilon)} + e^{-ik(x-i\epsilon)}) \,dk \nonumber\\ \amp = \lim_{\epsilon\to0^+} \left[ \frac{e^{ik(x+i\epsilon)}}{i(x+i\epsilon)} + \frac{e^{-ik(x-i\epsilon)}}{-i(x-i\epsilon)} \right]_0^\infty \nonumber\\ \amp = \lim_{\epsilon\to0^+} \left( 0 + 0 - \frac1{i(x+i\epsilon)} - \frac1{-i(x-i\epsilon)} \right) \nonumber\\ \amp = \lim_{\epsilon\to0^+} \left( \frac{i}{x+i\epsilon} - \frac{i}{x-i\epsilon} \right) \nonumber\\ \amp = \lim_{\epsilon\to0^+} \frac{2\epsilon}{x^2+\epsilon^2} = \cases{0 \amp $x\ne0$\\ \infty \amp $x=0$\\ } \end{align*}

where we have used the fact that, for \(\epsilon>0\text{,}\) \(e^{-k\epsilon}\) goes to \(0\) as \(k\) goes to \(\infty\text{.}\) 1 

The use of such regularization factors is quite common. Rigorous mathematical justification can be given, but informal arguments along the lines above are usually sufficient—if the informal argument works, the formal derivation should also, for an appropriate, reasonable class of functions.

It remains to show that the final expression has the correct normalization. But

\begin{equation} \int_{-\infty}^\infty \frac{2\epsilon}{x^2+\epsilon^2} \,dx = 2 \arctan\left(\frac{x}{\epsilon}\right) \Bigg|_{-\infty}^\infty = 2\pi \text{,}\tag{11.6.2} \end{equation}

which is independent of \(\epsilon\text{.}\)